Trigonometric Ratios and Identities

29 questions

Question 1Question

In right triangle PQRPQR, the measure of angle QQ is 9090^\circ. If tan(P)=43\tan(P) = \frac{4}{3}, what is the value of sin(P)\sin(P)?

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Answer: 45\frac{4}{5}

Answer

45\frac{4}{5}
The tangent of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the adjacent side. Given that tan(P)=43\tan(P) = \frac{4}{3}, we can set the length of the side opposite to angle P as 44 and the adjacent side as 33. Using the Pythagorean theorem, the hypotenuse is 32+42=5\sqrt{3^2 + 4^2} = 5. Since the sine of an angle is the ratio of the opposite side to the hypotenuse, sin(P)=45\sin(P) = \frac{4}{5}.

Step-by-Step Solution

1
Identify the relationship between the tangent ratio and the sides of the right triangle.
The tangent of angle P is defined as the opposite side divided by the adjacent side. Since tan(P)=43\tan(P) = \frac{4}{3}, we can set the opposite side to 44 and the adjacent side to 33.
This allows us to establish the relative lengths of the sides of the right triangle.
2
Calculate the length of the hypotenuse using the Pythagorean theorem.
The hypotenuse is 32+42=25=5\sqrt{3^2 + 4^2} = \sqrt{25} = 5.
The hypotenuse is required to calculate the sine ratio.
3
Determine the sine of angle P.
The sine of angle P is defined as the opposite side divided by the hypotenuse, which is 45\frac{4}{5}.
This yields the final requested trigonometric ratio.

Key Concept

Using basic trigonometric ratios (SOH CAH TOA) and the Pythagorean theorem to evaluate trigonometric ratios in a right triangle.
Estimated Time:45s
Question 2Question

In right triangle ABCABC, the measure of angle CC is 9090^\circ. If sin(A)cos(A)=15\sin(A) - \cos(A) = \frac{1}{5}, what is the value of sin(A)sin(B)\sin(A)\sin(B)?

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Answer: 1225\frac{12}{25}

Answer

1225\frac{12}{25}
By the co-function identity for complementary angles in a right triangle, sin(B)=cos(A)\sin(B) = \cos(A), which means the expression sin(A)sin(B)\sin(A)\sin(B) is equivalent to sin(A)cos(A)\sin(A)\cos(A). Squaring the equation sin(A)cos(A)=15\sin(A) - \cos(A) = \frac{1}{5} gives sin2(A)2sin(A)cos(A)+cos2(A)=125\sin^2(A) - 2\sin(A)\cos(A) + \cos^2(A) = \frac{1}{25}. Substituting the Pythagorean identity sin2(A)+cos2(A)=1\sin^2(A) + \cos^2(A) = 1 simplifies this to 12sin(A)cos(A)=1251 - 2\sin(A)\cos(A) = \frac{1}{25}. Solving for sin(A)cos(A)\sin(A)\cos(A) yields 2sin(A)cos(A)=24252\sin(A)\cos(A) = \frac{24}{25}, or sin(A)cos(A)=1225\sin(A)\cos(A) = \frac{12}{25}.

Step-by-Step Solution

1
Express the target expression in terms of a single angle using complementary relationships.
Since angle C=90C = 90^\circ, angles AA and BB are complementary, so sin(B)=cos(A)\sin(B) = \cos(A). Thus, sin(A)sin(B)=sin(A)cos(A)\sin(A)\sin(B) = \sin(A)\cos(A).
In any right triangle, the sine of one acute angle equals the cosine of the other acute angle.
2
Square both sides of the given equation sin(A)cos(A)=15\sin(A) - \cos(A) = \frac{1}{5}.
(sin(A)cos(A))2=(15)2    sin2(A)2sin(A)cos(A)+cos2(A)=125(\sin(A) - \cos(A))^2 = \left(\frac{1}{5}\right)^2 \implies \sin^2(A) - 2\sin(A)\cos(A) + \cos^2(A) = \frac{1}{25}.
Squaring the difference allows us to introduce the product term sin(A)cos(A)\sin(A)\cos(A) and the squared terms.
3
Apply the Pythagorean identity to simplify the equation.
Since sin2(A)+cos2(A)=1\sin^2(A) + \cos^2(A) = 1, the equation becomes 12sin(A)cos(A)=1251 - 2\sin(A)\cos(A) = \frac{1}{25}.
The sum of the squares of sine and cosine of the same angle is always equal to 1.
4
Solve for the product sin(A)cos(A)\sin(A)\cos(A).
2sin(A)cos(A)=1125=2425    sin(A)cos(A)=12252\sin(A)\cos(A) = 1 - \frac{1}{25} = \frac{24}{25} \implies \sin(A)\cos(A) = \frac{12}{25}.
Subtracting 1/251/25 from 11 and dividing the resulting fraction by 22 isolates the target product.

