Chapter 2.1, Problem 11PSB

To determine

To calculate: To find measure of $\angle 1,\angle 2,\angle 3$ .

Answer to Problem 11PSB

  $\angle 1={15}^{\circ },\angle 2={30}^{\circ },\angle 3={45}^{\circ }$

Explanation of Solution

Given information: $AB\perp BC$ and ratio of. $\angle 1,\angle 2$ and $\angle 3$ is 1:2:3

Let $\angle 1=x,\angle 2=2x,\angle 3=3x.$

  $\therefore AB\perp BC$ Hence $\angle ABC={90}^{\circ }$

  $\begin{array}{l}\therefore \angle 1+\angle 2+\angle 3={90}^{\circ }\\ \therefore x+2x+3x={90}^{\circ }\\ \therefore x={15}^{\circ }\\ \therefore \angle 1={15}^{\circ }\\ \angle 2=2\times {15}^{\circ }={30}^{\circ }\\ \angle 3=3\times {15}^{\circ }={45}^{\circ }\end{array}$