Answer
$\sin \theta=\dfrac{7}{15}$
$\cos \theta=\dfrac{4 \sqrt{11}}{15}$
$\tan \theta=\dfrac{7\sqrt{11}}{44}$
$\csc \theta=\dfrac{15}{7}$
$\sec \theta=\dfrac{15\sqrt{11}}{44}$
$\cot \theta=\dfrac{4\sqrt{11}}{7}$
Work Step by Step
We will have to use the Pythagorean Theorem to find the last side. Let $a$ be the last side, then $a=\sqrt {15^2-7^2}=\sqrt {225-49}=4 \sqrt{11}$
$\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{7}{15}$
$\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{4 \sqrt{11}}{15}$
$\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{7}{4\sqrt{11}}=\dfrac{7\sqrt{11}}{44}$
$\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{15}{7}$
$\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{15\sqrt{11}}{44}$
$\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{4\sqrt{11}}{7}$
