Using the exterior angle property, find the value of $y$ from the figure.

80^{\circ}

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100^{\circ}

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20^{\circ}

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50^{\circ}

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Solution

The correct option is B $100^{\circ}$

$∠ A P Q = ∠ P Q R$
[Alternate angles are equal]
$\Rightarrow 50^{\circ} = ∠ P Q R$

By exterior angle property,
$\Rightarrow ∠ P Q R + ∠ Q P R = ∠ P R D$
$\Rightarrow 50^{\circ} + y = 150^{\circ}$
$\Rightarrow y = 150^{\circ} - 50^{\circ} = 100^{\circ}$
$\Rightarrow y = 100^{\circ}$

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Derive the angle sum property of the triangle using the exterior angle property of the triangle. [4 MARKS]

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(b) AE is the angular bisector. $∠ 4 = 140^{\circ}, ∠ 2 = 60^{\circ}, ∠ E A C = x$ . Find $x$ .

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Q. Find the value of

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Using the exterior angle property, find the value of y from the figure.

The correct option is B 100∘ ∠APQ=∠PQR [Alternate angles are equal] ⇒50∘ =∠PQR By exterior angle property, ⇒∠PQR+∠QPR=∠PRD ⇒ 50∘ + y=150∘ ⇒ y=150∘−50∘=100∘ ⇒ y=100∘

Using the exterior angle property, find the value of $y$ from the figure.