Geometry and Trigonometry

184 soru

Soru 101Soru

A rhombus has an area of 120 cm2120\text{ cm}^2. If the length of one of its diagonals exceeds the length of the other diagonal by 14 cm14\text{ cm}, what is the perimeter of the rhombus in centimeters?

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Cevap: 52

Cevap

The perimeter of the rhombus is 52 cm.
The area of a rhombus is given by A = \frac{1}{2} d_1 d_2. Setting \frac{1}{2} d_1(d_1 + 14) = 120$ yields the quadratic equation d_1^2 + 14d_1 - 240 = 0, which factors to (d_1 - 10)(d_1 + 24) = 0. Taking the positive solution d_1 = 10\text{ cm} gives d_2 = 24\text{ cm}. The diagonals intersect at right angles, dividing the rhombus into four congruent right triangles with legs of 5 cm and 12 cm. The hypotenuse (side length s) is \sqrt{5^2 + 12^2} = 13\text{ cm}. Therefore, the perimeter is 4 \times 13\text{ cm} = 52\text{ cm}.

Adım Adım Çözüm

1
Set up the area formula for a rhombus in terms of its diagonals
d1d2=240d_1 \cdot d_2 = 240
The area of a rhombus is given by A = \frac{1}{2} d_1 d_2, so \frac{1}{2} d_1 d_2 = 120.
2
Substitute d_2 = d_1 + 14 into the area equation
d_1^2 + 14d_1 - 240 = 0
The difference between the diagonal lengths is 14 cm.
3
Solve the quadratic equation for d_1
d_1 = 10\text{ cm} \text{ and } d_2 = 24\text{ cm}
Factoring gives (d_1 - 10)(d_1 + 24) = 0. Discarding the negative root yields d_1 = 10 cm.
4
Calculate the side length s using the right-angled triangle formed by the perpendicular bisecting diagonals
s = 13\text{ cm}
s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2} = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\text{ cm}.
5
Multiply side length by 4 to get the total perimeter
P = 52\text{ cm}
All four sides of a rhombus are equal, so Perimeter = 4 \times s = 4 \times 13 = 52 cm.

Anahtar Kavram

Perimeter and Area of a Rhombus using Diagonals and Pythagorean Theorem
Soru 102Soru

A solid metal trophy is formed by mounting a right circular cone of slant height 5 cm5\text{ cm} on top of a right circular cylinder of height 8 cm8\text{ cm}. Both the cone and the cylinder share a common base radius of 3 cm3\text{ cm}. If the trophy is completely melted down and recast to form a solid pyramid with a square base of side length 6 cm6\text{ cm}, what is the vertical height of the pyramid?

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Cevap: 7π cm7\pi\text{ cm}

Cevap

7π cm7\pi\text{ cm}
First, find the height of the cone using the Pythagorean theorem: hcone=5232=4 cmh_{\text{cone}} = \sqrt{5^2 - 3^2} = 4\text{ cm}. The volume of the conical section is 13π(32)(4)=12π cm3\frac{1}{3}\pi (3^2)(4) = 12\pi\text{ cm}^3, and the volume of the cylindrical section is π(32)(8)=72π cm3\pi (3^2)(8) = 72\pi\text{ cm}^3. The total volume melted is 12π+72π=84π cm312\pi + 72\pi = 84\pi\text{ cm}^3. For the recast pyramid with square base area 62=36 cm26^2 = 36\text{ cm}^2, its volume is 13(36)h=12h\frac{1}{3}(36)h = 12h. Equating the volumes (12h=84π12h = 84\pi) yields a vertical height of 7π cm7\pi\text{ cm}.

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1
Calculate the vertical height of the conical top
hcone=5232=4 cmh_{\text{cone}} = \sqrt{5^2 - 3^2} = 4\text{ cm}
The slant height ll, radius rr, and vertical height hconeh_{\text{cone}} form a right-angled triangle where hcone=l2r2h_{\text{cone}} = \sqrt{l^2 - r^2}.
2
Compute the total volume of the composite metal solid
Vtotal=12π+72π=84π cm3V_{\text{total}} = 12\pi + 72\pi = 84\pi\text{ cm}^3
The total volume is the sum of the cone volume Vcone=13π(32)(4)=12π cm3V_{\text{cone}} = \frac{1}{3}\pi (3^2)(4) = 12\pi\text{ cm}^3 and the cylinder volume Vcylinder=π(32)(8)=72π cm3V_{\text{cylinder}} = \pi (3^2)(8) = 72\pi\text{ cm}^3.
3
Equate the total volume to the volume of the square pyramid and solve for its height
h=7π cmh = 7\pi\text{ cm}
The base area of the square pyramid is A=62=36 cm2A = 6^2 = 36\text{ cm}^2. Its volume is Vpyramid=13(36)h=12hV_{\text{pyramid}} = \frac{1}{3}(36)h = 12h. Setting 12h=84π12h = 84\pi gives h=84π12=7π cmh = \frac{84\pi}{12} = 7\pi\text{ cm}.

