In the given figure- if - BCA - RP Q -15-then - P QR is -A- 45-B- 75- 15-

The correct option is A 45^∘In Δ ABC and Δ PQR, ∠ BCA = ∠ RPQ = 15^∘ .... (Given) BC = PR = 1 km ....... nbsp;(Given) AC = PQ = 1.3 km .... nbsp;(Give ...

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Standard VII

Mathematics

Congruent Triangles

In the given ...

Question

In the given figure, if $∠ B C A = ∠ R P Q = 15^{\circ},$
then $∠ P Q R$ is .

45^{\circ}

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75^{\circ}

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15^{\circ}

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Solution

The correct option is A $45^{\circ}$
In $Δ A B C$ and $Δ P Q R$ ,

$∠ B C A = ∠ R P Q$ = $15^{\circ}$ .... (Given)

$B C = P R = 1 k m$ ....... (Given)

$A C = P Q = 1.3 k m$ .... (Given)

Thus, $Δ A C B ≅ Δ Q P R$ by SAS congruency.
$\Rightarrow ∠ P Q R = ∠ B A C$ .... (CPCT)

Along the horizontal line through point A, we have
$90^{\circ} + ∠ B A C + 45^{\circ} = 180^{\circ}$
$∠ B A C = 180^{\circ} - 45^{\circ} - 90^{\circ}$
$\Rightarrow ∠ P Q R = ∠ B A C = 45^{\circ}$

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