In quadrilateral ABCD- the bisectors of angle A and angle B meet in a point P- If - C-105- and - D -55- find the measure of angle APB-A- 60- B- 70-C- 80-D- 90-

The correct option is C 80^∘Given that, ∠ C=105^∘ and ∠ D=55^∘ We know that, ∠ A+∠ B+∠ C+∠ D=360^∘ ⇒∠ A+∠ B+105^∘+55^∘=360^∘ ⇒∠ A+∠ B=360^∘ 160^∘ ⇒∠ A+∠ B=200^ ...

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Byju's Answer

Standard IX

Mathematics

Introduction to Quadrilaterals

In quadrilate...

Question

In quadrilateral ABCD, the bisectors of angle A and angle B meet in a point P. If $∠ C = 105^{\circ}$ and $∠ D = 55^{\circ}$ , find the measure of angle APB.

60^{\circ}

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70^{\circ}

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80^{\circ}

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90^{\circ}

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Solution

The correct option is C $80^{\circ}$
Given that,
$∠ C = 105^{\circ} a n d ∠ D = 55^{\circ}$
We know that,
$∠ A + ∠ B + ∠ C + ∠ D = 360^{\circ}$
$\Rightarrow ∠ A + ∠ B + 105^{\circ} + 55^{\circ} = 360^{\circ}$
$\Rightarrow ∠ A + ∠ B = 360^{\circ} - 160^{\circ}$
$\Rightarrow ∠ A + ∠ B = 200^{\circ}$
$\Rightarrow \frac{1}{2} ∠ A + \frac{1}{2} ∠ B = 100^{\circ}$ ...(i)

Since, AP and PB are the bisectors of $∠ B A D a n d ∠ A B C$ .
Therefore, equation (i) can be written as,
$\Rightarrow ∠ P A B + ∠ P B A = 100^{\circ}$ ...(ii)

In $△ A P B$ ,
$∠ P A B + ∠ P B A + ∠ A P B = 180^{\circ}$
From (ii),
$100^{\circ} + ∠ A P B = 180^{\circ}$
$∠ A P B = 80^{\circ}$

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In quadrilateral ABCD, the bisectors of angle A and angle B meet in a point P. If ∠ C=105∘ and ∠ D=55∘, find the measure of angle APB.

In quadrilateral ABCD, the bisectors of angle A and angle B meet in a point P. If $∠ C = 105^{\circ}$ and $∠ D = 55^{\circ}$ , find the measure of angle APB.