# Math Help - question on properties of circle 7

1. ## question on properties of circle 7

This is yet to be solved.
AB and DC are tangents to the circle.
angle $APC = 125^o$ and angle $PDC = 30^o$
Find angle PAB

2. AP intersects the curcle at E and DP intersects the circle at F.

$\frac{m(arcCF)+m(arcBE)}{2}=55$

$\frac{m(arcEC)+m(arcBF)}{2}=125$

$\frac{m(arcEC)-m(arcCF)}{2}=30$

From the second equality substract the sum of the other two.

3. not quite understand.
...what does m stand for?

4. Is the measure of the arc.

5. 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

6. Originally Posted by orange gold
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
No, BC is not the diameter in the circle, so AB is not perpendicular to BP.

7. 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 $40^o$ but i cannot quite understand Red Dog's method...

8. Originally Posted by ukorov
This is yet to be solved.
AB and DC are tangents to the circle.
angle $APC = 125^o$ and angle $PDC = 30^o$
Find angle PAB
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.

9. Originally Posted by Opalg
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.
argh...that make sense.....
but what is the formal name of the theorem for the fact "the angles between a chord and the two tangents at the ends of the chord are equal" ?

10. Originally Posted by ukorov
what is the formal name of the theorem for the fact "the angles between a chord and the two tangents at the ends of the chord are equal" ?
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.