JAMB - Mathematics (1988 - No. 40)
PQR is a triangle in which PQ = 10cm and QPR = 60oS is a point equidistant from P and Q. Also S is a point equidistant from PQ and PR. If U is the foot of the perpendicular from S on PR, find the length SU in cm to one decimal place
2.7
4.33
3.1
3.3
Explanation
\(\bigtriangleup\)PUS is right angled
\(\frac{US}{5}\) = sin60o
US = 5 x \(\frac{\sqrt{3}}{2}\)
= 2.5\(\sqrt{3}\)
= 4.33cm
\(\frac{US}{5}\) = sin60o
US = 5 x \(\frac{\sqrt{3}}{2}\)
= 2.5\(\sqrt{3}\)
= 4.33cm
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