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In a ∆PQR, I is incentre and O is the orthocenter, ∠QIR : ∠QOR is 61 : 58. ∠Q is 44° more than ∠R and X is the external angle bisector of ∠P and ∠R, then what is the ratio of ∠PXR to ∠PCQ where C is the circum-center of ∆PQR?

In a ∆PQR, I is incentre and O is the orthocenter, ∠QIR : ∠QOR is 61 : 58. ∠Q is 44° more than ∠R and X is the external angle bisector of ∠P and ∠R, then what is the ratio of ∠PXR to ∠PCQ where C is the circum-center of ∆PQR? Correct Answer 25 : 36

Give:

∠QIR : ∠QOR = 61 : 58

Calculation :

Let ∠QIR and ∠QOR be ‘61x’ and ‘58x’ respectively.

∠QIR = 90° + (∠P/2)

⇒ 122x = 180° + ∠P     ----(i)

∠QOR = 180° – ∠P

⇒ 58x = 180° – ∠P      ----(ii)

From (i) and (ii), we get

x = 2 and ∠P = 64°

now, 64° + ∠Q + ∠Q – 44° = 180°

⇒ ∠Q = 80°

⇒ ∠R = 80° - 44° = 36°

∠PCQ = 2 × ∠R = 72°

∠PXR = 90° – (∠Q/2) = 90° – 40° = 50°

∴ Required ratio = 50 : 72

= 25 : 36

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