Geometry and Trigonometry

178 questions

Question 61Question

In triangle ABCABC, point DD lies on side BCBC such that the ratio of BDBD to DCDC is 11 to 22. Point EE lies on segment ADAD such that the ratio of AEAE to EDED is 33 to 11. A line passing through EE and parallel to ACAC intersects side ABAB at FF and side BCBC at GG. If the area of triangle ABCABC is 128128, what is the area of triangle BFGBFG?

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Answer: 32

Answer

32
The correct answer is 32. By establishing a coordinate system, we find that the similarity ratio of triangle BFGBFG to triangle ABCABC is exactly 1/21/2 regardless of the triangle's shape. Since the ratio of the areas of similar triangles is the square of the similarity ratio, the area of triangle BFGBFG is (1/2)2=1/4(1/2)^2 = 1/4 of the area of triangle ABCABC, which is 128×1/4=32128 \times 1/4 = 32.

Step-by-Step Solution

1
Set up a coordinate system to represent the triangle's vertices.
Let B=(0,0)B = (0, 0) and C=(3,0)C = (3, 0). Since DD lies on side BCBC and BD:DC=1:2BD:DC = 1:2, the coordinates of DD are (1,0)(1, 0). Let A=(a,b)A = (a, b).
Setting up coordinates simplifies the proof by allowing algebraic verification of the ratio.
2
Calculate the coordinates of point EE on segment ADAD.
Using the section formula with ratio AE:ED=3:1AE:ED = 3:1, E=(1(a)+3(1)4,1(b)+3(0)4)=(a+34,b4)E = \left(\frac{1(a) + 3(1)}{4}, \frac{1(b) + 3(0)}{4}\right) = \left(\frac{a+3}{4}, \frac{b}{4}\right).
Finding the coordinates of EE is necessary to determine the equation of line FGFG.
3
Find the equation of line FGFG which is parallel to ACAC and passes through EE.
The slope of ACAC is m=b3am = \frac{-b}{3-a}. The equation of FGFG is yb4=b3a(xa+34)y - \frac{b}{4} = \frac{-b}{3-a}\left(x - \frac{a+3}{4}\right).
The line FGFG is parallel to ACAC, meaning they share the same slope.
4
Find the coordinates of GG by setting y=0y = 0 in the equation of FGFG.
Setting y=0y = 0 yields b4=b3a(xGa+34)    xG=1.5-\frac{b}{4} = \frac{-b}{3-a}\left(x_G - \frac{a+3}{4}\right) \implies x_G = 1.5. Thus, G=(1.5,0)G = (1.5, 0).
Point GG is the intersection of the line FGFG with side BCBC (the x-axis).
5
Determine the similarity ratio and calculate the area of triangle BFGBFG.
Since GG is the midpoint of BCBC, the similarity ratio of BFG\triangle BFG to BAC\triangle BAC is k=12k = \frac{1}{2}. The ratio of their areas is k2=14k^2 = \frac{1}{4}. The area of BFG\triangle BFG is 128×14=32128 \times \frac{1}{4} = 32.
Similar triangles have area ratios equal to the square of their similarity ratio.

Key Concept

The ratio of the areas of similar triangles is equal to the square of their similarity ratio.

Alternative Method

Using Menelaus's Theorem on triangle ABDABD and transversal FEGFEG can also establish the midpoint relations directly without using coordinates.
Estimated Time:3m 0s
Question 62Question

An aerospace company designs two similar solar panels. The smaller panel is in the shape of a parallelogram and has an area of 8080 square centimeters. If the perimeter of the larger panel is 1.51.5 times the perimeter of the smaller panel, what is the area, in square centimeters, of the larger panel?

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Answer: 180

Answer

180
Since the two solar panels are similar, the ratio of their perimeters is equal to their linear scale factor, k=1.5k = 1.5. The ratio of their areas is equal to the square of the linear scale factor, k2=(1.5)2=2.25k^2 = (1.5)^2 = 2.25. Therefore, the area of the larger panel is the product of the smaller panel's area and the area scale factor: 80×2.25=18080 \times 2.25 = 180 square centimeters.

Step-by-Step Solution

1
Determine the linear scale factor between the two similar solar panels.
The linear scale factor is k=1.5k = 1.5.
For similar figures, the ratio of any corresponding linear measurements (such as perimeters) is equal to the linear scale factor. Here, the perimeter of the larger panel is 1.51.5 times that of the smaller panel.
2
Calculate the area scale factor by squaring the linear scale factor.
The area scale factor is k2=(1.5)2=2.25k^2 = (1.5)^2 = 2.25.
The ratio of the areas of two similar two-dimensional shapes is equal to the square of their linear scale factor.
3
Multiply the area of the smaller solar panel by the area scale factor to find the area of the larger panel.
The area of the larger solar panel is 80×2.25=18080 \times 2.25 = 180 square centimeters.
Multiplying the original area by the area scale factor yields the scaled area of the larger similar figure.

Key Concept

Area scaling of similar figures
Question 63Question

An acute angle θ\theta satisfies the equation sin(θ)=0.6\sin(\theta) = 0.6. What is the value of cos(θ)\cos(\theta)?

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Answer: 0.8

Answer

The value of cos(θ)\cos(\theta) is 0.8.
Using the fundamental Pythagorean trigonometric identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, we substitute the given value sin(θ)=0.6\sin(\theta) = 0.6 to obtain (0.6)2+cos2(θ)=1(0.6)^2 + \cos^2(\theta) = 1, which simplifies to 0.36+cos2(θ)=10.36 + \cos^2(\theta) = 1. Subtracting 0.360.36 from both sides yields cos2(θ)=0.64\cos^2(\theta) = 0.64. Taking the positive square root because θ\theta is an acute angle gives cos(θ)=0.8\cos(\theta) = 0.8.

Step-by-Step Solution

1
State the Pythagorean trigonometric identity
sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1
This identity relates the sine and cosine of any angle.
2
Substitute the value of sin(θ)\sin(\theta) into the identity
(0.6)2+cos2(θ)=1(0.6)^2 + \cos^2(\theta) = 1
We are given that sin(θ)=0.6\sin(\theta) = 0.6.
3
Solve for cos2(θ)\cos^2(\theta)
cos2(θ)=0.64\cos^2(\theta) = 0.64
Subtracting 0.360.36 from 11 isolates the squared cosine term.
4
Take the square root of both sides
cos(θ)=0.8\cos(\theta) = 0.8
Since θ\theta is an acute angle, the value of cos(θ)\cos(\theta) must be positive.

