Question

In Figure 2, the bisectors of ZA and ZB meet at point P. If ZC = 50° and ZD = 100°, find the measure of ZAPB.

if he or she gives the right answer I will give them points​

Answer 1

$\large\text{\underline{First to know?}}$

The sum of angles in a quadrilateral is 360°.

The sum of angles in a triangle is 180°.

$\large\text{\underline{Solution}}$

In the quadrilateral ABCD, the sum of all angles is 360°.

Let the bisected angles be $x=\angle PAB$ and $y=\angle PBA$.

$\implies 2x+2y+50^{\circ}+100^{\circ}=360^{\circ}$

$\implies 2x+2y=210^{\circ}$

$\implies x+y=105^{\circ}$

Since $\angle APB$ is the remaining angle in the $\triangle ABP$, the measure of the angle will be $180^{\circ}-(x+y)$.

$\implies \angle APB=180^{\circ}-(x+y)$

$\implies \angle APB=75^{\circ}$

$\large\text{\underline{Conclusion}}$

$\therefore\angle APB=75^{\circ}$

Answer 2

Answer:

◐ The measure of ∠APB $\mapsto\:{\boxed{\tt{\purple{75^{\circ}}}}}$

Explanation:

Given information,

In given figure, the bisectors of ∠A and ∠B meet at point P. If ∠C = 50° and ∠D = 100°, find the measure of ∠APB.

• ∠C = 50°
• ∠D = 100°
• ∠BAP = ½∠A
• ∠PBA = ½∠B

Let,

• ∠A = 2m
• ∠B = 2n

Hence,

• ∠BAP = ½∠A = ½ × 2x = m
• ∠PBA = ½∠B = ½ × 2y = n

Concepts used,

$\odot\:\boxed{\sf{\red{Sum\:of\:all\:angles\:of\:Quadrilateral = 360^{\circ}}}}$

$\odot\:\boxed{\sf{\pink{Sum\:of\:all\:angles\:of\:\triangle = 180^{\circ}}}}$

★ In Quadrilateral ABCD ::

➻ ∠A + ∠B + ∠C + ∠D = 360°

Putting all values,

➻ 2m + 2n + 50° + 100° = 360°

➻ 2m + 2n + 150° = 360°

➻ 2m + 2n = 360° - 150°

➻ 2m + 2n = 210°

➻ 2(m + n) = 210°

➻ m + n = (210/2)°

➻ m + n = 105° $\tt\qquad- (1)$

Now,

★ In ∆PAB ::

➻ ∠BAP + ∠PBA + ∠APB = 180°

Putting all values,

➻ m + n + ∠APB = 180°

➻ ∠APB = 180° - (m + n) $\tt\qquad- (2)$

From (1) putting in (2),

➻ ∠APB = 180° - 105°

➻ ∠APB = 75°

• Henceforth, the measure of ∠APB is 75°.

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