# basic circle geometry

• Jul 20th 2010, 11:58 PM
sammy28
basic circle geometry
Hi all,

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

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

$\angle PAB = 90 -14 = 76^{\circ}$

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

BUT....

Angle sum of quadrilateral is $360^{\circ}$

In which case

$\angle PBA = 360 -152 - 90 - 71 = 47^{\circ}$

(Punch)

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

thanks.
• Jul 21st 2010, 01:24 AM
earboth
Quote:

Originally Posted by sammy28
Hi all,

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

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

$\angle PAB = 90 -14 = 76^{\circ}$

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

BUT....

Angle sum of quadrilateral is $360^{\circ}$

In which case

$\angle PBA = 360 -152 - 90 - 71 = 47^{\circ}$

(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.
• Jul 21st 2010, 01:58 AM
sammy28
thanks earboth. i should pay more attention (Rock)