Key Concept

Pythagorean Identity
Question 64Question

A right triangle in the xyxy-plane has vertices at the coordinates (0,0)(0, 0), (15,0)(15, 0), and (15,8)(15, 8). What is the perimeter of this triangle?

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Answer: 40

Answer

The perimeter of the triangle is 40.
The perimeter of the triangle is the sum of its three side lengths. The horizontal leg has a length of 1515 (from (0,0)(0,0) to (15,0)(15,0)), and the vertical leg has a length of 88 (from (15,0)(15,0) to (15,8)(15,8)). Using the Pythagorean theorem, the hypotenuse has a length of 152+82=225+64=17\sqrt{15^2 + 8^2} = \sqrt{225 + 64} = 17. Summing these three lengths gives a perimeter of 15+8+17=4015 + 8 + 17 = 40.

Step-by-Step Solution

1
Identify the lengths of the two perpendicular legs of the right triangle from the given coordinates.
The horizontal leg lies on the x-axis from (0,0)(0, 0) to (15,0)(15, 0), which has a length of 150=1515 - 0 = 15. The vertical leg runs parallel to the y-axis from (15,0)(15, 0) to (15,8)(15, 8), which has a length of 80=88 - 0 = 8.
The length of horizontal and vertical segments in a coordinate plane can be found by taking the absolute difference of their non-matching coordinates.
2
Use the Pythagorean theorem to calculate the length of the hypotenuse.
The hypotenuse length is 152+82=225+64=289=17\sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17.
The Pythagorean theorem states that a2+b2=c2a^2 + b^2 = c^2 for any right triangle with legs aa and bb and hypotenuse cc.
3
Calculate the perimeter of the triangle by summing the lengths of all three sides.
15+8+17=4015 + 8 + 17 = 40.
The perimeter of a triangle is the total distance around the boundary, which equals the sum of its three side lengths.

Key Concept

Calculating the perimeter of a right triangle in the coordinate plane by using the Pythagorean theorem to find the hypotenuse.
Question 65Question

In right triangle ABCABC, the measure of angle CC is 9090^\circ. If cos(A)=3sin(A)\cos(A) = 3\sin(A), what is the value of tan(B)\tan(B)?

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Answer: 3

Answer

The correct answer is 33.
Because angle CC is 9090^\circ in right triangle ABCABC, the acute angles AA and BB are complementary (A+B=90A + B = 90^\circ). By the co-function identities, sin(B)=cos(A)\sin(B) = \cos(A) and cos(B)=sin(A)\cos(B) = \sin(A). The tangent of BB is defined as tan(B)=sin(B)cos(B)\tan(B) = \frac{\sin(B)}{\cos(B)}. Substituting the co-function identities gives tan(B)=cos(A)sin(A)\tan(B) = \frac{\cos(A)}{\sin(A)}. Since we are given that cos(A)=3sin(A)\cos(A) = 3\sin(A), we substitute this expression into the numerator to get tan(B)=3sin(A)sin(A)=3\tan(B) = \frac{3\sin(A)}{\sin(A)} = 3.

Step-by-Step Solution

1
Determine the relationship between the acute angles in right triangle ABCABC.
sin(B)=cos(A)\sin(B) = \cos(A) and cos(B)=sin(A)\cos(B) = \sin(A)
Since angle CC is 9090^\circ, the other two angles AA and BB must sum to 9090^\circ (they are complementary angles).
2
Express tan(B)\tan(B) in terms of the trigonometric ratios of angle AA.
tan(B)=cos(A)sin(A)\tan(B) = \frac{\cos(A)}{\sin(A)}
By definition, the tangent of angle BB is the ratio of its sine to its cosine, which yields cos(A)sin(A)\frac{\cos(A)}{\sin(A)} after substituting the complementary angle relations.
3
Substitute the given relation cos(A)=3sin(A)\cos(A) = 3\sin(A) into the expression for tan(B)\tan(B).
tan(B)=3sin(A)sin(A)\tan(B) = \frac{3\sin(A)}{\sin(A)}
Substituting the value of cos(A)\cos(A) allows us to simplify the fraction by expressing both terms with sin(A)\sin(A).
4
Simplify the fraction to get the final numerical value.
tan(B)=3\tan(B) = 3
The term sin(A)\sin(A) cancels out from the numerator and denominator since AA is an acute angle and sin(A)0\sin(A) \neq 0.

Key Concept

Co-function identities and trigonometric definitions in a right triangle.

Alternative Method

Alternatively, construct a right triangle where the side opposite to angle AA has length 11. Since cos(A)=3sin(A)\cos(A) = 3\sin(A), the ratio of the adjacent side to the hypotenuse is 33 times the ratio of the opposite side to the hypotenuse, meaning the side adjacent to angle AA must have length 33. Because angle BB is the complement of angle AA, the side opposite to angle BB is the side adjacent to angle AA (which is 33), and the side adjacent to angle BB is the side opposite to angle AA (which is 11). Therefore, tan(B)=oppositeadjacent=31=3\tan(B) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{1} = 3.
Estimated Time:1m 30s
Question 66Question

In rectangle ABCDABCD, the length of side ABAB is 1212 and the length of side BCBC is 99. Point PP lies on the diagonal ACAC such that AP=13ACAP = \frac{1}{3} AC. What is the area of triangle BPDBPD?

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Answer: 18

Answer

18
The correct answer is 18. The area of the right triangle ABDABD is half of the area of rectangle ABCDABCD, which is 12×92=54\frac{12 \times 9}{2} = 54. Since point PP lies on diagonal ACAC such that AP=13ACAP = \frac{1}{3} AC, triangle ABPABP has a base of APAP along line ACAC and shares vertex BB with triangle ABCABC. Thus, its area is 13\frac{1}{3} of the area of triangle ABCABC, which is 543=18\frac{54}{3} = 18. Similarly, triangle ADPADP shares vertex DD with triangle ADCADC and has base APAP, so its area is 13\frac{1}{3} of the area of triangle ADCADC, which is 543=18\frac{54}{3} = 18. Because AP=13AC<12ACAP = \frac{1}{3} AC < \frac{1}{2} AC, point PP lies inside triangle ABDABD. Therefore, the area of triangle BPDBPD is the area of triangle ABDABD minus the areas of triangles ABPABP and ADPADP, which is 541818=1854 - 18 - 18 = 18.