Key Concept

Pythagorean identity and co-function relationships in right triangles
Question 3Question

In right triangle ABCABC, the measure of angle BB is 9090^\circ. The acute angles AA and CC satisfy sin(A)=k3\sin(A) = \frac{k}{3} and cos(C)=4k+4\cos(C) = \frac{4}{k+4} for some positive constant kk. What is the value of kk?

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Answer: 2

Answer

2
In right triangle ABCABC, the measure of angle BB is 9090^\circ, which means the acute angles AA and CC are complementary. Therefore, the co-function identity states that sin(A)=cos(C)\sin(A) = \cos(C). Equating the given expressions yields k3=4k+4\frac{k}{3} = \frac{4}{k+4}. Cross-multiplying results in k(k+4)=12k(k+4) = 12, which simplifies to k2+4k12=0k^2 + 4k - 12 = 0. Factoring this equation gives (k+6)(k2)=0(k+6)(k-2) = 0. This quadratic has solutions k=6k = -6 and k=2k = 2. Since kk must be positive, the value of kk is 22.

Step-by-Step Solution

1
Identify the relationship between the acute angles in a right triangle.
Since the measure of angle BB is 9090^\circ, the sum of angles AA and CC is 9090^\circ, which means they are complementary angles. By the co-function identity, sin(A)=cos(C)\sin(A) = \cos(C).
In a right triangle, the sine of one acute angle equals the cosine of the other acute angle.
2
Equate the expressions for sin(A)\sin(A) and cos(C)\cos(C) and set up the equation for kk.
k3=4k+4\frac{k}{3} = \frac{4}{k+4}
This sets up the algebraic relation to solve for kk using the given trigonometric expressions.
3
Solve the algebraic equation for kk.
Cross-multiplying gives k(k+4)=12k(k+4) = 12, which expands to k2+4k=12k^2 + 4k = 12. Subtracting 1212 from both sides results in the quadratic equation k2+4k12=0k^2 + 4k - 12 = 0. Factoring this equation yields (k+6)(k2)=0(k+6)(k-2) = 0.
Cross-multiplication eliminates the denominators, converting the equation to a quadratic form that can be solved by factoring.
4
Determine the valid positive value for kk.
The solutions to the equation (k+6)(k2)=0(k+6)(k-2) = 0 are k=6k = -6 and k=2k = 2. Since kk must be a positive constant, we select k=2k = 2.
The problem specifies that kk is a positive constant, so the negative solution is discarded.

Key Concept

Co-function identity relating sine and cosine of complementary angles
Estimated Time:1m 30s
Question 4Question

In right triangle ABCABC, the measure of angle CC is 9090^\circ. If cos(A)=ksin(A)\cos(A) = k \sin(A) for some positive constant kk, which of the following expressions represents cos(B)\cos(B) in terms of kk?

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Answer: 11+k2\frac{1}{\sqrt{1 + k^2}}

Answer

The expression that represents cos(B)\cos(B) is 11+k2\frac{1}{\sqrt{1 + k^2}}.
In right triangle ABCABC with right angle CC, the acute angles AA and BB are complementary (A+B=90A + B = 90^\circ). By the co-function identity, cos(B)=sin(A)\cos(B) = \sin(A). Using the given relation cos(A)=ksin(A)\cos(A) = k \sin(A) and substituting it into the Pythagorean identity sin2(A)+cos2(A)=1\sin^2(A) + \cos^2(A) = 1, we get sin2(A)+(ksin(A))2=1    sin2(A)(1+k2)=1\sin^2(A) + (k \sin(A))^2 = 1 \implies \sin^2(A)(1 + k^2) = 1. Solving for sin(A)\sin(A) gives sin(A)=11+k2\sin(A) = \frac{1}{\sqrt{1 + k^2}} because sin(A)\sin(A) must be positive for an acute angle. Since cos(B)=sin(A)\cos(B) = \sin(A), we conclude that the correct expression is 11+k2\frac{1}{\sqrt{1 + k^2}}.

