1.

In ΔPQR, PS is the bisector of ∠P and PT ⊥ OR, then ∠TPS is equal to:

A. Q + ∠R
B. 0° + $$\frac{1}{2}$$ ∠Q
C. 0° - $$\frac{1}{2}$$ ∠R
D. $\frac{1}{2}$$ (∠Q - ∠R)
Answer» E.


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