Question

plz...give me fast ans

Answer 1

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Hey Mate !!

Here's your answer !!

Given: OT = 7 cm, OP = 25 cm

Consider Δ POT

= OT² + PT² = OP²

= OP² - OT² = PT²

= PT² = 25² - 7²

= PT² = 625 - 49

= PT² = 576

= PT = √576 = 24 cm

Since Tangents drawn from an external point are equal,

= PT = PT¹ = 24 cm

Consider Δ TOP

= ∠ TOP = 90°   ( Since Tangents are perpendicular to the Radius )

= ∠ TPO = 30°

Therefore by Angle sum property of Δ TOP we get,

= ∠ TOP + ∠ TPO + ∠ POT = 180°

= 90° + 30° + ∠ POT = 180°

= 120° + ∠ POT = 180°

= ∠ POT = 180° - 120°

= ∠ POT = 60°

Hope this helps !!

Cheers !!
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# Kalpesh :)
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Answer 2

Hey !!

HERE is your solution....,,

Given :-
OT = OT' = 7cm. .......( radius )
OP = 25cm
angle OPT = 30°
PT = PT' = ?
angle POT = ?

As we know, Tangent of circle form right angle with radius.
so, angle OTP = 90°
Hence ∆OPT is a right angled triangle.

OP^2 = OT^2+TP^2
TP^2 = OP^2 - OT^2
= (25)^2 - (7)^2
= 625 - 49
= 567
TP = √576
TP = 24
PT = PT' = 24 cm
PT'= 24cm

Angle OPT' = 30
Angle OPT' = Angle OPT = 30
Angle TPT' = Angle OPT + Angle OPT'
= 30 + 30
= 60
Angle TPT' + Angle TOT' = 180
60 + Angle TOT' = 180
Angle TOT' = 180 - 60
Angle TOT' = 120°

Angle TOP = Angle T'OP
Angle TOP + Angle T'OP = Angle TOT'
2Angle TOP = 120
Angle TOP = 120/2
Angle TOP = 60°


HOPE IT HELPS YOU....

THANKS ^-^