JAMB - Mathematics (1998 - No. 45)

In the diagram, QTR is a straight line and < PQT = 30o. find the sin of < PTR
\(\frac{8}{15}\)
\(\frac{2}{3}\)
\(\frac{3}{4}\)
\(\frac{15}{16}\)

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

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