Chapter 9.8, Problem 7E

To determine

Tofind:Sine and cosine ratios

Answer to Problem 7E

  $\sin A=\frac{12}{13}$

  $\sin B=\frac{5}{13}$

  $\cos A=\frac{5}{13}$

  $\cos B=\frac{12}{13}$

Explanation of Solution

Given information:

Concept Used:

In a right-angled triangle ABC, sine for acute angle $\angle A$ is the ratio of length of side opposite to angle $\angle A$ to the length of hypotenuse and cosine of acute angle $\angle A$ is the ratio of length of side adjacent to angle $\angle A$ to the length of hypotenuse.

Calculation:

Basis definition we can say-

  $\begin{array}{l}\sin A=\frac{Opposite}{Hypotenuse}\\ \Rightarrow \sin A=\frac{BC}{AB}\\ \Rightarrow \sin A=\frac{12}{13}\end{array}$

  $\begin{array}{l}\sin B=\frac{Opposite}{Hypotenuse}\\ \Rightarrow \sin B=\frac{AC}{AB}\\ \Rightarrow \sin B=\frac{5}{13}\end{array}$

  $\begin{array}{l}\cos A=\frac{Adjacent}{Hypotenuse}\\ \Rightarrow \cos A=\frac{AC}{AB}\\ \Rightarrow \cos A=\frac{5}{13}\end{array}$

  $\begin{array}{l}\cos B=\frac{Adjacent}{Hypotenuse}\\ \Rightarrow \cos B=\frac{BC}{AB}\\ \Rightarrow \cos B=\frac{12}{13}\end{array}$