Question

In the figure angle p = 60 If QO,RO are bisectors of angle Q and angle R respectively . Find the of angle QOR.

Answer 1

GIVEN:

∠P = 60°

∠OQP = ∠OQR   [ QO , OR are

∠PRO = ∠ORQ    the bisectors of ∠Q and ∠R]

TO FIND:

∠QOR

SOLUTION:

In ΔPQR

∠OQP = ∠OQR  = x

∴∠PQR = ∠PQO + ∠OQR = x+x =2x

∠PRO = ∠ORQ  = x

∴∠PRQ = ∠PRO + ∠ORQ = x+x = 2x

∠PQR +∠PRQ + ∠QPR = 180° [ SUM OF Δ]

2x+2x + 60° = 180°

4x = 180°- 60° = 120°

x = $\frac{120}{4}$ = 30°

∠PRO = 30°

∠ORQ = 30°

∠PQO = 30°

∠OQR  = 30°

∠PRQ  = x+x = 60°

∠PQR = x+x = 60°

In Δ OQR

∠PQO +∠OQR +∠QOR = 180° [ SUM OF Δ]

30° +30° +∠QOR  = 180°

∠QOR = 180° - 60° = 120°

                                             (or)

In ΔPQR

Let ∠PQR and ∠PRQ  be x

∠PQR +∠PRQ + ∠QPR = 180° [ SUM OF Δ]

x+x+60° = 180°

2x = 180°-60° = 120°

x = $\frac{120}{2}$ = 60°

∠PQR  =   60°

∠PRQ =    60°

PQR = ∠OQR

 2

60 = 30° = ∠OQR

2

∠PRQ  =  ∠ORQ

   2

60 =  30° = ∠ORQ

2

In ΔOQR

∠PQO +∠OQR +∠QOR = 180° [ SUM OF Δ]

30° +30° +∠QOR  = 180°

∠QOR = 180° - 60° = 120°

Answer 2

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