Step-by-Step Solution

1
Use the relationship between the acute angles in right triangle ABCABC.
Since angle CC is 9090^\circ, angles AA and BB are complementary. Thus, cos(B)=sin(A)\cos(B) = \sin(A).
This allows us to convert the target term cos(B)\cos(B) into sin(A)\sin(A), which can be related to the given equation.
2
Substitute the given relation cos(A)=ksin(A)\cos(A) = k \sin(A) into the Pythagorean identity.
Using sin2(A)+cos2(A)=1\sin^2(A) + \cos^2(A) = 1, we substitute to get sin2(A)+(ksin(A))2=1\sin^2(A) + (k \sin(A))^2 = 1, which simplifies to sin2(A)(1+k2)=1\sin^2(A)(1 + k^2) = 1.
This sets up a single equation containing only sin(A)\sin(A) and the constant kk.
3
Solve for sin(A)\sin(A) and substitute back to find cos(B)\cos(B).
Solving for sin(A)\sin(A) gives sin(A)=11+k2\sin(A) = \frac{1}{\sqrt{1 + k^2}} (since sin(A)>0\sin(A) > 0 for acute angle AA). Since cos(B)=sin(A)\cos(B) = \sin(A), we have cos(B)=11+k2\cos(B) = \frac{1}{\sqrt{1 + k^2}}.
This yields the final value of cos(B)\cos(B) in terms of kk.

Key Concept

Applying complementary angle trigonometric identities (co-functions) and the Pythagorean identity in a right triangle.

Alternative Method

Alternatively, you can model this by setting up a right triangle. Since cos(A)=ksin(A)\cos(A) = k \sin(A), we can divide both sides by sin(A)\sin(A) to get cot(A)=k\cot(A) = k, which means tan(A)=1k\tan(A) = \frac{1}{k}. In a right triangle ABCABC with right angle CC, tan(A)=oppositeadjacent=BCAC=1k\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AC} = \frac{1}{k}. Let BC=1BC = 1 and AC=kAC = k. By the Pythagorean theorem, the hypotenuse AB=12+k2=1+k2AB = \sqrt{1^2 + k^2} = \sqrt{1 + k^2}. Then, cos(B)=adjacent to Bhypotenuse=BCAB=11+k2\cos(B) = \frac{\text{adjacent to } B}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{1}{\sqrt{1 + k^2}}.
Estimated Time:2m 0s
Question 5Question

If cos(θ)=725\cos(\theta) = \frac{7}{25} and 0<θ<π20 < \theta < \frac{\pi}{2}, what is the value of sin(π2θ)\sin\left(\frac{\pi}{2} - \theta\right)?

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Answer: 725\frac{7}{25}

Answer

725\frac{7}{25}
The correct answer is 725\frac{7}{25} because the co-function identity states that for any angle θ\theta, the sine of its complement, π2θ\frac{\pi}{2} - \theta, is equal to the cosine of θ\theta. Given that cos(θ)=725\cos(\theta) = \frac{7}{25}, it follows that sin(π2θ)=725\sin\left(\frac{\pi}{2} - \theta\right) = \frac{7}{25}.

Step-by-Step Solution

1
Identify the relevant trigonometric identity.
Use the co-function identity sin(π2x)=cos(x)\sin\left(\frac{\pi}{2} - x\right) = \cos(x).
Since the angle is given in radians, the complementary angle identity relates the sine of the complement to the cosine of the original angle.
2
Substitute the given value.
Since cos(θ)=725\cos(\theta) = \frac{7}{25}, then sin(π2θ)=725\sin\left(\frac{\pi}{2} - \theta\right) = \frac{7}{25}.
Direct substitution of the given cosine value into the identity yields the final answer.

Key Concept

Co-function identities relate the sine and cosine of complementary angles.
Question 6Question

In right triangle ABCABC, the measure of angle BB is 9090^\circ. If sin(A)=513\sin(A) = \frac{5}{13}, what is the value of cos(C)\cos(C)?

