MCQOPTIONS
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| 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. | |