Chapter 9.8, Problem 3E

To determine

Tofind:Sine and cosine ratios

Answer to Problem 3E

  $\sin J=\frac{40}{41}$

  $\sin L=\frac{9}{41}$

  $\cos J=\frac{9}{41}$

  $\cos L=\frac{40}{41}$

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-

  $\begin{array}{l}\sin J=\frac{Opposite}{Hypotenuse}\\ \Rightarrow \sin J=\frac{KL}{JL}\\ \Rightarrow \sin J=\frac{40}{41}\end{array}$

  $\begin{array}{l}\sin L=\frac{Opposite}{Hypotenuse}\\ \Rightarrow \sin L=\frac{KJ}{JL}\\ \Rightarrow \sin L=\frac{9}{41}\end{array}$

  $\begin{array}{l}\cos J=\frac{Adjacent}{Hypotenuse}\\ \Rightarrow \cos J=\frac{KJ}{JL}\\ \Rightarrow \cos J=\frac{9}{41}\end{array}$

  $\begin{array}{l}\cos L=\frac{Adjacent}{Hypotenuse}\\ \Rightarrow \cos L=\frac{KL}{JL}\\ \Rightarrow \cos L=\frac{40}{41}\end{array}$