Question

ABCD is a quadrilateral in which the bisectors of ∠B and ∠C meet at O. If ∠A = 100° and ∠D = 90°, find the measure of ∠BOC.​

Answer 1

       $\underline{\bold{Solution:}}$

       $\text{It is given that: }$

       $\angle{A} = 100^{o}$

       $\angle{D} = 90^{o}$

       $\text{Let }\angle{B}\text{ and }\angle{C}\text{ be }2x \text{ and }2y \text{ respectively}$

       $\text{In }ABCD$

       $\text{We know that, sum of angles of a quadrilateral} = 360^{o}$

$\implies \angle{A} + \angle{B} + \angle{C} + \angle{D} = 360^{o}$

       $\text{Substituting the values of the angles}$

$\implies 100 + 2x + 2y + 90 = 360^{o}$

       $\text{Simplifying...}$

$\implies 190 + 2x + 2y = 360^{o}$

       $\text{Transposing 190 to R.H.S}$

$\implies 2x + 2y = 360 - 190$

       $\text{Simplifying...}$

$\implies 2x + 2y = 170^{o}$

       $\text{Simplifying...}$

$\implies 2(x+y) = 170$

       $\text{Transposing 2 to R.H.S}$

$\implies x + y = \dfrac{170}{2}$

       $\text{Simplifying...}$

$\implies x + y = 85^{o} \text{ ------ (i)}$

       $\text{In }\triangle{BOC}$

       $\angle{OBC} = \dfrac12\angle{B} = x \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(BO \text{ is the angle bisector of }\angle{B})$

       $\angle{OCB} = \dfrac12\angle{C} = y \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(CO \text{ is the angle bisector of }\angle{C})$

       $\text{We know that, sum of angles of a triangle} = 180^{o}$

$\implies \angle{OBC}+\angle{BOC} + \angle {OCB} = 180^{o}$

       $\text{Substituting the values of the angles}$

$\implies x +\angle{BOC}+ y= 180^{o}$

       $\text{From (i)}$

$\implies \angle{BOC}+85 = 180^{o}$

       $\text{Transposing 85 to R.H.S}$

$\implies \angle{BOC}= 180 - 85$

       $\text{Simplifying...}$

$\implies \boxed{\angle{BOC}= 95^{o} }$