Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
3. Right Triangles and Trigonometric Ratios
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Exercise 53 Page 926

G

Practice makes perfect

We know that in △ XYZ, the right angle is ∠ Z. We are also given the following trigonometric expression. tan X = 8 15 Recall that in a right triangle, the tangent of an acute angle is defined as the ratio of the opposite side to the adjacent side. tan θ = Opposite/AdjacentWe will compare the given expression with this definition. tan θ = Opposite/Adjacent ⇒ tan X = 8/15 Next, we can assume that the length of the opposite side to X is 8 and the length of the adjacent is 15.

Now that we have the length of the both sides, we can use the Pythagorean Theorem to find the length of the hypotenuse. c^2 = a^2 + b^2 We will substitute a= 15 and b= 8 into the formula and solve it for c.

c^2 = a^2 +b^2
c^2 = 15^2 + 8^2
c^2 = 225 +64
c^2 = 289
c=17

Next, we will use the trigonometric ratio for sine to find the value of sin Y. sin θ = Opposite/Hypotenuse In this case, the length of the opposite side to Y is 15 and the length of the hypotenuse is 17.

Then, we can rewrite the trigonometric ratio for sine by using these values. sin θ = Opposite/Hypotenuse ⇒ sin Y = 15/17 Therefore, the correct option is G.