Step-by-Step Solution

1
Calculate the area of triangle ABDABD.
Area(ABD)=12×92=54\text{Area}(\triangle ABD) = \frac{12 \times 9}{2} = 54
The diagonal BDBD divides the rectangle ABCDABCD into two congruent right triangles, each with an area equal to half of the rectangle's total area.
2
Find the areas of triangles ABPABP and ADPADP using the ratio of APAP to ACAC.
Area(ABP)=13×Area(ABC)=18\text{Area}(\triangle ABP) = \frac{1}{3} \times \text{Area}(\triangle ABC) = 18 and Area(ADP)=13×Area(ADC)=18\text{Area}(\triangle ADP) = \frac{1}{3} \times \text{Area}(\triangle ADC) = 18.
Triangles ABPABP and ABCABC share the same altitude from vertex BB to the line containing diagonal ACAC. Therefore, the ratio of their areas is equal to the ratio of their bases, which is APAC=13\frac{AP}{AC} = \frac{1}{3}. The same logic applies to triangles ADPADP and ADCADC with vertex DD.
3
Subtract the areas of triangles ABPABP and ADPADP from the area of triangle ABDABD to find the area of triangle BPDBPD.
Area(BPD)=541818=18\text{Area}(\triangle BPD) = 54 - 18 - 18 = 18
Since AP=13ACAP = \frac{1}{3} AC, which is less than half the length of the diagonal, point PP lies strictly within the interior of triangle ABDABD. Thus, the area of triangle ABDABD is partitioned into the areas of triangles ABPABP, ADPADP, and BPDBPD.

Key Concept

Partitioning the area of a polygon and using the ratio of bases for triangles sharing a vertex to compute sub-areas.

Alternative Method

Alternatively, place the rectangle in a coordinate system with BB at the origin (0,0)(0,0), CC at (12,0)(12,0), AA at (0,9)(0,9), and DD at (12,9)(12,9). The coordinates of point PP on diagonal ACAC (from (0,9)(0,9) to (12,0)(12,0)) at one-third of the distance from AA to CC are x=0+13(120)=4x = 0 + \frac{1}{3}(12 - 0) = 4 and y=9+13(09)=6y = 9 + \frac{1}{3}(0 - 9) = 6. The area of triangle BPDBPD with vertices B(0,0)B(0,0), P(4,6)P(4,6), and D(12,9)D(12,9) can be found using the shoelace formula: Area=120(69)+4(90)+12(06)=123672=18\text{Area} = \frac{1}{2} |0(6 - 9) + 4(9 - 0) + 12(0 - 6)| = \frac{1}{2} |36 - 72| = 18.
Estimated Time:2m 0s
Question 67Question

In right triangle RSTRST, the measure of angle TT is 9090^\circ and the length of the hypotenuse RSRS is 1010. If tan(R)+tan(S)=52\tan(R) + \tan(S) = \frac{5}{2}, what is the area of triangle RSTRST?

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Answer: 20

Answer

20
Let the lengths of the legs of right triangle RSTRST be RT=xRT = x and ST=yST = y. The tangent ratios are tan(R)=yx\tan(R) = \frac{y}{x} and tan(S)=xy\tan(S) = \frac{x}{y}. Substituting these into the given equation gives yx+xy=52\frac{y}{x} + \frac{x}{y} = \frac{5}{2}, which simplifies to x2+y2xy=52\frac{x^2 + y^2}{xy} = \frac{5}{2}. By the Pythagorean theorem, x2+y2=102=100x^2 + y^2 = 10^2 = 100. Thus, 100xy=52\frac{100}{xy} = \frac{5}{2}, which solves to xy=40xy = 40. The area of the triangle is 12xy=12(40)=20\frac{1}{2}xy = \frac{1}{2}(40) = 20.

Step-by-Step Solution

1
Express the tangent of angles RR and SS in terms of the leg lengths RTRT and STST.
tan(R)=STRT\tan(R) = \frac{ST}{RT} and tan(S)=RTST\tan(S) = \frac{RT}{ST}.
By definition, the tangent of an acute angle in a right triangle is the ratio of the opposite leg to the adjacent leg.
2
Substitute these expressions into the given equation and simplify using the Pythagorean theorem.
STRT+RTST=ST2+RT2RTST=100RTST=52\frac{ST}{RT} + \frac{RT}{ST} = \frac{ST^2 + RT^2}{RT \cdot ST} = \frac{100}{RT \cdot ST} = \frac{5}{2}.
Finding a common denominator yields the sum of the squares of the legs in the numerator, which equals the square of the hypotenuse (RS2=102=100RS^2 = 10^2 = 100).
3
Solve for the product of the legs RTSTRT \cdot ST and calculate the area of the triangle.
RTST=40RT \cdot ST = 40, so Area=12(RTST)=12(40)=20\text{Area} = \frac{1}{2}(RT \cdot ST) = \frac{1}{2}(40) = 20.
The area of a right triangle is half the product of its perpendicular legs.

Key Concept

Using trigonometric ratios, complementary angles, and the Pythagorean theorem to calculate the area of a right triangle.
Question 68Question

In right triangle ABCABC, the hypotenuse ACAC has length 2525, and leg ABAB has length 2020. What is the length of leg BCBC?

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Answer: 15

Answer

15
For any right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse (a2+b2=c2a^2 + b^2 = c^2). In right triangle ABCABC, the hypotenuse is AC=25AC = 25 and one of the legs is AB=20AB = 20. Substituting these values into the theorem gives 202+BC2=25220^2 + BC^2 = 25^2, which simplifies to 400+BC2=625400 + BC^2 = 625. Subtracting 400400 from both sides results in BC2=225BC^2 = 225. Taking the square root of both sides gives the length of leg BCBC as 1515.

