In the diagram, QTR is a straight line and < PQT = 30^o. find the sin of < PTR

$\frac{8}{15}$

$\frac{2}{3}$

$\frac{3}{4}$

$\frac{15}{16}$

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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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In the diagram, QTR is a straight line and < PQT = 30o. find the sin of < PTR

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In the diagram, QTR is a straight line and < PQT = 30^o. find the sin of < PTR