Question

In the adjoining figure , the bisecter of angle A and B meet at a point P . if angle C =100and D=50find the measure of angle APB
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Answer 1

Step-by-step explanation:

100+50+x=360

250+x=360. (say APB =x)

x=360-250

x=110

APB angle is 110

Answer 2

$\star\;\normalsize{\underline{\underline{\frak \pink{GivEn:-}}}}$

• the bisecter of angle ∠A and ∠B meet at a point P.

• ∠C =100 and ∠D=50.

$\star\;\normalsize{\underline{\underline{\frak \pink{To\;Find:-}}}}$

• ∠APB = ?

$\star\;\normalsize{\underline{\underline{\frak \red{Solution:-}}}}$

★ Let ∠A and ∠B be x.

★ $\normalsize\sf {\underline{In\; Quadrilateral\;ABCD:-}}$

$:\implies\normalsize\sf ∠A + ∠B + ∠C + ∠D = 360^\circ \\\\ :\implies\normalsize\sf x + x + 100^\circ + 50^\circ = 360^\circ \\\\ :\implies\normalsize\sf 2x + 150^\circ = 360^\circ \\\\ :\implies\normalsize\sf 2x = 360^\circ - 150^\circ \\\\ :\implies\normalsize\sf 2x = 210^\circ \\\\ :\implies\normalsize\sf x = \cancel{ \dfrac{210^\circ}{2}} \\\\ :\implies\normalsize\sf x = \bf{105^\circ}$

$\therefore\normalsize\sf ∠A \; and \; ∠B = \bf{105^\circ}$

★ $\normalsize\sf {\underline{In\;\triangle\;APB:-}}$

• $\normalsize\sf ∠BAP = \dfrac{1}{2} ∠A = \cancel{ \dfrac{105^\circ}{2}} = \bf{52.5^\circ}$

• $\normalsize\sf ∠PBA = \dfrac{1}{2} ∠B = \cancel{ \dfrac{105^\circ}{2}} = \bf{52.5^\circ}$

$\leadsto$ ∠APB + ∠BAP + ∠PBA = 180°

$\leadsto$ ∠APB + ∠52.5° + ∠52.5° = 180°

$\leadsto$ ∠APB + ∠105° = 180°

$\leadsto$ ∠APB = 180° - 105°

$\leadsto$ ∠APB = $\bf{75^\circ}$

$\normalsize{\underline{\underline{\sf{\purple{\dag\;Hence\;Solved!!}}}}}$

$\rule{200}{3}$