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Answer: 513\frac{5}{13}

Answer

513\frac{5}{13}
The correct answer is 513\frac{5}{13}. In any right triangle ABCABC where the right angle is at BB, the two acute angles AA and CC are complementary, meaning A+C=90A + C = 90^\circ. The co-function identity states that the sine of an acute angle is equal to the cosine of its complement, or sin(A)=cos(90A)=cos(C)\sin(A) = \cos(90^\circ - A) = \cos(C). Given that sin(A)=513\sin(A) = \frac{5}{13}, the value of cos(C)\cos(C) must also be 513\frac{5}{13}.

Step-by-Step Solution

1
Identify the relationship between the acute angles in a right triangle.
Since the measure of angle BB is 9090^\circ, the sum of the measures of angles AA and CC must be 9090^\circ. Thus, angles AA and CC are complementary.
The sum of angles in any triangle is 180180^\circ.
2
Apply the co-function identity for complementary angles.
For any two complementary angles AA and CC, the identity cos(C)=sin(A)\cos(C) = \sin(A) holds true.
The sine of an angle is the ratio of the opposite side to the hypotenuse, which is the same as the ratio of the adjacent side of its complement to the hypotenuse (cosine of the complement).
3
Substitute the given value to find cos(C)\cos(C).
Since sin(A)=513\sin(A) = \frac{5}{13}, it follows that cos(C)=513\cos(C) = \frac{5}{13}.
Direct substitution into the identity cos(C)=sin(A)\cos(C) = \sin(A).

Key Concept

Co-function identities for complementary angles in a right triangle
Question 7Question

If sin(x)=cos(38)\sin(x^\circ) = \cos(38^\circ), where 0<x<900 < x < 90, what is the value of xx?

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Answer: 52

Answer

The value of xx is 52.
Applying the co-function identity sin(θ)=cos(90θ)\sin(\theta) = \cos(90^\circ - \theta) allows us to set the argument of the sine function as the complement of the cosine's argument. Thus, x=9038=52x = 90 - 38 = 52.

Step-by-Step Solution

1
Identify the relevant trigonometric identity.
The co-function identity sin(x)=cos(90x)\sin(x^\circ) = \cos(90^\circ - x^\circ) is appropriate here.
We need to relate the sine of one angle to the cosine of another angle.
2
Equate the angle arguments using the identity.
sin(x)=cos(90x)=cos(38)\sin(x^\circ) = \cos(90^\circ - x^\circ) = \cos(38^\circ), which implies 90x=3890 - x = 38.
For acute angles, if their cosine values are equal, the angles themselves must be equal.
3
Solve the linear equation for xx.
x=9038=52x = 90 - 38 = 52.
Isolating xx gives the final answer.

Key Concept

Co-function identities relate the sine of an angle to the cosine of its complement: sin(θ)=cos(90θ)\sin(\theta) = \cos(90^\circ - \theta).
Question 8Question

In right triangle XYZXYZ, the measure of angle YY is 9090^\circ, XY=12XY = 12, and YZ=5YZ = 5. What is the value of cos(X)sin(X)\cos(X) - \sin(X)?

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Answer: 713\frac{7}{13}

Answer

713\frac{7}{13}
The length of the hypotenuse XZXZ is first found using the Pythagorean theorem: XZ=122+52=13XZ = \sqrt{12^2 + 5^2} = 13. Using the definitions of the trigonometric ratios, cos(X)=adjacenthypotenuse=1213\cos(X) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13} and sin(X)=oppositehypotenuse=513\sin(X) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}. The difference is cos(X)sin(X)=1213513=713\cos(X) - \sin(X) = \frac{12}{13} - \frac{5}{13} = \frac{7}{13}.

Step-by-Step Solution

1
Find the length of the hypotenuse XZXZ using the Pythagorean theorem.
XZ=XY2+YZ2=122+52=144+25=169=13XZ = \sqrt{XY^2 + YZ^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13.
The hypotenuse is needed as the denominator for the sine and cosine ratios.
2
Determine the values of cos(X)\cos(X) and sin(X)\sin(X).
cos(X)=XYXZ=1213\cos(X) = \frac{XY}{XZ} = \frac{12}{13} and sin(X)=YZXZ=513\sin(X) = \frac{YZ}{XZ} = \frac{5}{13}.
By definition, cos(θ)=adjacenthypotenuse\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} and sin(θ)=oppositehypotenuse\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}.
3
Calculate the difference cos(X)sin(X)\cos(X) - \sin(X).
cos(X)sin(X)=1213513=713\cos(X) - \sin(X) = \frac{12}{13} - \frac{5}{13} = \frac{7}{13}.
This is the final subtraction requested by the question.

