Question

Q13 and Q14 plz give this answers to me

Answer 1

ques -13
Given
OP= 2r
∠OTP = 90° (radius drawn at the point of contact is perpendicular to the tangent)
Now ,
In ΔOTP,
Sin∠ OPT = OT/OP = 1/2 = Sin 30o
⇒ ∠ OPT = 30o
therefore,
∠ TOP=60o
∴ ΔOTP is a 30o-60o-90o, right triangle.

In ΔOTS,
OT = OS … (Radii of the same circle)
therefore,
ΔOTS is an isosceles triangle.
∴∠OTS = ∠OST … (Angles opposite to equal sides of an isosceles triangle are equal)

In ΔOTQ and ΔOSQ

OS = OT … (Radii of the same circle)
OQ = OQ ...(side common to both triangles)
∠OTQ = ∠OSQ … (angles opposite to equal sides of an isosceles triangle are
equal)

∴ ΔOTQ ≅ ΔOSQ … (By S.A.S)
∴ ∠TOQ = ∠SOQ = 60° … (C.A.C.T)
∴ ∠TOS = 120° … (∠TOS = ∠TOQ + ∠SOQ = 60° + 60° = 120°)
∴ ∠OTS + ∠OST = 180° – 120° = 60°
∴ ∠OTS = ∠OST = 60° ÷ 2 = 30°

que - 14

Angle ABQ=1/2 angle AOQ
=1/2*58
=29
Angle A =90(AT is a tangent)
angle BAT+angle ABT +angle ATQ=180(angle sum
property of triangle.)
90+29+angle ATQ= 180
angle ATQ=180-119
angle ATQ=61

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Answer 2

the attachment is required answer