In Δ PQR, PS is the bisector of ∠P ...
In Δ PQR, PS is the bisector of ∠ P and PT ⊥ OR, then ∠ TPS is equal to:
• a)
∠Q + ∠ R
• b)
900 + 1/2 ∠Q
• c)
900 - 1/2 ∠R
• d)
1/2 (∠ Q - ∠ R)
• e)
None of These
In Δ PQR, PS is the bisector of ∠P and PT ⊥ OR, then &a...
∠1 + ∠2 = ∠3 [PS is bisector.] ------ (1)
∠Q = 900 - ∠1
∠R = 900 -∠2 - ∠3
So,
∠Q - ∠R = (900 - ∠1) - (900 - ∠2 - ∠3)
∠Q - ∠R = ∠2 + ∠3 - ∠1
∠Q - ∠ R = ∠2 + (∠1 + ∠2) -∠1[using equation 1]
∠Q - ∠R = 2 ∠2
1/2 * (∠Q - ∠R) = ∠TPS.
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In Δ PQR, PS is the bisector of ∠P and PT ⊥ OR, then ∠TPS is equal to:a)∠Q + ∠ Rb)900+ 1/2 ∠Qc)900- 1/2 ∠Rd)1/2 (∠ Q - ∠ R)e)None of TheseCorrect answer is option 'D'. Can you explain this answer?
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