Chapter 3.6, Problem 14PSC

To determine

To Find: The $m\angle 2\text{ or }m\angle ABC$ from the given figure

Answer to Problem 14PSC

The $m\angle 2\text{ or }m\angle ABC\text{ = 60}°$

Explanation of Solution

Given information: In the given figure $\overline{AB}\text{ , }\overline{AC}$ are legs of isosceles $\Delta ABC\text{ and }m\angle 1\text{ = 5}x\text{ , }m\angle 3\text{ = 2}x\text{ +12}$

Concept used: vertically opposite angles, adjacent angles forming linear pair.

Calculation: We know that the legs of an isosceles triangle represents its equal sides and the angles containing those legs are also equal. Thus, in the given question we have:

  $\begin{array}{l}\overline{AB}=\overline{AC}\mathrm{....................................}\text{(given)}\\ \\ \Rightarrow \angle ABC\text{ = }m\angle 2=\angle ACB\mathrm{.......................................}(\text{i)}\\ \\ \text{also }m\angle 1=5x\\ \\ \text{and }m\angle 3\text{ = 2}x\text{ + 12 }\mathrm{.......................}\text{(ii)}\\ \end{array}$

Also, in the given figure we have a pair of vertically opposite angles and we know that vertically opposite angles are always equal thus we can say that:

  $m\angle 3=\angle ACB=2x+12\mathrm{.............................................}\text{(iii)}$

Thus from (i) and (ii) we can write this relation as:

  $m\angle 3\text{ = }m\angle 2=2x+\mathrm{12.............................}(\text{iv)}$

From figure we are given that $\angle 1\text{ and }\angle 2$ are adjacent angles standing on a straight line so they form a linear pair which means:

  $\begin{array}{l}m\angle 1+m\angle 2=180°\\ \\ \text{ using values of }m\angle 1andm\angle 2\text{ we get :}\\ \\ 5x\text{ + 2}x\text{ + 12 = 180}°\\ \\ \Rightarrow \text{ 7}x\text{ = 180}°\text{ - 12}\\ \\ \Rightarrow \text{ 7}x\text{ = 168}°\\ \\ \Rightarrow x\text{ = 24}°\end{array}$

Substituting this value in (iv) we get:

  $m\angle 2=2x+12\text{ = (2}\times \text{24) + 12 = 48 +12 =60}°$

Conclusion:

The $m\angle 2\text{ or }m\angle ABC\text{ = 60}°$