Given, 'p' is the exterior angle of $△ P Q R$ and $^{'} q + r^{'}$ is the sum of interior angles opposite to 'p'. Which statement is true?

p + r = q

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p = q + r

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p + q = r

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r = q - p

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Solution

The correct option is D $p = q + r$
As $∠ θ + ∠ Q + ∠ R = 180^{\circ}$ ( $∵ P Q R$ is a triangle )
$\Rightarrow ∠ Q + ∠ R = 180^{\circ} - 20^{\circ}$
$\Rightarrow ∠ Q + ∠ R = ∠ P$

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Given, 'p' is the exterior angle of △PQR and ′q+r′ is the sum of interior angles opposite to 'p'. Which statement is true?

The correct option is D p=q+rAs ∠θ+∠Q+∠R=180∘ (∵PQR is a triangle ) ⇒∠Q+∠R=180∘−20∘ ⇒∠Q+∠R=∠P

Given, 'p' is the exterior angle of $△ P Q R$ and $^{'} q + r^{'}$ is the sum of interior angles opposite to 'p'. Which statement is true?

The correct option is D $p = q + r$
As $∠ θ + ∠ Q + ∠ R = 180^{\circ}$ ( $∵ P Q R$ is a triangle )
$\Rightarrow ∠ Q + ∠ R = 180^{\circ} - 20^{\circ}$
$\Rightarrow ∠ Q + ∠ R = ∠ P$