Answer
$\sin\theta=\dfrac{7}{8}$
$\cos\theta=\dfrac{\sqrt{15}}{8}$
$\tan\theta=\dfrac{7\sqrt{15}}{15}$
$\csc\theta=\dfrac{8}{7}$
$\sec\theta=\dfrac{8\sqrt{15}}{15}$
$\cot\theta=\dfrac{\sqrt{15}}{7}$
Work Step by Step
The triangle is shown in the attached image below.
Let $c$ be the unknown cathetus of the triangle. Find it using the Pythagorean Theorem:
$c=\sqrt{8^{2}-7^{2}}=\sqrt{64-49}=\sqrt{15}$
$\sin\theta=\dfrac{opposite}{hypotenuse}=\dfrac{7}{8}$
$\cos\theta=\dfrac{adjacent}{hypotenuse}=\dfrac{\sqrt{15}}{8}$
$\tan\theta=\dfrac{opposite}{adjacent}=\dfrac{7}{\sqrt{15}}=\dfrac{7\sqrt{15}}{15}$
$\csc\theta=\dfrac{hypotenuse}{opposite}=\dfrac{8}{7}$
$\sec\theta=\dfrac{hypotenuse}{adjacent}=\dfrac{8}{\sqrt{15}}=\dfrac{8\sqrt{15}}{15}$
$\cot\theta=\dfrac{adjacent}{opposite}=\dfrac{\sqrt{15}}{7}$
