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sagar kumar

· started a discussion

· 1 Months ago

explain (Question: 186120.0 )

Question:
Two circles touch externally at P. QR is a common tangent of the circles touching the circles at Q and R. Then measure of \(\angle \)QPR  is –
Options:
A) 60º
B) 30º
C) 90º
D) 45º
Solution:
Ans: (c) 

\(\because \) SQ = SP (Tangents from an exterior 

point to a circle are equal)

\(\Rightarrow\) \(\angle \)1 = \(\angle \)2 (Angles opp. to equal sides 

of a are equal) 

Similarly \(\angle \)3 = \(\angle \)4

Now, \(\angle \)1 + \(\angle \)2 + \(\angle \)3 + \(\angle \)4 = 180° (ASP)

\(\Rightarrow\) \(\angle \)2 + \(\angle \)2 + \(\angle \)4 + \(\angle \)4 = 180°

\(\Rightarrow\) 2(\(\angle \)2 + \(\angle \)4) = 180°

\(\Rightarrow\) \(\angle \)2 + \(\angle \)4 = 90°

Knowledge Expert

· commented

· 1 Months ago

Hi, Please click on report and view detailed solution...........
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