Question on: JAMB Mathematics - 2006
A
10 \(\sqrt{3}\)cm
B
10cm
C
45cm
D
14cm
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Correct Option: D
Draw a line perpendicular to |SR| to form |PT| in (\bigtriangleup) PST, cos 60 = (\frac{|ST|}{8}(
|PT| = 8 cos 6 = 4cm
Since |TR| = |PQ|
SR = ST + TR
= (4 + 10)cm
= 14cm
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