Step-by-Step Solution

1
Set up the Pythagorean theorem equation for the right triangle.
AB2+BC2=AC2AB^2 + BC^2 = AC^2
The Pythagorean theorem relates the lengths of the legs and the hypotenuse of any right triangle.
2
Substitute the known lengths of side ABAB and hypotenuse ACAC.
202+BC2=25220^2 + BC^2 = 25^2
We plug in 2020 for leg ABAB and 2525 for hypotenuse ACAC.
3
Evaluate the squared terms.
400+BC2=625400 + BC^2 = 625
Squaring 2020 gives 400400, and squaring 2525 gives 625625.
4
Isolate the variable term BC2BC^2.
BC2=225BC^2 = 225
Subtracting 400400 from both sides of the equation yields 225225.
5
Find the length of side BCBC.
BC=15BC = 15
Taking the square root of 225225 gives the positive length of leg BCBC.

Key Concept

Applying the Pythagorean theorem to calculate a missing leg length of a right triangle when the hypotenuse and one leg length are given.
Question 69Question

An angle has a measure of 7272^\circ. What is the measure of this angle in radians?

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Answer: 2π5\frac{2\pi}{5}

Answer

The correct answer is 2π5\frac{2\pi}{5} radians.
To convert an angle from degrees to radians, multiply the degree measure by π180\frac{\pi}{180}. Doing so for 7272^\circ yields 72π180\frac{72\pi}{180}. Dividing the numerator and denominator by their greatest common divisor, 36, simplifies the fraction to 2π5\frac{2\pi}{5}.

Step-by-Step Solution

1
Set up the conversion from degrees to radians by multiplying the given degree measure by π180\frac{\pi}{180}.
θ=72×π180=72π180\theta = 72 \times \frac{\pi}{180} = \frac{72\pi}{180}
The conversion factor from degrees to radians is π radians180\frac{\pi \text{ radians}}{180^\circ}.
2
Simplify the fraction by dividing the numerator and the denominator by their greatest common divisor, 36.
72÷36180÷36π=2π5\frac{72 \div 36}{180 \div 36}\pi = \frac{2\pi}{5}
Simplifying the fraction expresses the angle in standard reduced radian form.

Key Concept

Converting degrees to radians
Question 70Question

In right triangle ABCABC, the measure of angle CC is 9090^\circ, AC=15AC = 15, and BC=20BC = 20. Altitude CDCD is drawn to the hypotenuse ABAB. From point DD, perpendicular line segments DEDE and DFDF are drawn to sides ACAC and BCBC, respectively, where EE lies on ACAC and FF lies on BCBC. What is the ratio of the area of triangle CEFCEF to the area of triangle ABCABC?

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Answer: 144625\frac{144}{625}

Answer

The ratio of the area of triangle CEFCEF to the area of triangle ABCABC is 144625\frac{144}{625}.
The correct answer is the fraction representing the ratio of the area of the smaller triangle to the larger triangle. This ratio can be computed by calculating the lengths of the legs of the right triangle CEFCEF using similarity relationships, yielding CE=9.6CE = 9.6 and CF=7.2CF = 7.2, which gives an area of 34.5634.56. Dividing this by the area of triangle ABCABC (150150) yields the fraction 144625\frac{144}{625}. Alternatively, the ratio is equal to the square of the ratio of the altitude to the hypotenuse, (CDAB)2=(1225)2=144625(\frac{CD}{AB})^2 = (\frac{12}{25})^2 = \frac{144}{625}.

Step-by-Step Solution

1
Use the Pythagorean theorem to calculate the hypotenuse ABAB of the right triangle ABCABC.
AB=AC2+BC2=152+202=225+400=625=25AB = \sqrt{AC^2 + BC^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25
Knowing the hypotenuse allows us to calculate the altitude and set up similarity ratios.
2
Calculate the area of triangle ABCABC.
Area(ABC)=12×AC×BC=12×15×20=150\text{Area}(ABC) = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times 15 \times 20 = 150
This establishes the base area to compare with the area of the smaller triangle.
3
Calculate the length of altitude CDCD by relating the area using base ABAB.
150=12×AB×CD150=12×25×CDCD=12150 = \frac{1}{2} \times AB \times CD \Rightarrow 150 = \frac{1}{2} \times 25 \times CD \Rightarrow CD = 12
The altitude is a key dimension that relates the nested geometry to the main triangle.
4
Identify the properties of the quadrilateral DECFDECF and the triangle CEFCEF.
DECFDECF is a rectangle with diagonals CD=EF=12CD = EF = 12, and CEF\triangle CEF is a right triangle with legs CECE and CFCF.
Since three angles are 9090^\circ (at CC, EE, and FF), the figure is a rectangle, meaning its diagonals are congruent.
5
Apply similarity properties to find the lengths of CECE and CFCF.
CE=9.6CE = 9.6 and CF=7.2CF = 7.2.
Using similar right triangles, CE=ACAE=155.4=9.6CE = AC - AE = 15 - 5.4 = 9.6 and CF=BCBF=2012.8=7.2CF = BC - BF = 20 - 12.8 = 7.2.
6
Calculate the area of right triangle CEFCEF and find the ratio.
Area(CEF)=12×9.6×7.2=34.56\text{Area}(CEF) = \frac{1}{2} \times 9.6 \times 7.2 = 34.56. The ratio is 34.56150=144625\frac{34.56}{150} = \frac{144}{625}. Alternatively, the ratio is (CDAB)2=(1225)2=144625\left(\frac{CD}{AB}\right)^2 = \left(\frac{12}{25}\right)^2 = \frac{144}{625}.
Comparing the two areas gives the final required ratio.

Key Concept

Area ratios and side lengths in similar right triangles using perpendicular projections.

