In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP.
(i) Show that: ∠P : ∠R = 1 : 3
(ii) Find the value of ∠Q.

Question

In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP. (i) Show that: ∠P : ∠R = 1 : 3 (ii) Find the value of ∠Q.

KnowledgeBoat · Accepted Answer

Δ PQR such that PQ = PR. A and B are the two points on PQ and PR, so that QR = RA = AB = BP. (i) In Δ PQR, PQ = PR If two sides of a triangle are equal, then two opposite angles are always equal. So, ∠Q = ∠R Similarly, in Δ PAB, AB = PB So, ∠BAP = ∠BPA ⇒ ∠BAP = ∠P Using exterior angle property, ⇒ ∠ABR = ∠P + ∠BAP = ∠P + ∠P = 2∠P In Δ ABR, AB = AR So, ∠ABR = ∠ARB = 2 ∠P Sum of all angles in a triangle ABR is 180°. ⇒ ∠ABR + ∠ARB + ∠RAB = 180° ⇒ 2∠P + 2∠P + ∠RAB = 180° ⇒ 4∠P + ∠RAB = 180° ⇒ ∠RAB = 180° - 4∠P Sum of angles on a straight line = 180° ⇒ ∠PAB + ∠BAR + ∠RAQ = 180° ⇒ ∠P + 180° - 4∠P + ∠RAQ = 180° ⇒ - 3∠P + ∠RAQ = 0 ⇒ ∠RAQ = 3∠P In Δ ARQ, AR = QR So, ∠RQA = ∠QAR = 3∠P And, we already proved, ∠Q = ∠R So, ∠R = 3∠P ⇒ Hence, ∠P : ∠R = 1 : 3. (ii) From (i), ∠P : ∠R = 1 : 3 Let ∠P be x and ∠R be 3x. As we know ∠R = ∠Q = 3x. In Δ PQR, sum of all angles is 180°. ∠P + ∠R + ∠Q = 180° ⇒ x + 3x + 3x = 180° ⇒ 7x = 180° ⇒ x = ⇒ x = 25 So, ∠Q = 3x = 3 x = = 77 Hence, the value of ∠Q = 77 .

Mathematics

In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP.
(i) Show that: ∠P : ∠R = 1 : 3
(ii) Find the value of ∠Q.

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In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP.(i) Show that: ∠P : ∠R = 1 : 3(ii) Find the value of ∠Q.

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(i) Show that: ∠P : ∠R = 1 : 3
(ii) Find the value of ∠Q.

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If ∠BCA = 2∠BAC, show that AC = 2BC.

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