Chapter 2, Problem 34RP

To determine

To find the magnitude of the required angle

Answer to Problem 34RP

The magnitude of the required angle is $\text{32}°$ , i.e., $\text{m}\angle \text{MOE}=\text{32}°$

Explanation of Solution

Given information :

  $\begin{array}{l}\overrightarrow{\text{OP}}\text{ bisects }\angle \text{MOE}\\ \text{m}\angle \text{MOP}=10-3\text{x}\\ \text{m}\angle {\text{POE=x}}^2-6\text{x}\end{array}$

Formula used :

Angle bisector divides the angle in two equal part

Negative value of angles are ignored

Calculation :

  $\begin{array}{l}\text{Since, }\overrightarrow{\text{OP}}\text{ bisects }\angle \text{MOE}\\ \therefore \angle \text{MOP}=\angle \text{POE}\mathrm{.......}\left(\text{as per definition}\right)\mathrm{........}\left(i\right)\\ \text{It is given that}\\ \text{m}\angle \text{MOP}=10-3\text{x and m}\angle {\text{POE=x}}^2-6\text{x}\end{array}$

  $\begin{array}{l}\text{On substituting values of }\angle \text{MOP and }\angle \text{POE in}\left(i\right),\text{we get}\\ 10-3{\text{x =x}}^2-6\text{x}\\ \Rightarrow {\text{x}}^2-3\text{x}-10=0\\ \Rightarrow {\text{x}}^2-5\text{x}+2\text{x}-10=0\\ \Rightarrow \text{x}\left(\text{x}-5\right)+2\left(\text{x}-5\right)=0\\ \Rightarrow \left(x-5\right)\left(x+2\right)=0\\ \Rightarrow \text{x}=5\text{ or x }=-2\end{array}$

  $\begin{array}{l}\text{Now, for x = 5, we have}\\ \therefore \text{m}\angle \text{MOP}\\ =10-3\text{x}\\ \text{= 10}-3\times 5\\ =10-15\\ =-5\left(n\text{ot possible, as magnitude of angle can not be negative}\right)\\ \text{and }\\ \text{m}\angle \text{POE}\\ {\text{=x}}^2-6\text{x}\\ {\text{=5}}^2-6\times 5\\ =25-30\\ =-5\left(n\text{ot possible, as magnitude of angle can not be negative}\right)\\ \therefore \text{x = }5,\text{ will be ignored}\end{array}$

  $\begin{array}{l}\text{Now, for x = }-\text{2, we have}\\ \therefore \text{m}\angle \text{MOP}\\ =10-3\text{x}\\ \text{= 10}-3\times \left(-2\right)\\ =10+6\\ =16\\ \text{and }\\ \text{m}\angle \text{POE}\\ {\text{=x}}^2-6\times \left(-2\right)\\ \text{=}{\left(-2\right)}^2+12\\ =4+12\\ =16\\ \therefore \angle \text{MOE}=\angle \text{MOP+}\angle \text{POE}\\ \Rightarrow \angle \text{MOE}=16+16=32°\end{array}$