Alternative Method

Using trigonometric ratios, we can express the lengths in terms of θ=A\theta = \angle A. Since CD=ABsinθcosθCD = AB \sin\theta \cos\theta, we have CE=CDsinθCE = CD \sin\theta and CF=CDcosθCF = CD \cos\theta. The area of triangle CEFCEF is 12CECF=12CD2sinθcosθ=CD2Area(ABC)AB2\frac{1}{2} CE \cdot CF = \frac{1}{2} CD^2 \sin\theta \cos\theta = CD^2 \frac{\text{Area}(ABC)}{AB^2}, which gives the area ratio as (CDAB)2\left(\frac{CD}{AB}\right)^2. Since AB=25AB = 25 and CD=12CD = 12, the ratio is (1225)2=144625\left(\frac{12}{25}\right)^2 = \frac{144}{625}.
Estimated Time:3m 0s
Question 71Question

In a right triangle, one of the acute angles is θ\theta. If tan(θ)=xy\tan(\theta) = \frac{x}{y}, where xx and yy are positive constants, what is the value of cos(π2θ)\cos\left(\frac{\pi}{2} - \theta\right) in terms of xx and yy?

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Answer: xx2+y2\frac{x}{\sqrt{x^2 + y^2}}

Answer

xx2+y2\frac{x}{\sqrt{x^2 + y^2}}
Using the cofunction identity, cos(π2θ)\cos\left(\frac{\pi}{2} - \theta\right) is equivalent to sin(θ)\sin(\theta). For an angle θ\theta in a right triangle, tan(θ)=xy\tan(\theta) = \frac{x}{y} defines the ratio of the opposite side (xx) to the adjacent side (yy). Applying the Pythagorean theorem, the hypotenuse is x2+y2\sqrt{x^2 + y^2}. Therefore, sin(θ)\sin(\theta), which is the ratio of the opposite side to the hypotenuse, is equal to xx2+y2\frac{x}{\sqrt{x^2 + y^2}}.

Step-by-Step Solution

1
Apply the cofunction identity to rewrite the target expression.
cos(π2θ)=sin(θ)\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta)
By the cofunction identities, the sine of an acute angle is equal to the cosine of its complementary angle.
2
Relate the given tangent ratio to the side lengths of a right triangle.
tan(θ)=oppositeadjacent=xy\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{y}, so the opposite side can be represented as xx and the adjacent side as yy.
The tangent of an angle in a right triangle is the ratio of the length of the opposite side to the length of the adjacent side.
3
Apply the Pythagorean theorem to calculate the hypotenuse of the triangle.
Hypotenuse=x2+y2\text{Hypotenuse} = \sqrt{x^2 + y^2}
The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
4
Express the sine of the angle as a ratio of the opposite side to the hypotenuse.
sin(θ)=xx2+y2\sin(\theta) = \frac{x}{\sqrt{x^2 + y^2}}
The sine of an angle in a right triangle is the ratio of the length of the opposite side to the length of the hypotenuse.

Key Concept

Cofunction identities and right triangle trigonometric ratios
Estimated Time:1m 30s
Question 72Question

In the figure, triangle RSTRST is a right triangle with the right angle at vertex SS. If the length of side RSRS is 88 and the length of side STST is 1515, what is the length of the hypotenuse RTRT?

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Answer: 17

Answer

17
According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the legs: RT2=RS2+ST2RT^2 = RS^2 + ST^2. Substituting the given values yields RT2=82+152=64+225=289RT^2 = 8^2 + 15^2 = 64 + 225 = 289. Taking the square root of 289289 gives RT=17RT = 17.

Step-by-Step Solution

1
Set up the Pythagorean theorem for right triangle RSTRST.
RS2+ST2=RT2RS^2 + ST^2 = RT^2
In any right triangle, the sum of the squares of the legs equals the square of the hypotenuse.
2
Substitute the given side lengths into the equation.
82+152=RT28^2 + 15^2 = RT^2
The legs of the triangle are RS=8RS = 8 and ST=15ST = 15.
3
Calculate the squares and sum them.
64+225=289=RT264 + 225 = 289 = RT^2
82=648^2 = 64 and 152=22515^2 = 225. Adding them gives 289289.
4
Take the square root of both sides to find the length of the hypotenuse.
RT=289=17RT = \sqrt{289} = 17
The length must be a positive value, and 289\sqrt{289} simplifies to 1717.

Key Concept

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs: a2+b2=c2a^2 + b^2 = c^2.
Estimated Time:45s
Question 73Question

A wheel rotates through a central angle of 11π18\frac{11\pi}{18} radians. What is the measure of this angle in degrees?

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Answer: 110

Answer

110
To convert radians to degrees, multiply the angle by 180π\frac{180}{\pi}. Thus, 11π18×180π=110\frac{11\pi}{18} \times \frac{180}{\pi} = 110. The measure of the angle is 110 degrees.

Step-by-Step Solution

1
Identify the conversion relationship between radians and degrees.
Multiply the angle in radians by 180π\frac{180}{\pi} to convert to degrees.
Since π\pi radians is equal to 180180 degrees, the conversion factor is 180π\frac{180}{\pi}.
2
Multiply the given radian measure of 11π18\frac{11\pi}{18} by the conversion factor.
110
Applying the conversion factor simplifies the expression by canceling π\pi and dividing 180180 by 1818 to get 1010, which is then multiplied by 1111 to get the final degree measure.

Key Concept

To convert an angle from radians to degrees, multiply the angle in radians by 180π\frac{180}{\pi}.
Estimated Time:45s
Question 74Question

In triangle ABCABC, the angle at vertex BB is a right angle. The lengths of sides ABAB and BCBC are 1212 and 1616, respectively. A point DD is chosen on the hypotenuse ACAC such that AD=5AD = 5. A line drawn through DD perpendicular to ACAC intersects the line passing through BB and CC at point GG, such that BB lies between GG and CC. What is the length of segment GDGD?

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Answer: 11.25

Answer

The length of segment GDGD is 11.25.
By the Pythagorean theorem, the hypotenuse ACAC of right triangle ABCABC is 122+162=20\sqrt{12^2 + 16^2} = 20. Subtracting the length of ADAD from ACAC gives DC=205=15DC = 20 - 5 = 15. Because the line GDGD is perpendicular to ACAC, the angle GDC\angle GDC is 9090^\circ. The triangles GDC\triangle GDC and ABC\triangle ABC share the angle at vertex CC and both have a right angle, which means they are similar by Angle-Angle (AA) similarity: GDCABC\triangle GDC \sim \triangle ABC. Using the ratio of corresponding sides, we have GDAB=DCBC\frac{GD}{AB} = \frac{DC}{BC}, which translates to GD12=1516\frac{GD}{12} = \frac{15}{16}. Solving for GDGD yields GD=12×1516=11.25GD = 12 \times \frac{15}{16} = 11.25.

