Question

*crying*In the given figure AB is a chord of a circle with Centre O such that AB equal to 16 cm and radius of a circle is 10 cm. Tangents at A and B intersect each other at P. Find the length of PA. :^3

Answer 1

OA= 10 cm
We know that perpendicular from centre to the chord bisects the chord. 

Thus, AM= MB= 8cm          (Since AB= 16 cm)

By Pythagoras theorem

OM= √ (10² - 8²)
OM= √ 100- 64
OM= 6 cm


Now  In triangle OAM, 
Tan ∠AOM = 8/6 = 4/3

In triangle OAP 
Tan ∠AOM = PA/OA
4/3 = PA/10

PA= 40/3 cm

Answer 2

Answer:

Let LP be x

AB ➳ AL = LB = 8 cm

OA = OB = 10 cm

Now, In ∆ OAL, where ∠ L = 90°

By Phythagoras theorem :

➳(Hypotenuse)² = (Perpendicular)² + (Base)²

➳ (OA)² = (OL)² + (AL)²

➳ 10² = (OL)² + 8²

➳ 100 = (OL)² + 64

➳ OL² = 100 - 64

➳ OL² = 36

➳ OL = √36

➳ OL = 6 cm

___________________

In ∆ LAP, where ∠ L = 90°

By Phythagoras theorem :

⇢(AP)² = (AL)² + (LP)²

⇢(AP)² = 8² + (LP)²

⇢AP² = 8² + x².........[Equation (i)]

___________________

In ∆ OAP, where ∠ A = 90°

By Phythagoras theorem :

➙ (OP)² = (OA)² + (AP)²

➙ (OL + LP)² = 10² + (AP)²

➙ AP² = -100 + (6 + x)²

➙ AP² = -100 + 36 + x² + 12x........[Equation (ii)]

Now,from equation (i) and (ii) we get,

➙ 64 + x² = -100 + 36 + x² + 12x

➙ x² - x² = -100 + 36 - 64 + 12x

➙ 12x = 128

➙ x = 128/12

➙ x = 32/3 cm

So,

➥ AP² = 64 + x²

➥ AP² = 64 + (32/3)²

➥ AP² = 576 + 1024/9

➥ AP = √1600/√9

➥ AP = 40/3 cm

Therefore, AP is 40/3 cm