Anahtar Kavram

Conservation of volume during recasting of 3D solids and combined solid geometry
Tahmini Süre:2m 0s
Soru 103Soru

In a circle of radius 13 cm13\text{ cm}, two parallel chords ABAB and CDCD are drawn on the same side of the center OO. If AB=24 cmAB = 24\text{ cm} and CD=10 cmCD = 10\text{ cm}, calculate the perpendicular distance between the two chords in centimeters.

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Cevap: 7

Cevap

The perpendicular distance between the two chords is 7 cm7\text{ cm}.
The perpendicular line from the center OO to a chord bisects the chord. Applying the Pythagorean theorem to the right triangles formed by the radius (13 cm13\text{ cm}) and half-chords (12 cm12\text{ cm} and 5 cm5\text{ cm}) yields distances of 5 cm5\text{ cm} and 12 cm12\text{ cm} from the center, respectively. Since both parallel chords are on the same side of the center, the distance between them is 12 cm5 cm=7 cm12\text{ cm} - 5\text{ cm} = 7\text{ cm}.

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1
Find the perpendicular distance from center OO to chord ABAB
d1=5 cmd_1 = 5\text{ cm}
A line drawn from the center of a circle perpendicular to a chord bisects the chord. Using the right triangle formed by the radius, half-chord (12 cm12\text{ cm}), and perpendicular distance: d1=132122=5 cmd_1 = \sqrt{13^2 - 12^2} = 5\text{ cm}.
2
Find the perpendicular distance from center OO to chord CDCD
d2=12 cmd_2 = 12\text{ cm}
Using the perpendicular bisector property for chord CDCD (half-chord is 5 cm5\text{ cm}): d2=13252=12 cmd_2 = \sqrt{13^2 - 5^2} = 12\text{ cm}.
3
Calculate the distance between the parallel chords
7 cm7\text{ cm}
Because both chords are on the same side of the center OO, the distance between them is the difference of their individual distances from the center: 12 cm5 cm=7 cm12\text{ cm} - 5\text{ cm} = 7\text{ cm}.

Anahtar Kavram

Perpendicular from the center of a circle to a chord bisects the chord
Soru 104Soru

Two boats depart simultaneously from a port PP. Boat AA travels along a straight path for 8 km8\text{ km}, while Boat BB travels along another straight path for 15 km15\text{ km}. If the angle between their paths at the port is 6060^\circ, what is the distance between the two boats in kilometers?

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Cevap: 13

Cevap

The distance between the two boats is 13 km.
The distance between the two boats forms the third side of a triangle where two side lengths (8 km8\text{ km} and 15 km15\text{ km}) and the included angle (6060^\circ) are known. By the Cosine Rule, d2=82+1522(8)(15)cos(60)=64+225120=169d^2 = 8^2 + 15^2 - 2(8)(15)\cos(60^\circ) = 64 + 225 - 120 = 169, so d=169=13 kmd = \sqrt{169} = 13\text{ km}.

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1
Formulate the geometric model
A triangle with two sides of length 8 km8\text{ km} and 15 km15\text{ km}, and an included angle of 6060^\circ.
The paths of the two boats and the distance between them form a triangle with two given side lengths and the included angle.
2
Set up the Cosine Rule formula
d2=82+1522(8)(15)cos(60)d^2 = 8^2 + 15^2 - 2(8)(15)\cos(60^\circ)
The Cosine Rule (c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C) is used when two sides and the included angle are known (SAS configuration).
3
Evaluate the trigonometric term and simplify
d2=64+225240(0.5)=289120=169d^2 = 64 + 225 - 240(0.5) = 289 - 120 = 169
Since cos(60)=0.5\cos(60^\circ) = 0.5, the subtraction term reduces to 120120.
4
Calculate the principal square root
d=13d = 13
Taking the positive square root gives the distance in kilometers.

Anahtar Kavram

Applying the Cosine Rule to calculate the unknown side of a triangle given two sides and the included angle (SAS).
Soru 105Soru

A trapezium has parallel sides of length 12 cm12\text{ cm} and 18 cm18\text{ cm}, and a perpendicular height of 7 cm7\text{ cm}. What is the area of the trapezium?

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Cevap: 105 cm2105\text{ cm}^2

Cevap

The area of the trapezium is 105 cm2105\text{ cm}^2.
The area of a trapezium is calculated using the formula Area=12(a+b)h\text{Area} = \frac{1}{2}(a + b)h, where aa and bb are the lengths of the parallel sides, and hh is the perpendicular height. Substituting a=12 cma = 12\text{ cm}, b=18 cmb = 18\text{ cm}, and h=7 cmh = 7\text{ cm} yields Area=12(12+18)×7=12(30)×7=105 cm2\text{Area} = \frac{1}{2}(12 + 18) \times 7 = \frac{1}{2}(30) \times 7 = 105\text{ cm}^2.

