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
The correct answer is: B
Explanation
\(\bigtriangleup\)PUS is right angled\(\frac{US}{5}\) = sin60o
US = 5 x \(\frac{\sqrt{3}}{2}\)
= 2.5\(\sqrt{3}\)
= 4.33cm