Question

question 8 of 10
\sin \theta = \frac{20}{29}. find \tan \theta.
image of a right triangle with \theta at the bottom right vertex
a. \frac{20}{21}
b. \frac{21}{20}
c. \frac{21}{29}
d. \frac{29}{21}

Explanation:

Step1: Recall trigonometric definitions

In a right triangle, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$, and $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$. Given $\sin\theta = \frac{20}{29}$, so opposite side $= 20$, hypotenuse $= 29$.

Step2: Find adjacent side

Use Pythagorean theorem: $a^2 + b^2 = c^2$. Let adjacent be $a$, opposite $b = 20$, hypotenuse $c = 29$. Then $a = \sqrt{c^2 - b^2} = \sqrt{29^2 - 20^2} = \sqrt{841 - 400} = \sqrt{441} = 21$.

Step3: Calculate $\tan\theta$

$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{20}{21}$.

Answer:

A. $\frac{20}{21}$