*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
Answers
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:
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
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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