Step-by-Step Solution

1
Calculate the length of the hypotenuse ACAC using the Pythagorean theorem in right triangle ABCABC.
AC=122+162=20AC = \sqrt{12^2 + 16^2} = 20
The length of ACAC is required to find the segment lengths on the hypotenuse.
2
Determine the length of segment DCDC.
DC=ACAD=205=15DC = AC - AD = 20 - 5 = 15
The segment DCDC is a side of the similar triangle GDC\triangle GDC that corresponds to side BCBC in ABC\triangle ABC.
3
Establish the similarity between triangles GDC\triangle GDC and ABC\triangle ABC.
GDCABC\triangle GDC \sim \triangle ABC by AA similarity
Both triangles share the angle at vertex CC, and both have a right angle (GDC=ABC=90\angle GDC = \angle ABC = 90^\circ).
4
Set up the ratio of corresponding sides and solve for GDGD.
GDAB=DCBCGD=12×1516=11.25\frac{GD}{AB} = \frac{DC}{BC} \Rightarrow GD = 12 \times \frac{15}{16} = 11.25
The ratio of corresponding sides in similar triangles is equal.

Key Concept

Using right triangle similarity and the Pythagorean theorem to solve for unknown side lengths.
Question 75Question

If the measure of an angle is 5π4\frac{5\pi}{4} radians, what is its measure in degrees?

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Answer: 225225^\circ

Answer

225225^\circ
To convert an angle from radians to degrees, multiply the radian measure by 180π\frac{180^\circ}{\pi}. Multiplying 5π4\frac{5\pi}{4} by 180π\frac{180^\circ}{\pi} simplifies to 54×180=5×45=225\frac{5}{4} \times 180^\circ = 5 \times 45^\circ = 225^\circ. Therefore, the correct degree measure of the angle is 225225^\circ.

Step-by-Step Solution

1
Set up the conversion from radians to degrees by multiplying the given radian measure by 180π\frac{180^\circ}{\pi}.
5π4×180π\frac{5\pi}{4} \times \frac{180^\circ}{\pi}
Since π\pi radians is equal to 180180^\circ, multiplying by the conversion factor 180π\frac{180^\circ}{\pi} converts the units from radians to degrees.
2
Simplify the expression by canceling the π\pi terms and performing the arithmetic.
225225^\circ
The π\pi in the numerator and denominator cancel out, leaving 54×180\frac{5}{4} \times 180^\circ. Dividing 180180 by 44 gives 4545, and multiplying 55 by 4545 yields 225225.

Key Concept

To convert an angle from radians to degrees, multiply the angle in radians by 180π\frac{180^\circ}{\pi}.

Alternative Method

Another way to convert is to substitute 180180^\circ directly for π\pi in the expression: 5π4=5(180)4=5×45=225\frac{5\pi}{4} = \frac{5(180^\circ)}{4} = 5 \times 45^\circ = 225^\circ.
Estimated Time:45s
Question 76Question

For an acute angle θ\theta, cos(θ)=513\cos(\theta) = \frac{5}{13}. What is the value of 5tan(θ)+13sin(θ)5\tan(\theta) + 13\sin(\theta)?

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Answer: 24

Answer

24
Using the Pythagorean identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 with cos(θ)=513\cos(\theta) = \frac{5}{13} gives sin(θ)=1213\sin(\theta) = \frac{12}{13} because θ\theta is an acute angle. The quotient identity gives tan(θ)=sin(θ)cos(θ)=125\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{12}{5}. Substituting these ratios into the expression 5tan(θ)+13sin(θ)5\tan(\theta) + 13\sin(\theta) gives 5(125)+13(1213)=12+12=245\left(\frac{12}{5}\right) + 13\left(\frac{12}{13}\right) = 12 + 12 = 24.

Step-by-Step Solution

1
Find the value of sin(θ)\sin(\theta) using the Pythagorean identity.
sin(θ)=1213\sin(\theta) = \frac{12}{13}
Since sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and θ\theta is an acute angle, the sine value is positive: sin(θ)=1(513)2=1213\sin(\theta) = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}.
2
Find the value of tan(θ)\tan(\theta) using the quotient identity.
tan(θ)=125\tan(\theta) = \frac{12}{5}
By definition, tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}. Substituting the known values yields tan(θ)=12/135/13=125\tan(\theta) = \frac{12/13}{5/13} = \frac{12}{5}.
3
Substitute the trigonometric ratios into the given expression and simplify.
24
Substituting the values of tan(θ)\tan(\theta) and sin(θ)\sin(\theta) into 5tan(θ)+13sin(θ)5\tan(\theta) + 13\sin(\theta) gives 5(125)+13(1213)=12+12=245\left(\frac{12}{5}\right) + 13\left(\frac{12}{13}\right) = 12 + 12 = 24.

Key Concept

Trigonometric ratios and identities, specifically the Pythagorean identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and the definition of tangent as sin(θ)cos(θ)\frac{\sin(\theta)}{\cos(\theta)}.

Alternative Method

Alternatively, draw a right triangle with an acute angle θ\theta. Since cos(θ)=513=adjacenthypotenuse\cos(\theta) = \frac{5}{13} = \frac{\text{adjacent}}{\text{hypotenuse}}, label the adjacent side as 5 and the hypotenuse as 13. By the Pythagorean theorem, the opposite side is 13252=12\sqrt{13^2 - 5^2} = 12. From this triangle, sin(θ)=oppositehypotenuse=1213\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13} and tan(θ)=oppositeadjacent=125\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{5}. Substituting these values into the expression gives 5(125)+13(1213)=245\left(\frac{12}{5}\right) + 13\left(\frac{12}{13}\right) = 24.
Estimated Time:1m 30s
Question 77Question

For an acute angle θ\theta, the sum of its sine and cosine is 72\frac{\sqrt{7}}{2}. What is the value of sin3(θ)+cos3(θ)\sin^3(\theta) + \cos^3(\theta)?

