basic circle geometry

Feb 2009
81
10
Hi all,

\(\displaystyle \angle OAP = 90^{\circ}\), tangent is perpendicular to radius at point of contact

\(\displaystyle \angle OAB = \frac{180-152}{2}=14^{\circ}\), angle sum of triangle is 180, triangle OAB is isosceles (equal angles opposite equal sides)

\(\displaystyle \angle PAB = 90 -14 = 76^{\circ}\)

\(\displaystyle \angle PBA = 180 - 71 - 76 = 33^{\circ}\), angle sum triangle is 180

BUT....

Angle sum of quadrilateral is \(\displaystyle 360^{\circ}\)

In which case

\(\displaystyle \angle PBA = 360 -152 - 90 - 71 = 47^{\circ}\)

(Punch)

is this a poorly drawn up question or is the reasoning flawed?

thanks.
 

Attachments

Last edited:

earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
Hi all,

\(\displaystyle \angle OAP = 90^{\circ}\), tangent is perpendicular to radius at point of contact

\(\displaystyle \angle OAB = \frac{180-152}{2}=14^{\circ}\), angle sum of triangle is 180, triangle OAB is isosceles (equal angles opposite equal sides)

\(\displaystyle \angle PAB = 90 -14 = 76^{\circ}\)

\(\displaystyle \angle PBA = 180 - 71 - 76 = 33^{\circ}\), angle sum triangle is 180

BUT....

Angle sum of quadrilateral is \(\displaystyle 360^{\circ}\)

In which case

\(\displaystyle \angle PBA = 360 -152 - 90 - 71 = 47^{\circ}\)

(Punch)

is this a poorly drawn up question or is the reasoning flawed?

thanks.
What you've calculated is the angle indicated in red.

Subtract the base angle of the isosceles triangle and you'll get your first result.
 

Attachments

  • Like
Reactions: sammy28
Feb 2009
81
10
thanks earboth. i should pay more attention (Rock)