Question

in the given figure ap is the tangent to the circle with Centre O OA =16 AP=30 QP= X find x

Answer 1

$angle \: o \: = {90}^{.} \: \\ angle \: p = angle \:= {30}^{.} ...........by \: pythagoras \: \: \\ \cos(30) = \frac{base}{hypo} \\ \\ \cos(30) = \frac{pr}{ao} \\ \\ \ \frac{ \sqrt{3} }{2} = \frac{pr}{16} \\ \\ \frac{16 \sqrt{3} }{2} = pr \\ \\ 8 \sqrt{3} = pr = rq........divided \: into \: 2 \: equal \: parts \\ \\ pq = pr + rq \\ pq = 2 \times pr \\ pq = 2 \times 8 \sqrt{3} \\ \\ pq = 16 \sqrt{3} units.$
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