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Answer: 5716\frac{5\sqrt{7}}{16}

Answer

The correct answer is 5716\frac{5\sqrt{7}}{16}.
The correct answer is 5716\frac{5\sqrt{7}}{16}. Squaring the given equation sin(θ)+cos(θ)=72\sin(\theta) + \cos(\theta) = \frac{\sqrt{7}}{2} gives sin2(θ)+2sin(θ)cos(θ)+cos2(θ)=74\sin^2(\theta) + 2\sin(\theta)\cos(\theta) + \cos^2(\theta) = \frac{7}{4}. Substituting the Pythagorean identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 gives 1+2sin(θ)cos(θ)=741 + 2\sin(\theta)\cos(\theta) = \frac{7}{4}, which simplifies to sin(θ)cos(θ)=38\sin(\theta)\cos(\theta) = \frac{3}{8}. Factoring the sum of cubes yields sin3(θ)+cos3(θ)=(sin(θ)+cos(θ))(1sin(θ)cos(θ))\sin^3(\theta) + \cos^3(\theta) = (\sin(\theta) + \cos(\theta))(1 - \sin(\theta)\cos(\theta)). Substituting the values results in (72)(138)=5716\left(\frac{\sqrt{7}}{2}\right)\left(1 - \frac{3}{8}\right) = \frac{5\sqrt{7}}{16}.

Step-by-Step Solution

1
Square both sides of the given equation sin(θ)+cos(θ)=72\sin(\theta) + \cos(\theta) = \frac{\sqrt{7}}{2}.
(sin(θ)+cos(θ))2=sin2(θ)+2sin(θ)cos(θ)+cos2(θ)=74(\sin(\theta) + \cos(\theta))^2 = \sin^2(\theta) + 2\sin(\theta)\cos(\theta) + \cos^2(\theta) = \frac{7}{4}.
This sets up the expression to apply the Pythagorean trigonometric identity.
2
Substitute the Pythagorean identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and solve for the product term.
1+2sin(θ)cos(θ)=74    2sin(θ)cos(θ)=34    sin(θ)cos(θ)=381 + 2\sin(\theta)\cos(\theta) = \frac{7}{4} \implies 2\sin(\theta)\cos(\theta) = \frac{3}{4} \implies \sin(\theta)\cos(\theta) = \frac{3}{8}.
This determines the value of the product of the sine and cosine ratios.
3
Apply the sum of cubes factorization formula to sin3(θ)+cos3(θ)\sin^3(\theta) + \cos^3(\theta).
sin3(θ)+cos3(θ)=(sin(θ)+cos(θ))(sin2(θ)sin(θ)cos(θ)+cos2(θ))=(sin(θ)+cos(θ))(1sin(θ)cos(θ))\sin^3(\theta) + \cos^3(\theta) = (\sin(\theta) + \cos(\theta))(\sin^2(\theta) - \sin(\theta)\cos(\theta) + \cos^2(\theta)) = (\sin(\theta) + \cos(\theta))(1 - \sin(\theta)\cos(\theta)).
This expresses the target quantity in terms of known sum and product values.
4
Substitute the values of the sum and product into the factored expression and simplify.
sin3(θ)+cos3(θ)=(72)(138)=(72)(58)=5716\sin^3(\theta) + \cos^3(\theta) = \left(\frac{\sqrt{7}}{2}\right)\left(1 - \frac{3}{8}\right) = \left(\frac{\sqrt{7}}{2}\right)\left(\frac{5}{8}\right) = \frac{5\sqrt{7}}{16}.
This yields the final simplified numerical ratio.

Key Concept

Pythagorean Identity and Algebraic Manipulation of Trigonometric Ratios
Question 78Question

For an acute angle θ\theta, the equation 2cos2(θ)5sin(θ)+1=02\cos^2(\theta) - 5\sin(\theta) + 1 = 0 is true. What is the value of sin(θ)\sin(\theta)?

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Answer: 0.5

Answer

The value of sin(θ)\sin(\theta) is 0.50.5 (or 12\frac{1}{2})
By applying the Pythagorean identity cos2(θ)=1sin2(θ)\cos^2(\theta) = 1 - \sin^2(\theta), the equation 2cos2(θ)5sin(θ)+1=02\cos^2(\theta) - 5\sin(\theta) + 1 = 0 can be rewritten entirely in terms of sin(θ)\sin(\theta), yielding 2sin2(θ)5sin(θ)+3=0-2\sin^2(\theta) - 5\sin(\theta) + 3 = 0. Multiplying by 1-1 gives the standard quadratic equation 2sin2(θ)+5sin(θ)3=02\sin^2(\theta) + 5\sin(\theta) - 3 = 0, which factors as (2sin(θ)1)(sin(θ)+3)=0(2\sin(\theta) - 1)(\sin(\theta) + 3) = 0. Solving for sin(θ)\sin(\theta) yields sin(θ)=0.5\sin(\theta) = 0.5 or sin(θ)=3\sin(\theta) = -3. Since the sine value of any angle must be in the range [1,1][-1, 1] and the sine of an acute angle must be positive, sin(θ)=0.5\sin(\theta) = 0.5 is the only valid solution.

Step-by-Step Solution

1
Apply the Pythagorean identity to rewrite the cosine term.
2(1sin2(θ))5sin(θ)+1=02(1 - \sin^2(\theta)) - 5\sin(\theta) + 1 = 0
The equation contains both cos2(θ)\cos^2(\theta) and sin(θ)\sin(\theta). Substituting cos2(θ)=1sin2(θ)\cos^2(\theta) = 1 - \sin^2(\theta) allows the equation to be expressed in terms of a single trigonometric function, sin(θ)\sin(\theta).
2
Distribute and simplify the equation into standard quadratic form.
2sin2(θ)+5sin(θ)3=02\sin^2(\theta) + 5\sin(\theta) - 3 = 0
Expanding the equation yields 22sin2(θ)5sin(θ)+1=02 - 2\sin^2(\theta) - 5\sin(\theta) + 1 = 0, which simplifies to 2sin2(θ)5sin(θ)+3=0-2\sin^2(\theta) - 5\sin(\theta) + 3 = 0. Multiplying the entire equation by 1-1 puts it into standard quadratic form as2+bs+c=0as^2 + bs + c = 0.
3
Factor the quadratic expression.
(2sin(θ)1)(sin(θ)+3)=0(2\sin(\theta) - 1)(\sin(\theta) + 3) = 0
Finding two numbers that multiply to 6-6 (from 2×32 \times -3) and add to 55 leads to the factors 66 and 1-1. Splitting the middle term and factoring by grouping yields (2sin(θ)1)(sin(θ)+3)=0(2\sin(\theta) - 1)(\sin(\theta) + 3) = 0.
4
Determine the valid solution based on the angle's constraints.
sin(θ)=0.5\sin(\theta) = 0.5
Setting each factor to zero gives sin(θ)=0.5\sin(\theta) = 0.5 or sin(θ)=3\sin(\theta) = -3. Since the sine of any real angle must be between 1-1 and 11, sin(θ)=3\sin(\theta) = -3 is undefined. Furthermore, because θ\theta is an acute angle (0<θ<900^\circ < \theta < 90^\circ), the sine value must be positive, confirming sin(θ)=0.5\sin(\theta) = 0.5.