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1
Identify the given dimensions of the trapezium
Parallel sides a=12 cma = 12\text{ cm}, b=18 cmb = 18\text{ cm}, and height h=7 cmh = 7\text{ cm}.
These are the necessary parameters for the area formula of a trapezium.
2
Sum the lengths of the parallel sides
a+b=12+18=30 cma + b = 12 + 18 = 30\text{ cm}.
The area formula requires the sum of the bases.
3
Multiply by the perpendicular height and divide by 2
\text{Area} = \frac{1}{2} \times 30 \times 7 = 15 \times 7 = 105\text{ cm}^2.
Applying the standard formula Area=12(a+b)h\text{Area} = \frac{1}{2}(a + b)h gives the plane region's total area.

Anahtar Kavram

Area of a Trapezium
Soru 106Soru

In a right-angled triangle PQRPQR, where Q=90\angle Q = 90^\circ and tanP=512\tan P = \frac{5}{12}, what is the exact value of sinP+cosP\sin P + \cos P?

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Cevap: 1713\frac{17}{13}

Cevap

1713\frac{17}{13}
Using tanP=512\tan P = \frac{5}{12}, the triangle has opposite side 55, adjacent side 1212, and hypotenuse 52+122=13\sqrt{5^2+12^2} = 13. Therefore, sinP=513\sin P = \frac{5}{13} and cosP=1213\cos P = \frac{12}{13}, giving a sum of 1713\frac{17}{13}.

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1
Identify the side lengths of the right triangle PQRPQR using the given tangent ratio.
Opposite side to P=5P = 5, adjacent side to P=12P = 12.
By definition, tanP=OppositeAdjacent=512\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{5}{12}.
2
Calculate the length of the hypotenuse PRPR using the Pythagorean theorem.
Hypotenuse PR=52+122=25+144=169=13PR = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13.
In a right triangle, hypotenuse2=opposite2+adjacent2\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2.
3
Determine sinP\sin P and cosP\cos P and compute their sum.
sinP=513\sin P = \frac{5}{13}, cosP=1213\cos P = \frac{12}{13}, so sinP+cosP=513+1213=1713\sin P + \cos P = \frac{5}{13} + \frac{12}{13} = \frac{17}{13}.
sinP=OppositeHypotenuse\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}} and cosP=AdjacentHypotenuse\cos P = \frac{\text{Adjacent}}{\text{Hypotenuse}}.

Anahtar Kavram

Basic Trigonometric Ratios and Pythagorean Triples
Tahmini Süre:1m 30s
Soru 107Soru

A forest ranger station YY is located on a bearing of 115115^\circ from an observation tower XX. What is the bearing of the observation tower XX from the forest ranger station YY?

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Cevap: 295295^\circ

Cevap

295295^\circ
The bearing of YY from XX is 115115^\circ. To find the bearing of XX from YY (the back bearing), add 180180^\circ to the forward bearing because 115<180115^\circ < 180^\circ. Calculating 115+180115^\circ + 180^\circ yields 295295^\circ.

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1
Identify the given forward bearing
The bearing of YY from XX is θ=115\theta = 115^\circ.
This is the initial directional angle measured clockwise from True North at point XX.
2
Calculate the back bearing of XX from YY
Back bearing =115+180=295= 115^\circ + 180^\circ = 295^\circ.
Because the forward bearing is less than 180180^\circ, the back bearing is obtained by adding 180180^\circ to find the opposite direction.

Anahtar Kavram

Back Bearing Calculation
Soru 108Soru

A point P(x,y)P(x, y) moves such that its perpendicular distance from the straight line L1:4x3y+5=0L_1: 4x - 3y + 5 = 0 is equal to its perpendicular distance from the straight line L2:3x+4y10=0L_2: 3x + 4y - 10 = 0. Which of the following equations represents one of the straight lines constituting the locus of PP?

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Cevap: x7y+15=0x - 7y + 15 = 0

Cevap

The equation x7y+15=0x - 7y + 15 = 0 represents one of the lines constituting the locus.
The locus of a point equidistant from two intersecting straight lines is the pair of angle bisectors between those lines. Setting the perpendicular distance formulas equal yields 4x3y+5=±(3x+4y10)4x - 3y + 5 = \pm(3x + 4y - 10). Solving the positive branch yields x7y+15=0x - 7y + 15 = 0, which correctly represents one of the component lines of the locus.

