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