In a circle with centre O, AB is a chord of length 10 cm. Tangents at points A and B intersect outside the circle at P. If OP = 2 OA, then find the length (in cm) of AP.

10

Let the radius of the circle be R cm.

So, OA = OB = R

then, OP = 2 x OA = 2R

Tangent segments AP, BP and radii OA, OB from a kite

So, OP bisects the chord AB.

Thn, AQ = BQ = 5 cm

Now, $\Delta$OAP & $\Delta$OBP are two right angled triangles.

From $\Delta$OAP,

cos $\angle$AOP = $\frac{OA}{OP}$

= cos $\angle$AOP = $\frac{R}{2R}$

= cos $\angle$AOP = $\frac{1}{2}$

= $\angle$AOP = cos 60

So, $\angle$OPA = 180 - (90 + 60) = 30 

Also, $\Delta$AQO & $\Delta$BQO are two right-angled triangles.

So, $\angle$OAQ = 180 - 90 - 60 = 30

From $\Delta$AQO,

cos $\angle$QAO = $\frac{AQ}{OA}$

= cos 30 = $\frac{5}{OA}$

= $\sqrt {3 }$/2 = $\frac{5}{OA}$

= OA = 10$\sqrt {3 }$

So, AP = (10$\sqrt {3 }$) x $\sqrt {3 }$ = 10 cm

Therefore, AP is 10 cm.