In the diagram, QTR is a straight line and < PQT = 30o. find the sin of < PTR
- \(\frac{8}{15}\)
- \(\frac{2}{3}\)
-
\(\frac{3}{4}\)
- \(\frac{15}{16}\)
The correct answer is: C
Explanation
\(\frac{10}{\sin 30^o} = \frac{15}{\sin x} = \frac{10}{0.5} = \frac{15}{\sin x}\)\(\frac{15}{20} = \sin x\)
sin x = \(\frac{15}{20} = \frac{3}{4}\)
N.B x = < PRQ