In a circle with centre O, AB is a chord of length 10 cm. Tangents at…

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.