In the figure, P is hte centre of an excircle of ▲ ABC. If $∠ A B C = 40 °$ and $∠ B A C = 80 °$ , find the angle measure of triangle APC.

60°

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50°

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120°

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110°

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Solution

The correct option is A
60°

Given, $∠ B A C = 80 °$

Therefore, $∠ E A C = 100 °$ [Linear pair( $∠ B A C + ∠ E A C = 180 °$ )]

$∠ P A C = \frac{1}{2} \times 100 = 50 °$

Also, $∠ A P C = 90 ° - \frac{1}{2} ∠ A B C$ [ $∠ A B C$ = 40°, given]

= 90° - 20° = 70°

$∠ P C A = 180 ° - (50 ° + 70 °) = 60 °$

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In the figure, P is hte centre of an excircle of ▲ ABC. If ∠ABC=40° and ∠BAC=80°, find the angle measure of triangle APC.

The correct option is A 60° Given, ∠BAC=80° Therefore, ∠EAC=100° [Linear pair(∠BAC+∠EAC=180°)] ∠PAC=12×100=50° Also, ∠APC=90°−12∠ABC [∠ABC = 40°, given] = 90° - 20° = 70° ∠PCA=180°−(50°+70°)=60°

In the figure, P is hte centre of an excircle of ▲ ABC. If $∠ A B C = 40 °$ and $∠ B A C = 80 °$ , find the angle measure of triangle APC.