Question

In triangle PQR , PQ=PR , A is a point on PQ and B is a point in PR , so that QR=RA=AB=BP. Prove that:-. 1) angle P : angle R = 1:3 2) angle Q = ?​

Answer 1

Answer:  (i) Hence Proved

(ii) Angle Q = 77.15

Step-by-step explanation:

(I) In ΔPAB

AB =PB

so ∠PAB = ∠P

∠ABR = ∠PAB + ∠P  [ Exterior angle of triangle]

∠ABR = 2∠P....(i)

In ΔABR

∠ABR = ∠ARB = 2∠P......[From (i)]

∠BAR + ∠ARB +∠ABR = 180°......[Sum of linear pair angles]

∠BAR = 180° - 4∠P......(ii)

∠PAB + ∠BAR + ∠QAR = 180°

From (i) and (ii)

∠P + 180° - 4∠P + ∠QAR = 180°

∠QAR = 3∠P.......(iii)

In ΔQAR

AR = QR

∠QAR = ∠Q = 3∠P.....(iv)

Hence Proved

2. In ΔPQR

PQ = PR

∠Q =∠R

∠P +∠Q +∠R = 180°

1/3 ∠Q + ∠Q + ∠Q = 180°

∠Q = 77.15°

Answer 2

Answer:

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