This is yet to be solved.
AB and DC are tangents to the circle.
angle $\displaystyle APC = 125^o$ and angle $\displaystyle PDC = 30^o$
Find angle PAB
AP intersects the curcle at E and DP intersects the circle at F.
$\displaystyle \frac{m(arcCF)+m(arcBE)}{2}=55$
$\displaystyle \frac{m(arcEC)+m(arcBF)}{2}=125$
$\displaystyle \frac{m(arcEC)-m(arcCF)}{2}=30$
From the second equality substract the sum of the other two.
angle bpa + cpa = 180 (we know because all lines are 180 degrees)
cpa = 125
bpa = 55
pba + pab + bpa = 180 (we know because all real triangles are 180 degrees)
bpa = 55
pba = 90 (we know because its a tangent of a radius)
90 + pab + 55 = 180
pab = 35
? i think i did it right
Be careful all of you, something is not given here:
BPC is not diameter and P is not radius.
BPC is straight line but APD may be NOT.
yes AB and DC are tangents to the circle. However, angles B and C would be 90 degrees BUT ONLY IF BP and CP are radius....
The correct answer is $\displaystyle 40^o$ but i cannot quite understand Red Dog's method...
In triangle PCD the angles are 30º at D, 55º at P, and therefore 95º at C. Now extend the tangent DC to a point E say, on the far side of C from D. Then angle BCE is 85º. But the angles between a chord and the two tangents at the ends of the chord are equal. Therefore angle ABC is also 85º. Then in triangle ABP the angles are 85º at B, 55º at P, and hence 40º at A.
I don't know that it is a theorem with a formal name. But if you draw the two radii from the centre of the circle to the ends of the chord, and also the line from the centre of the circle to the midpoint of the chord, then you get two congruent triangles. That, together with the fact that the radii are perpendicular to the tangents, gives you the result.