1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord, such that ∠QPT = 50°, then ∠PQT = ? (a) 100° (b) 90° (c) 80° (d) 75°

Open in App
Solution

## $\left(\mathrm{a}\right)100°\phantom{\rule{0ex}{0ex}}\mathrm{Given},\angle QPT={50}^{0}\phantom{\rule{0ex}{0ex}}\mathrm{and}\angle OPT={90}^{0}\left(\text{T}\mathrm{angents}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{}\mathrm{an}\mathrm{}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{are}\mathrm{}\mathrm{perpendicular}\phantom{\rule{0ex}{0ex}}\mathrm{to}\mathrm{}\mathrm{the}\mathrm{}\mathrm{radius}\mathrm{}\mathrm{at}\mathrm{}\mathrm{the}\mathrm{}\mathrm{point}\mathrm{}\mathrm{of}\mathrm{}\mathrm{contact}\right)\phantom{\rule{0ex}{0ex}}\therefore \angle OPQ=\left(\angle OPT-\angle QPT\right)=\left({90}^{0}-{50}^{0}\right)={40}^{0}\phantom{\rule{0ex}{0ex}}OP=OQ\left(\text{R}\mathrm{adius}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{same}\mathrm{}\mathrm{circle}\right)\phantom{\rule{0ex}{0ex}}⇒\angle OQP=\angle OPQ={40}^{0}\phantom{\rule{0ex}{0ex}}\mathrm{In}∆POQ,\angle POQ+\angle OQP+\angle OPQ={180}^{0}\phantom{\rule{0ex}{0ex}}\therefore \angle POQ={180}^{0}-\left({40}^{0}+{40}^{0}\right)={100}^{0}\phantom{\rule{0ex}{0ex}}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Tangents Drawn from an External Point
MATHEMATICS
Watch in App
Join BYJU'S Learning Program