Question

Who can solve the questions given in the attachment? i will give 100 thanks + follow you.

Answer 1

$\huge\mathcal{\underbrace{SOLUTION:-}}$

✍️ See the attachment figure .

GIVEN :-

• A triangle ABC such that the bisectors of ⟨ABC and ⟨ACB meet at a point O .

• $\rm{\angle{ABO}\:=\:\angle{1}\:}$

• $\rm{\angle{ACO}\:=\:\angle{2}\:}$

• $\rm{\red{\implies\:\angle{B}\:=\:2(\angle{1})\:}}$

• $\rm{\red{\implies\:\angle{C}\:=\:2(\angle{2})\:}}$

To Prove :-

• ⟨BOC = 90° + ⟨A/2 .

Proof :-

✍️ In ∆BOC, we have

• ⟨1 + ⟨2 + ⟨BOC = 180° -----(1)

✍️ In ∆ABC,

• ⟨A + ⟨B + ⟨C = 180°

$\rm{\implies\:\angle{A}}$ + 2($\rm{\angle{1}}$) + 2($\rm{\angle{2}}$) = 180°

$\rm{\implies\:\dfrac{\angle{A}}{2}\:+\:\angle{1}\:+\:\angle{2}\:=\:90^{\degree}\:}$

$\rm{\implies\:\angle{1}\:+\:\angle{2}\:=\:90\:-\:\dfrac{\angle{A}}{2}\:}$

✍️ Now put the value of the above equation in the equation (1), we get

$\rm{\implies\:90\:-\:\dfrac{\angle{A}}{2}\:+\:\angle{BOC}\:=\:180^{\degree}\:}$

$\rm{\purple{\implies\:\angle{BOC}\:=\:90^{\degree}\:+\:\dfrac{\angle{A}}{2}\:}}$

✍️ Hence, Proved .