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1
Write the perpendicular distance formulas from point P(x,y)P(x, y) to both given lines.
d1=4x3y+542+(3)2=4x3y+55d_1 = \frac{|4x - 3y + 5|}{\sqrt{4^2 + (-3)^2}} = \frac{|4x - 3y + 5|}{5} and d2=3x+4y1032+42=3x+4y105d_2 = \frac{|3x + 4y - 10|}{\sqrt{3^2 + 4^2}} = \frac{|3x + 4y - 10|}{5}.
The locus of points equidistant from two intersecting lines consists of the angle bisectors of the angles between the lines.
2
Set the two perpendicular distances equal to each other.
\frac{|4x - 3y + 5|}{5} = \frac{|3x + 4y - 10|}{5} \implies |4x - 3y + 5| = |3x + 4y - 10|.
Since the point is equidistant from both lines, d1=d2d_1 = d_2.
3
Remove absolute values by considering both positive and negative cases.
4x3y+5=±(3x+4y10)4x - 3y + 5 = \pm(3x + 4y - 10).
Absolute value equality A=B|A| = |B| implies A=BA = B or A=BA = -B.
4
Evaluate Case 1 (positive sign) to find the first line equation.
4x3y+5=3x+4y10    (4x3x)+(3y4y)+(5+10)=0    x7y+15=04x - 3y + 5 = 3x + 4y - 10 \implies (4x - 3x) + (-3y - 4y) + (5 + 10) = 0 \implies x - 7y + 15 = 0.
Grouping like terms yields the linear equation for the first angle bisector.
5
Evaluate Case 2 (negative sign) to find the second line equation.
4x3y+5=(3x+4y10)    4x3y+5=3x4y+10    7x+y5=04x - 3y + 5 = -(3x + 4y - 10) \implies 4x - 3y + 5 = -3x - 4y + 10 \implies 7x + y - 5 = 0.
Grouping like terms yields the linear equation for the second angle bisector.

Anahtar Kavram

Locus equidistant from two intersecting lines (Angle Bisectors)
Tahmini Süre:2m 0s
Soru 109Soru

What is the exact numerical value of the trigonometric expression 6sin2602cos245tan230+sec245\frac{6\sin^2 60^\circ - 2\cos^2 45^\circ}{\tan^2 30^\circ + \sec^2 45^\circ}?

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Cevap: 1.5

Cevap

1.5
Substituting the exact special angle values gives a numerator of 6(34)2(12)=726\left(\frac{3}{4}\right) - 2\left(\frac{1}{2}\right) = \frac{7}{2} and a denominator of 13+2=73\frac{1}{3} + 2 = \frac{7}{3}. Dividing 72\frac{7}{2} by 73\frac{7}{3} yields 32=1.5\frac{3}{2} = 1.5.

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1
Substitute the exact values for the trigonometric ratios of the special angles.
sin60=32\sin 60^\circ = \frac{\sqrt{3}}{2}, cos45=12\cos 45^\circ = \frac{1}{\sqrt{2}}, tan30=13\tan 30^\circ = \frac{1}{\sqrt{3}}, and sec45=2\sec 45^\circ = \sqrt{2}.
Exact surd forms for special angles 3030^\circ, 4545^\circ, and 6060^\circ must be used.
2
Evaluate and simplify the numerator expression 6sin2602cos2456\sin^2 60^\circ - 2\cos^2 45^\circ.
6(34)2(12)=921=726\left(\frac{3}{4}\right) - 2\left(\frac{1}{2}\right) = \frac{9}{2} - 1 = \frac{7}{2}.
Square each trigonometric ratio first, multiply by the coefficients, and then subtract.
3
Evaluate and simplify the denominator expression tan230+sec245\tan^2 30^\circ + \sec^2 45^\circ.
(13)2+(2)2=13+2=73\left(\frac{1}{\sqrt{3}}\right)^2 + (\sqrt{2})^2 = \frac{1}{3} + 2 = \frac{7}{3}.
Square each trigonometric ratio and simplify the sum into a single improper fraction.
4
Divide the numerator result by the denominator result.
7/27/3=72×37=32=1.5\frac{7/2}{7/3} = \frac{7}{2} \times \frac{3}{7} = \frac{3}{2} = 1.5.
Dividing by a fraction is equivalent to multiplying by its reciprocal.

Anahtar Kavram

Evaluation of Trigonometric Expressions using Special Angles
Soru 110Soru

A point PP moves in a plane such that its distance from a fixed point OO is always 7 cm7\text{ cm}. What is the diameter, in cm\text{cm}, of the geometric locus traced out by point PP?

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Cevap: 14

Cevap

The diameter of the locus traced out by point PP is 14 cm14\text{ cm}.
The locus of a point that maintains a constant distance from a fixed point is a circle. The fixed point OO is the center of the circle, and the constant distance of 7 cm7\text{ cm} is its radius (rr). Since the diameter (DD) of a circle is twice its radius (D=2rD = 2r), the diameter is 2×7=14 cm2 \times 7 = 14\text{ cm}.

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1
Identify the shape of the geometric locus defined by the condition
A circle centered at point OO with radius r=7 cmr = 7\text{ cm}
By definition, the set of all points at a fixed distance from a single point forms a circle.
2
Calculate the diameter using the radius
D=2×7 cm=14 cmD = 2 \times 7\text{ cm} = 14\text{ cm}
The diameter of a circle is equal to twice its radius.

Anahtar Kavram

Locus of a point at a constant distance from a fixed point
Tahmini Süre:45s
Soru 111Soru

In ΔLMN\Delta LMN, the length of side l=10 cml = 10\text{ cm}, side m=103 cmm = 10\sqrt{3}\text{ cm}, and angle L=30\angle L = 30^\circ. Given that angle M\angle M is an obtuse angle, what is the measure of angle M\angle M?

