Given that $\cos \theta = \frac { 14 } { 15 }$, find $\csc \theta$. (Hint: Sketch a right triangle corresponding to the trigonometric function of the acute angle $\theta$; then use the Pythagorean Theorem to determine the third side.)

A) $\frac { \sqrt { 29 } } { 15 }$

B) $\frac { 15 } { 14 }$

C) $\frac { 14 } { \sqrt { 29 } }$

D) $15 \sqrt { 29 }$

E) $\frac { 15 } { \sqrt { 29 } }$

Question

Given that

\cos \theta = \frac { 14 } { 15 }

, find

\csc \theta

. (Hint: Sketch a right triangle corresponding to the trigonometric function of the acute angle

\theta

; then use the Pythagorean Theorem to determine the third side.)

A) $\frac { \sqrt { 29 } } { 15 }$

B) $\frac { 15 } { 14 }$

C) $\frac { 14 } { \sqrt { 29 } }$

D) $15 \sqrt { 29 }$

E) $\frac { 15 } { \sqrt { 29 } }$

Quizplus · Accepted Answer

Let $a$ be the adjacent side and $h$ be the hypotenuse. Then we have $\cos 	heta = \frac{a}{h} = \frac{14}{15}$. Using the Pythagorean Theorem, we have $a^2 + b^2 = h^2$. Plugging in $\frac{14}{15}h$ for $a$, we get $\left(\frac{14}{15}hight)^2 + b^2 = h^2$. Simplifying, we get $\frac{196}{225}h^2 + b^2 = h^2$, or $b^2 = \frac{29}{225}h^2$. Taking the square root of both sides, we get $b = \frac{\sqrt{29}}{15}h$. Therefore, $\csc 	heta = \frac{h}{b} = \frac{15}{\sqrt{29}}$. The answer is $\boxed{	ext{(E)}}$.

Given That cos⁡θ=1415\cos \theta = \frac { 14 } { 15 }cosθ=1514​

Given That $\cos \theta = \frac { 14 } { 15 }$