Key Concept

Pythagorean identity and quadratic trigonometric equations
Estimated Time:2m 0s
Question 79Question

For an acute angle xx measured in degrees, sin(x)sin(90x)=15\sin(x) - \sin(90^\circ - x) = \frac{1}{5}. What is the value of 12(tan(x)+tan(90x))12(\tan(x) + \tan(90^\circ - x))?

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Answer: 25

Answer

The value of the expression is 25.
Applying the co-function identity sin(90x)=cos(x)\sin(90^\circ - x) = \cos(x) allows the given equation to be written as sin(x)cos(x)=15\sin(x) - \cos(x) = \frac{1}{5}. Squaring both sides of this equation yields sin2(x)2sin(x)cos(x)+cos2(x)=125\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = \frac{1}{25}. Applying the Pythagorean identity sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1 simplifies this to 12sin(x)cos(x)=1251 - 2\sin(x)\cos(x) = \frac{1}{25}, which gives sin(x)cos(x)=1225\sin(x)\cos(x) = \frac{12}{25}. The expression to be evaluated is 12(tan(x)+tan(90x))12(\tan(x) + \tan(90^\circ - x)). Using the identity tan(90x)=cot(x)\tan(90^\circ - x) = \cot(x), this expression can be rewritten as 12(sin(x)cos(x)+cos(x)sin(x))=12(sin2(x)+cos2(x)sin(x)cos(x))=12(1sin(x)cos(x))12\left(\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}\right) = 12\left(\frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)}\right) = 12\left(\frac{1}{\sin(x)\cos(x)}\right). Substituting the value of sin(x)cos(x)\sin(x)\cos(x) gives 12×2512=2512 \times \frac{25}{12} = 25.

Step-by-Step Solution

1
Apply the co-function identity to rewrite the equation.
sin(x)cos(x)=15\sin(x) - \cos(x) = \frac{1}{5}
Since sin(90x)=cos(x)\sin(90^\circ - x) = \cos(x) for any angle xx, we can substitute cos(x)\cos(x) into the given equation.
2
Square both sides of the rewritten equation.
sin2(x)2sin(x)cos(x)+cos2(x)=125\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = \frac{1}{25}
Squaring both sides allows us to use the Pythagorean trigonometric identity to find the product of sine and cosine.
3
Substitute the Pythagorean identity and solve for sin(x)cos(x)\sin(x)\cos(x).
sin(x)cos(x)=1225\sin(x)\cos(x) = \frac{12}{25}
Substituting sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1 yields 12sin(x)cos(x)=1251 - 2\sin(x)\cos(x) = \frac{1}{25}, which simplifies to sin(x)cos(x)=1225\sin(x)\cos(x) = \frac{12}{25}.
4
Rewrite the target expression in terms of sine and cosine.
tan(x)+tan(90x)=1sin(x)cos(x)\tan(x) + \tan(90^\circ - x) = \frac{1}{\sin(x)\cos(x)}
Using the co-function identity tan(90x)=cot(x)\tan(90^\circ - x) = \cot(x) and expressing the tangent and cotangent functions as ratios of sine and cosine yields sin(x)cos(x)+cos(x)sin(x)=sin2(x)+cos2(x)sin(x)cos(x)=1sin(x)cos(x)\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} = \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)} = \frac{1}{\sin(x)\cos(x)}.
5
Substitute the value of sin(x)cos(x)\sin(x)\cos(x) and multiply by 12.
25
Substituting sin(x)cos(x)=1225\sin(x)\cos(x) = \frac{12}{25} into 12(tan(x)+tan(90x))12(\tan(x) + \tan(90^\circ - x)) gives 12×2512=2512 \times \frac{25}{12} = 25.

Key Concept

Applying co-function identities, the Pythagorean identity, and fundamental trigonometric relations to simplify expressions.
Estimated Time:2m 0s
Question 80Question

A circle has a radius of 1010. A shaded sector of the circle has an area of 10π10\pi. What is the measure, in degrees, of the central angle of the shaded sector?

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Answer: 36

Answer

The measure of the central angle of the shaded sector is 36 degrees.
The total area of the circle is A=π(10)2=100πA = \pi (10)^2 = 100\pi. The shaded sector's area is 10π10\pi, which represents 10π100π=110\frac{10\pi}{100\pi} = \frac{1}{10} of the total area of the circle. Since a circle consists of 360360^\circ, the central angle of the sector is 110×360=36\frac{1}{10} \times 360^\circ = 36^\circ.

Step-by-Step Solution

1
Calculate the total area of the circle.
Total area = 100π100\pi
The area of a circle is calculated using the formula A=πr2A = \pi r^2 where the radius r=10r = 10.
2
Determine the proportion of the circle covered by the shaded sector.
Proportion = 110\frac{1}{10}
Dividing the sector's area of 10π10\pi by the total area of 100π100\pi gives the fraction of the circle represented by the sector.
3
Find the central angle in degrees.
Central angle = 36
Since a full circle has a central angle of 360360^\circ, multiplying the proportion 110\frac{1}{10} by 360360^\circ yields the central angle of the sector.

Key Concept

The ratio of the area of a sector to the total area of a circle is equal to the ratio of the sector's central angle measure to the total degree measure of a circle (360360^\circ).
Estimated Time:45s
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