Chapter 14, Problem 6MCQ

To determine

The value of $\sin \theta$ for the given expression.

Answer to Problem 6MCQ

The value of the expression will be $\sin \theta =\frac{4}{5}$ .

Explanation of Solution

Given:

The expression is given as:

  $\cos \theta =\frac{3}{5};0^{\circ }<\theta <{90}^{\circ }$

Consider the expression.

  $\cos \theta =\frac{3}{5};0^{\circ }<\theta <{90}^{\circ }$

From identify:

  ${\sin }^2\theta +{\cos }^2\theta =1$

Therefore,

  $\begin{array}{l}{\sin }^2\theta =1-{\left(\frac{3}{5}\right)}^2\\ {\sin }^2\theta =1-\frac{9}{25}\\ {\sin }^2\theta =\frac{16}{25}\\ {\sin }^2\theta =\pm \frac{4}{5}\end{array}$

Since, expression is in the first quadrant, so the solution will be:

  $\sin \theta =\frac{4}{5}$