Key Concept

Calculating trigonometric ratios in a right triangle and applying basic operations.
Question 9Question

If sin(θ)=53\sin(\theta) = \frac{\sqrt{5}}{3} and θ\theta is an acute angle, what is the value of tan2(θ)\tan^2(\theta)?

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Answer: 1.25

Answer

The correct answer is 1.25 (or the equivalent fraction 5/4).
Using the Pythagorean identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, we find cos2(θ)=159=49\cos^2(\theta) = 1 - \frac{5}{9} = \frac{4}{9}. Since tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, it follows that tan2(θ)=sin2(θ)cos2(θ)=5/94/9=1.25\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{5/9}{4/9} = 1.25.

Step-by-Step Solution

1
Calculate the square of the cosine of the angle using the Pythagorean identity.
cos2(θ)=1sin2(θ)=1(53)2=159=49\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{\sqrt{5}}{3}\right)^2 = 1 - \frac{5}{9} = \frac{4}{9}
The Pythagorean identity states that sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 for any angle θ\theta.
2
Express the tangent squared in terms of sine squared and cosine squared.
tan2(θ)=sin2(θ)cos2(θ)\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}
By definition, the tangent function is the ratio of sine to cosine, so tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.
3
Substitute the values of sin2(θ)\sin^2(\theta) and cos2(θ)\cos^2(\theta) to solve for tan2(θ)\tan^2(\theta).
tan2(θ)=5/94/9=54=1.25\tan^2(\theta) = \frac{5/9}{4/9} = \frac{5}{4} = 1.25
Plugging in the squared ratios and simplifying gives the final evaluation.

Key Concept

Pythagorean trigonometric identity and quotient identity of tangent
Question 10Question

In right triangle JKLJKL, the measure of angle KK is 9090^\circ. If cos(J)=0.28\cos(J) = 0.28, what is the value of sin(L)\sin(L)?

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Answer: 0.28

Answer

The correct answer is 0.28 (or the equivalent fraction 7/25).
In a right triangle, the two acute angles are complementary, meaning they add up to 9090^\circ. The co-function identity states that the sine of an acute angle is equal to the cosine of its complement. Therefore, in right triangle JKLJKL with right angle KK, sin(L)=cos(J)\sin(L) = \cos(J). Given that cos(J)=0.28\cos(J) = 0.28, the value of sin(L)\sin(L) must also be 0.280.28.

Step-by-Step Solution

1
Determine the relationship between angles JJ and LL.
J+L=90J + L = 90^\circ
Since the sum of angles in a triangle is 180180^\circ and angle KK is 9090^\circ, the sum of the remaining two angles must be 18090=90180^\circ - 90^\circ = 90^\circ.
2
Use the co-function trigonometric identity.
sin(L)=cos(J)\sin(L) = \cos(J)
The sine of an acute angle in a right triangle is equal to the cosine of its complementary angle.
3
Substitute the given value of cos(J)\cos(J) into the identity.
sin(L)=0.28\sin(L) = 0.28
We are given that cos(J)=0.28\cos(J) = 0.28.

Key Concept

Complementary angle trigonometric identity (co-function identity)
Question 11Question

An acute angle θ\theta in a right triangle satisfies cos(θ)=35\cos(\theta) = \frac{3}{5}. If ϕ\phi is the other acute angle of the triangle, what is the value of sin(ϕ)tan(θ)\sin(\phi) \tan(\theta)?

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Answer: 45\frac{4}{5}

Answer

four-fifths
The correct answer is four-fifths. Because θ\theta and ϕ\phi are complementary angles in a right triangle, sin(ϕ)\sin(\phi) equals cos(θ)\cos(\theta), which is given as three-fifths. Using the Pythagorean identity, sin(θ)\sin(\theta) is four-fifths, making tan(θ)\tan(\theta) equal to four-thirds. Multiplying sin(ϕ)\sin(\phi) and tan(θ)\tan(\theta) gives three-fifths times four-thirds, which simplifies to four-fifths.

