PQR is a triangle in which PQ = 10cm and QPR = 60

^{o}S 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-
**A.**

2.7 -
**B.**

4.33 -
**C.**

3.1 -
**D.**

3.3

##### Correct Answer: Option B

##### Explanation

(bigtriangleup)PUS is right angled

(frac{US}{5}) = sin60^{o}

US = 5 x (frac{sqrt{3}}{2})

= 2.5(sqrt{3})

= 4.33cm