Chapter 9, Problem 4CR

To determine

Tofind: $\sin D,\cos D,\tan D$

Answer to Problem 4CR

  $\tan D=\frac{d}{e}$

  $\sin D=\frac{d}{f}$

  $\cos D=\frac{e}{f}$

Explanation of Solution

Given information:

Concept Used:

In a right-angled triangle ABC, tangent for acute angle $\angle A$ is the ratio of length of side opposite to angle $\angle A$ to the length of side adjacent to $\angle A$

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-

  $\begin{array}{l}\tan D=\frac{opposite}{adjacent}\\ \Rightarrow \tan D=\frac{FE}{DF}\\ \Rightarrow \tan D=\frac{d}{e}\end{array}$

  $\begin{array}{l}\sin D=\frac{opposite}{hypotenuse}\\ \Rightarrow \sin D=\frac{FE}{DE}\\ \Rightarrow \sin D=\frac{d}{f}\end{array}$

  $\begin{array}{l}\cos D=\frac{adjacent}{hypotenuse}\\ \Rightarrow \cos D=\frac{DF}{DE}\\ \Rightarrow \cos D=\frac{e}{f}\end{array}$