# Chapter 6 - Section 6.2 - Trigonometry of Right Triangles - 6.2 Exercises: 9

$a)$ $\sin\alpha=\cos\beta=\dfrac{3\sqrt{34}}{34}$ $b)$ $\tan\alpha=\cot\beta=\dfrac{3}{5}$ $c)$ $\sec\alpha=\csc\beta=\dfrac{\sqrt{34}}{5}$

#### Work Step by Step

The triangle is shown below. Let $h$ be the hypotenuse of the triangle. Find it using the Pythagorean Theorem: $h=\sqrt{5^{2}+3^{2}}=\sqrt{25+9}=\sqrt{34}$ $a)$ $\sin\alpha$ and $\cos\beta$ $\sin\alpha=\dfrac{opposite}{hypotenuse}=\dfrac{3}{\sqrt{34}}=\dfrac{3\sqrt{34}}{34}$ $\cos\beta=\dfrac{adjacent}{hypotenuse}=\dfrac{3}{\sqrt{34}}=\dfrac{3\sqrt{34}}{34}$ $b)$ $\tan\alpha$ and $\cot\beta$ $\tan\alpha=\dfrac{opposite}{adjacent}=\dfrac{3}{5}$ $\cot\beta=\dfrac{adjacent}{opposite}=\dfrac{3}{5}$ $c)$ $\sec\alpha$ and $\csc\beta$ $\sec\alpha=\dfrac{hypotenuse}{adjacent}=\dfrac{\sqrt{34}}{5}$ $\csc\beta=\dfrac{hypotenuse}{opposite}=\dfrac{\sqrt{34}}{5}$

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