Step-by-Step Solution

1
Use the complementary angle relationship to find sin(ϕ)\sin(\phi).
sin(ϕ)=cos(θ)=35\sin(\phi) = \cos(\theta) = \frac{3}{5}
The two acute angles in a right triangle, θ\theta and ϕ\phi, sum to 9090^\circ. Therefore, the sine of one angle equals the cosine of the other.
2
Find sin(θ)\sin(\theta) using the Pythagorean identity.
sin(θ)=45\sin(\theta) = \frac{4}{5}
Since sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, we have sin2(θ)+(35)2=1\sin^2(\theta) + \left(\frac{3}{5}\right)^2 = 1, which gives sin2(θ)=1625\sin^2(\theta) = \frac{16}{25}. Because θ\theta is acute, sin(θ)=45\sin(\theta) = \frac{4}{5}.
3
Calculate tan(θ)\tan(\theta) using the quotient identity.
tan(θ)=43\tan(\theta) = \frac{4}{3}
By definition, tan(θ)=sin(θ)cos(θ)=4/53/5=43\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{4/5}{3/5} = \frac{4}{3}.
4
Multiply sin(ϕ)\sin(\phi) by tan(θ)\tan(\theta).
sin(ϕ)tan(θ)=45\sin(\phi) \tan(\theta) = \frac{4}{5}
Substitute the found values: sin(ϕ)tan(θ)=35×43=45\sin(\phi) \tan(\theta) = \frac{3}{5} \times \frac{4}{3} = \frac{4}{5}.

Key Concept

Trigonometric co-function identities and quotient identities in right triangles
Question 12Question

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 13Question

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 14Question

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?

Show answer & explanation

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 15Question

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?

Show answer & explanation

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 16Question

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)?

Show answer & explanation

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 17Question

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)?

Show answer & explanation

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 18Question

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)?

Show answer & explanation

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 19Question

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))?

Show answer & explanation

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 20Question

A right triangle has acute angles PP and RR. The sine of angle PP is defined by a13\frac{a}{13} and the cosine of angle RR is 513\frac{5}{13}, where aa is a positive constant. What is the value of sin(R)\sin(R)?

Show answer & explanation

Answer: 1213\frac{12}{13}

Answer

1213\frac{12}{13}
The correct answer is the value 1213\frac{12}{13}. Since PP and RR are the acute angles of a right triangle, they are complementary, which means sin(P)=cos(R)=513\sin(P) = \cos(R) = \frac{5}{13}. By using the Pythagorean identity sin2(R)+cos2(R)=1\sin^2(R) + \cos^2(R) = 1, we can substitute the value of cos(R)\cos(R) to get sin2(R)+(513)2=1\sin^2(R) + \left(\frac{5}{13}\right)^2 = 1. Solving for sin(R)\sin(R) yields sin2(R)=125169=144169\sin^2(R) = 1 - \frac{25}{169} = \frac{144}{169}. Since RR is an acute angle, taking the positive square root yields sin(R)=1213\sin(R) = \frac{12}{13}.

Step-by-Step Solution

1
Apply the complementary angle identity to determine the value of the constant aa.
Since PP and RR are the acute angles of a right triangle, they are complementary, meaning P+R=90P + R = 90^\circ. The co-function identity states that sin(P)=cos(R)\sin(P) = \cos(R). Given cos(R)=513\cos(R) = \frac{5}{13}, we have sin(P)=a13=513\sin(P) = \frac{a}{13} = \frac{5}{13}, which means a=5a = 5.
This establishes the relationship between the trigonometric ratios of the two acute angles.
2
Use the Pythagorean identity to find sin(R)\sin(R).
Using the Pythagorean identity sin2(R)+cos2(R)=1\sin^2(R) + \cos^2(R) = 1, substitute cos(R)=513\cos(R) = \frac{5}{13} into the equation: sin2(R)+(513)2=1    sin2(R)+25169=1    sin2(R)=125169=144169\sin^2(R) + \left(\frac{5}{13}\right)^2 = 1 \implies \sin^2(R) + \frac{25}{169} = 1 \implies \sin^2(R) = 1 - \frac{25}{169} = \frac{144}{169}. Taking the positive square root because RR is an acute angle gives sin(R)=1213\sin(R) = \frac{12}{13}.
This determines the sine of the angle from its known cosine value.

Key Concept

Trigonometric ratios of complementary angles and the Pythagorean identity in right triangles.
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