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Cevap: 120120^\circ

Cevap

The measure of angle M\angle M is 120120^\circ.
Using the Sine Rule, we find sinM=103sin3010=32\sin M = \frac{10\sqrt{3} \cdot \sin 30^\circ}{10} = \frac{\sqrt{3}}{2}. The inverse sine gives a reference angle of 6060^\circ. Since the problem explicitly states that angle M\angle M is obtuse, we select 18060=120180^\circ - 60^\circ = 120^\circ.

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1
Apply the Sine Rule relating sides l,ml, m and their opposite angles L,ML, M.
lsinL=msinM\frac{l}{\sin L} = \frac{m}{\sin M}
The Sine Rule allows us to find an unknown angle given two sides and one non-included opposite angle.
2
Substitute the given values into the formula and solve for sinM\sin M.
\sin M = \frac{10\sqrt{3} \cdot \sin 30^\circ}{10} = \sqrt{3} \cdot 0.5 = \frac{\sqrt{3}}{2}
Since sin30=12\sin 30^\circ = \frac{1}{2}, simplifying the fraction gives 32\frac{\sqrt{3}}{2}.
3
Determine the obtuse angle solution for sinM=32\sin M = \frac{\sqrt{3}}{2}.
\angle M = 180^\circ - 60^\circ = 120^\circ
The principal value is 6060^\circ, but because M\angle M is specified to be obtuse (90<M<18090^\circ < \angle M < 180^\circ), we take the supplementary angle in the second quadrant.

Anahtar Kavram

Sine Rule and the Ambiguous Case (SSA Condition)
Tahmini Süre:1m 30s
Soru 112Soru

What is the numerical value of 4sin30+2cos60tan454\sin 30^\circ + 2\cos 60^\circ - \tan 45^\circ?

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Cevap: 2

Cevap

2
Substituting the exact trigonometric values for special angles into the expression gives 4(12)+2(12)1=2+11=24\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) - 1 = 2 + 1 - 1 = 2.

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1
Recall exact values for special trigonometric angles
sin30=12\sin 30^\circ = \frac{1}{2}, cos60=12\cos 60^\circ = \frac{1}{2}, and tan45=1\tan 45^\circ = 1
These are standard special angle values derived from standard right-angled triangles.
2
Substitute the values into the original expression
4(12)+2(12)1=2+114\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) - 1 = 2 + 1 - 1
Direct algebraic substitution.
3
Simplify the resulting numerical expression
2
Combining terms 2+112 + 1 - 1 gives 22.

Anahtar Kavram

Basic Trigonometric Ratios and Special Angles
Soru 113Soru

A solid right pyramid has a square base with a side length of 6 cm6\text{ cm} and a vertical height of 4 cm4\text{ cm}. What is the total surface area of the pyramid in square centimeters?

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Cevap: 96

Cevap

The total surface area of the pyramid is 96 cm296\text{ cm}^2.
To determine the total surface area of a square pyramid, sum the base area and the total area of the four triangular faces. The base area is 6×6=36 cm26 \times 6 = 36\text{ cm}^2. The slant height of each triangular face is found via the Pythagorean theorem using half the base side length (3 cm3\text{ cm}) and the vertical height (4 cm4\text{ cm}), yielding 32+42=5 cm\sqrt{3^2 + 4^2} = 5\text{ cm}. The area of one triangular face is 12×6×5=15 cm2\frac{1}{2} \times 6 \times 5 = 15\text{ cm}^2, making four faces equal to 60 cm260\text{ cm}^2. Adding the base area gives 36+60=96 cm236 + 60 = 96\text{ cm}^2.

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1
Calculate the area of the square base
Base area = 36 cm236\text{ cm}^2
The base is a square of side 6 cm6\text{ cm}, so Area=62=36 cm2\text{Area} = 6^2 = 36\text{ cm}^2.
2
Find the slant height of each triangular lateral face
Slant height l=5 cml = 5\text{ cm}
The slant height forms the hypotenuse of a right-angled triangle inside the pyramid with legs equal to the vertical height (4 cm4\text{ cm}) and half the base edge (3 cm3\text{ cm}): l=42+32=5 cml = \sqrt{4^2 + 3^2} = 5\text{ cm}.
3
Calculate the combined area of the four triangular faces
Lateral area = 60 cm260\text{ cm}^2
Each triangle has base 6 cm6\text{ cm} and height 5 cm5\text{ cm}, giving an area of 12×6×5=15 cm2\frac{1}{2} \times 6 \times 5 = 15\text{ cm}^2. For 4 identical faces, the total is 4×15=60 cm24 \times 15 = 60\text{ cm}^2.
4
Calculate total surface area
Total Surface Area = 96 cm296\text{ cm}^2
Sum the base area and the total lateral area: 36+60=96 cm236 + 60 = 96\text{ cm}^2.

Anahtar Kavram

Total Surface Area of a Right Pyramid
Soru 114Soru

In ΔABC\Delta ABC, sinA=35\sin A = \frac{3}{5}, sinB=45\sin B = \frac{4}{5}, and the side opposite angle AA has length a=15 cma = 15\text{ cm}. What is the length of side bb, in centimeters?

