Question

Find each trigonometric ratio. Give answer in the simplest form

Answer 1

Given the triangle below:

Required: To evaluate the trigonometric ratios for angles R and Q.

For angle R:

Step 1:

Name the sides of the above triangle with respect to the angle R.

$\begin{gathered} QR\Rightarrow hypotenuse\text{ (longest side of the triangle)} \\ QS\Rightarrow opposite\text{ (side opposite to the angle R)} \\ SR\Rightarrow adjacent \end{gathered}$

Step 2:

Evaluate the trigonometric ratios with respect to the angle R.

From the trigonometric ratios,

$\begin{gathered} \sin \text{ R = }\frac{opposite}{hypotenuse}\text{ = }\frac{QS}{QR} \\ \cos \text{ R = }\frac{adjacent}{hypotenuse}=\frac{SR}{QR} \\ \tan \text{ R = }\frac{opposite}{adjacent}=\text{ }\frac{QS}{SR} \end{gathered}$$\begin{gathered} \text{where} \\ QR\text{ = 34, QS =16, SR = 30} \end{gathered}$

thus, evaluating for angle R, we have

$\begin{gathered} \sin \text{ R = }\frac{opposite}{hypotenuse}\text{ = }\frac{QS}{QR}\text{ = }\frac{16}{34} \\ \Rightarrow\frac{8}{17} \\ \cos \text{ R = }\frac{adjacent}{hypotenuse}=\frac{SR}{QR}=\text{ }\frac{30}{34} \\ \Rightarrow\frac{15}{17} \\ \tan \text{ R = }\frac{opposite}{adjacent}=\text{ }\frac{QS}{SR}=\frac{16}{30} \\ \Rightarrow\frac{8}{15} \end{gathered}$

For angle Q:

Step 1:

Name the sides of the above triangle with respect to the angle Q.

$\begin{gathered} QR\Rightarrow hypotenuse\text{ (longest side of the triangle)} \\ QS\Rightarrow adjacent \\ SR\Rightarrow opposite\text{ (side opposite the angle Q)} \end{gathered}$

Step 2:

Evaluate the trigonometric ratios with respect to the angle Q.

From the trigonometric ratios,

$\begin{gathered} \sin \text{ Q = }\frac{opposite}{hypotenuse}\text{ = }\frac{SR}{QR} \\ \cos \text{ Q = }\frac{adjacent}{hypotenuse}=\frac{QS}{QR} \\ \tan \text{ Q = }\frac{opposite}{adjacent}=\text{ }\frac{SR}{QS} \\ \end{gathered}$$\begin{gathered} \text{where} \\ QR\text{ = 34, QS =16, SR = 30} \end{gathered}$

thus, evaluating for angle R, we have

$\begin{gathered} \sin \text{ Q = }\frac{opposite}{hypotenuse}\text{ = }\frac{SR}{QR}=\frac{30}{34} \\ \Rightarrow\frac{15}{17} \\ \cos \text{ Q = }\frac{adjacent}{hypotenuse}=\frac{QS}{QR}=\frac{16}{34} \\ \Rightarrow\frac{8}{17} \\ \tan \text{ Q = }\frac{opposite}{adjacent}=\text{ }\frac{SR}{QS}=\frac{30}{16} \\ \Rightarrow\frac{15}{8} \end{gathered}$

Hence,

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