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In the figure given above TP and TQ​ are tangent to the circle with center O at B and C, respectively.

If angle PBA=60 degrees and angle ACQ =70 degrees, what is angle BTC in degrees?

 

 May 15, 2020
 #1
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arc BA = 2* anglePBA  =  120°

arc CA = 2* angle ACQ = 140°

 

So  minor arc BC  =  360 - (120 + 140) =  100°

 

If we draw radii OB and OC  then angles TBO and TCO  = 90°

 

Then in quadrilateral TBOC the sum of the interior angles = 360°

 

And cental angle BOC = minor arc BC  =100°

 

So

 

( 90 + 90 + 100 + angle BTC)  = 360

 

  280  + angleTBC   =360

 

angle TBC  =  360 - 280   =  80°

 

 

cool cool cool

 May 15, 2020

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