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Cevap: 20

Cevap

The length of side bb is 20 cm.
Using the Sine Rule asinA=bsinB\frac{a}{\sin A} = \frac{b}{\sin B}, substitute a=15a = 15, sinA=35\sin A = \frac{3}{5}, and sinB=45\sin B = \frac{4}{5}. Evaluating 153/5\frac{15}{3/5} gives 2525. Multiplying 2525 by 45\frac{4}{5} yields 20 cm20\text{ cm}.

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1
Set up the Sine Rule relationship between sides aa, bb and their opposite angles AA, BB.
asinA=bsinB\frac{a}{\sin A} = \frac{b}{\sin B}
The Sine Rule relates the side lengths of a triangle to the sines of its angles.
2
Substitute a=15a = 15, sinA=35\sin A = \frac{3}{5}, and sinB=45\sin B = \frac{4}{5} into the Sine Rule equation.
153/5=b4/5\frac{15}{3/5} = \frac{b}{4/5}
Direct substitution of known values allows us to solve for the unknown side bb.
3
Simplify the left side of the equation.
15×53=2515 \times \frac{5}{3} = 25
Dividing 15 by 35\frac{3}{5} is equivalent to multiplying 15 by 53\frac{5}{3}.
4
Multiply both sides by 45\frac{4}{5} to find bb.
b=25×45=20 cmb = 25 \times \frac{4}{5} = 20\text{ cm}
Isolating bb gives the final length of side bb.

Anahtar Kavram

Sine Rule
Soru 115Soru

In ΔABC\Delta ABC, the lengths of the sides are a=x cma = x\text{ cm}, b=(x+3) cmb = (x + 3)\text{ cm}, and c=(x+2) cmc = (x + 2)\text{ cm}. If C=60\angle C = 60^\circ, what is the value of xx?

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Cevap: 5

Cevap

5
Applying the Cosine Rule c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C with a=xa = x, b=x+3b = x + 3, c=x+2c = x + 2, and cos60=12\cos 60^\circ = \frac{1}{2} gives (x+2)2=x2+(x+3)2x(x+3)(x+2)^2 = x^2 + (x+3)^2 - x(x+3). Expanding both sides leads to x2+4x+4=x2+3x+9x^2 + 4x + 4 = x^2 + 3x + 9. Subtracting x2x^2 and isolating xx yields x=5x = 5.

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1
Apply the Cosine Rule for side cc and angle CC.
c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C
The Cosine Rule relates all three sides of a triangle to the cosine of an included angle.
2
Substitute a=xa = x, b=x+3b = x + 3, c=x+2c = x + 2, and cos60=12\cos 60^\circ = \frac{1}{2} into the formula.
(x+2)2=x2+(x+3)22(x)(x+3)(12)(x + 2)^2 = x^2 + (x + 3)^2 - 2(x)(x + 3)\left(\frac{1}{2}\right)
Inserting the given algebraic side lengths and angle allows solving for xx.
3
Expand both sides of the equation.
x2+4x+4=x2+(x2+6x+9)(x2+3x)x^2 + 4x + 4 = x^2 + (x^2 + 6x + 9) - (x^2 + 3x)
Cancel the factor of 22 with 12\frac{1}{2} and expand (x+2)2(x+2)^2 and (x+3)2(x+3)^2.
4
Simplify the right-hand side.
x2+4x+4=x2+3x+9x^2 + 4x + 4 = x^2 + 3x + 9
Combine like terms: (x2+x2x2)+(6x3x)+9=x2+3x+9(x^2 + x^2 - x^2) + (6x - 3x) + 9 = x^2 + 3x + 9.
5
Subtract x2x^2 from both sides and isolate xx.
4x3x=94    x=54x - 3x = 9 - 4 \implies x = 5
Subtracting x2+3x+4x^2 + 3x + 4 from both sides gives the linear solution x=5x = 5.

Anahtar Kavram

Solving for unknown algebraic side lengths using the Cosine Rule.
Soru 116Soru

A vertical flagpole of height 15 m15\text{ m} standing on level ground casts a shadow of length 153 m15\sqrt{3}\text{ m}. Calculate the angle of elevation of the sun in degrees.

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Cevap: 30

Cevap

The angle of elevation of the sun is 30 degrees.
The tangent of the angle of elevation is given by the height divided by the shadow length, tanθ=15153=13\tan\theta = \frac{15}{15\sqrt{3}} = \frac{1}{\sqrt{3}}, which corresponds to an angle of 3030^\circ.

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1
Formulate the trigonometric relationship using the right triangle formed by the flagpole, the shadow, and the sunlight ray.
\tan\theta = \frac{\text{height of flagpole}}{\text{length of shadow}} = \frac{15}{15\sqrt{3}}
The tangent of an angle in a right-angled triangle is defined as the ratio of the opposite side to the adjacent side.
2
Simplify the ratio and evaluate the angle.
\tan\theta = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ
The standard acute angle whose tangent value is 13\frac{1}{\sqrt{3}} is 3030^\circ.

