Chapter 9.9, Problem 12PSB

To determine

To find:The value of $\cos \angle P$

Answer to Problem 12PSB

The value of $\cos \angle P=\frac{4}{5}$

Explanation of Solution

Given information:The following information is given in right triangle PQR

  $\sin \angle P=\frac{3}{5}$

  $PQ=10$

Formula used:Basic trigonometric identity has been used

  ${\sin }^2\angle P+{\cos }^2\angle P=1$

Calculation:We know that the value of $\cos \angle P$ will be

  $\cos \angle P=\sqrt{1-{\sin }^2\angle P}$

Hence, substituting the value of $\sin \angle P=\frac{3}{5}$ , we can say that

  $\begin{array}{l}\cos \angle P=\sqrt{1-{\left(\frac{3}{5}\right)}^2}\\ \cos \angle P=\sqrt{1-\frac{9}{25}}\end{array}$

  $\cos \angle P=\frac{4}{5}$