Anahtar Kavram

Angle of Elevation and Trigonometric Ratios
Soru 117Soru

If sinθ=45\sin \theta = \frac{4}{5}, where θ\theta is an acute angle, what is the exact value of the expression sec2θ1cotθ+cscθ\frac{\sec^2 \theta - 1}{\cot \theta + \csc \theta}?

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Cevap: 89\frac{8}{9}

Cevap

The exact value of the expression is 89\frac{8}{9}.
Given an acute angle θ\theta with sinθ=45\sin \theta = \frac{4}{5}, the adjacent side is 33. This gives tanθ=43\tan \theta = \frac{4}{3}, cotθ=34\cot \theta = \frac{3}{4}, and cscθ=54\csc \theta = \frac{5}{4}. Using the identity sec2θ1=tan2θ\sec^2 \theta - 1 = \tan^2 \theta, the numerator is (43)2=169\left(\frac{4}{3}\right)^2 = \frac{16}{9}. The denominator cotθ+cscθ=34+54=2\cot \theta + \csc \theta = \frac{3}{4} + \frac{5}{4} = 2. Dividing 169\frac{16}{9} by 22 yields 89\frac{8}{9}.

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1
Determine all required trigonometric ratios for the acute angle θ\theta.
Given sinθ=45\sin \theta = \frac{4}{5}, the opposite side is 44 and the hypotenuse is 55. By the Pythagorean theorem, the adjacent side is 5242=3\sqrt{5^2 - 4^2} = 3. Therefore, cosθ=35\cos \theta = \frac{3}{5}, tanθ=43\tan \theta = \frac{4}{3}, cotθ=34\cot \theta = \frac{3}{4}, and cscθ=54\csc \theta = \frac{5}{4}.
Defining the right-triangle side lengths allows direct evaluation of all six trigonometric ratios.
2
Simplify the numerator using standard trigonometric identities.
sec2θ1=tan2θ=(43)2=169\sec^2 \theta - 1 = \tan^2 \theta = \left(\frac{4}{3}\right)^2 = \frac{16}{9}.
Applying the fundamental identity 1+tan2θ=sec2θ1 + \tan^2 \theta = \sec^2 \theta simplifies the numerator.
3
Evaluate the denominator.
cotθ+cscθ=34+54=84=2\cot \theta + \csc \theta = \frac{3}{4} + \frac{5}{4} = \frac{8}{4} = 2.
Adding fractional values with a common denominator.
4
Divide the numerator by the denominator.
16/92=89\frac{16/9}{2} = \frac{8}{9}.
Dividing the simplified numerator by the simplified denominator yields the final result.

Anahtar Kavram

Evaluation of trigonometric expressions using fundamental identities and right-triangle ratio definitions.
Soru 118Soru

A composite plane figure is formed by constructing a semicircle externally on one side of a square of side length 14 cm14\text{ cm}. What is the perimeter of the resulting figure? (Take π=227\pi = \frac{22}{7})

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Cevap: 64 cm64\text{ cm}

Cevap

64 cm64\text{ cm}
The outer perimeter of the composite shape consists of three straight edges of the square and one curved semicircular arc. Three sides of length 14 cm14\text{ cm} yield 42 cm42\text{ cm}. The arc length of a semicircle with radius 7 cm7\text{ cm} is πr=227×7=22 cm\pi r = \frac{22}{7} \times 7 = 22\text{ cm}. Adding these together gives 42 cm+22 cm=64 cm42\text{ cm} + 22\text{ cm} = 64\text{ cm}.

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1
Identify the exposed straight sides of the square contributing to the perimeter.
The square has 4 sides, but 1 side is attached to the semicircle inside the figure. Thus, 3 sides are on the outer boundary: 3×14 cm=42 cm3 \times 14\text{ cm} = 42\text{ cm}.
Perimeter only includes the outer boundary of a composite shape.
2
Calculate the radius and arc length of the attached semicircle.
The diameter of the semicircle is equal to the side length of the square (d=14 cmd = 14\text{ cm}), so radius r=7 cmr = 7\text{ cm}. Semicircular arc length =πr=227×7=22 cm= \pi r = \frac{22}{7} \times 7 = 22\text{ cm}.
The curved boundary is half of the total circle circumference.
3
Sum the outer straight boundaries and the curved boundary.
Total perimeter =42 cm+22 cm=64 cm= 42\text{ cm} + 22\text{ cm} = 64\text{ cm}.
Combining all outer segment lengths gives the total perimeter.

Anahtar Kavram

Perimeter of composite plane figures involving straight line segments and circular arcs.
Tahmini Süre:1m 30s
Soru 119Soru

Given that cosxsinx=15\cos x - \sin x = \frac{1}{\sqrt{5}} for an acute angle xx, what is the exact value of cos3xsin3x\cos^3 x - \sin^3 x?

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Cevap: 7525\frac{7\sqrt{5}}{25}

Cevap

7525\frac{7\sqrt{5}}{25}
Squaring cosxsinx=15\cos x - \sin x = \frac{1}{\sqrt{5}} yields 12sinxcosx=151 - 2\sin x \cos x = \frac{1}{5}, which gives sinxcosx=25\sin x \cos x = \frac{2}{5}. Using the difference of cubes factorization, cos3xsin3x=(cosxsinx)(1+sinxcosx)=15(1+25)=755\cos^3 x - \sin^3 x = (\cos x - \sin x)(1 + \sin x \cos x) = \frac{1}{\sqrt{5}} \left(1 + \frac{2}{5}\right) = \frac{7}{5\sqrt{5}}. Rationalizing the denominator produces 7525\frac{7\sqrt{5}}{25}.

Adım Adım Çözüm

1
Square both sides of the given equation to find the product sinxcosx\sin x \cos x.
(cosxsinx)2=(15)2    cos2x2sinxcosx+sin2x=15(\cos x - \sin x)^2 = \left(\frac{1}{\sqrt{5}}\right)^2 \implies \cos^2 x - 2\sin x \cos x + \sin^2 x = \frac{1}{5}.
Squaring allows us to use the Pythagorean identity sin2x+cos2x=1\sin^2 x + \cos^2 x = 1 to isolate sinxcosx\sin x \cos x.
2
Simplify using cos2x+sin2x=1\cos^2 x + \sin^2 x = 1 to solve for sinxcosx\sin x \cos x.
12sinxcosx=15    2sinxcosx=115=45    sinxcosx=251 - 2\sin x \cos x = \frac{1}{5} \implies 2\sin x \cos x = 1 - \frac{1}{5} = \frac{4}{5} \implies \sin x \cos x = \frac{2}{5}.
Finding the product of sinx\sin x and cosx\cos x is necessary for the algebraic expansion of the difference of cubes.
3
Apply the difference of cubes algebraic identity to cos3xsin3x\cos^3 x - \sin^3 x.
cos3xsin3x=(cosxsinx)(cos2x+sinxcosx+sin2x)=(cosxsinx)(1+sinxcosx)\cos^3 x - \sin^3 x = (\cos x - \sin x)(\cos^2 x + \sin x \cos x + \sin^2 x) = (\cos x - \sin x)(1 + \sin x \cos x).
The identity a3b3=(ab)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2) breaks the target expression down into known terms.
4
Substitute the known values into the expression and rationalize the denominator.
(15)(1+25)=15×75=755=7525\left(\frac{1}{\sqrt{5}}\right) \left(1 + \frac{2}{5}\right) = \frac{1}{\sqrt{5}} \times \frac{7}{5} = \frac{7}{5\sqrt{5}} = \frac{7\sqrt{5}}{25}.
Evaluating the product and rationalizing 755\frac{7}{5\sqrt{5}} yields the final exact surd form.

Anahtar Kavram

Basic Trigonometric Ratios, Special Angles, and Identities
Tahmini Süre:2m 0s
Soru 120Soru

Given that θ\theta is an acute angle such that tanθ=43\tan \theta = \frac{4}{3}, calculate the numerical value of the expression 3sinθ+2cosθ3sinθcosθ\frac{3\sin \theta + 2\cos \theta}{3\sin \theta - \cos \theta}.

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Cevap: 2

Cevap

The exact numerical value of the given expression is 2.
Dividing both the numerator and denominator of 3sinθ+2cosθ3sinθcosθ\frac{3\sin \theta + 2\cos \theta}{3\sin \theta - \cos \theta} by cosθ\cos \theta gives 3tanθ+23tanθ1\frac{3\tan \theta + 2}{3\tan \theta - 1}. Substituting tanθ=43\tan \theta = \frac{4}{3} yields 3(4/3)+23(4/3)1=4+241=63=2\frac{3(4/3) + 2}{3(4/3) - 1} = \frac{4 + 2}{4 - 1} = \frac{6}{3} = 2.

Adım Adım Çözüm

1
Express sine and cosine terms in terms of tangent or find individual ratio values
Divide every term in the numerator and denominator by cosθ\cos \theta to obtain 3tanθ+23tanθ1\frac{3\tan \theta + 2}{3\tan \theta - 1}. Alternatively, using a right triangle with opposite side = 4 and adjacent side = 3 gives hypotenuse = 5, so sinθ=45\sin \theta = \frac{4}{5} and cosθ=35\cos \theta = \frac{3}{5}.
Converting to tanθ\tan \theta simplifies the calculation directly without evaluating square roots or hypotenuse.
2
Substitute the value of tanθ=43\tan \theta = \frac{4}{3} into the expression
Numerator: 3(43)+2=4+2=63\left(\frac{4}{3}\right) + 2 = 4 + 2 = 6. Denominator: 3(43)1=41=33\left(\frac{4}{3}\right) - 1 = 4 - 1 = 3.
Simplifies numerical fractions in both parts of the fraction.
3
Divide numerator by denominator
63=2.\frac{6}{3} = 2.
Yields the final integer solution.

Anahtar Kavram

Basic Trigonometric Ratios and Quotients
ÖncekiSayfa 6 / 10Sonraki
Geometry and Trigonometry Alıştırma Soruları — JAMB UTME — Sayfa 6 | Examkin