Question

AB is a diameter of a circle with centre O and AT is a tangent. If angle AOQ=58 find angle ATQ.

Answer 1

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

Answer 2

Answer:

61

Step-by-step explanation:

In ΔBOQ,

∠B + ∠Q = ∠AOQ                                            [Exterior Angle Property of Δs]

But, OB = OQ

∴ ∠OBQ = ∠OQB

So, 2∠OQB = 58°

→ ∠OQB = 29°

Now, ∠AOQ + ∠BOQ = 180°                                  [Linear Pair]

∴ ∠BOQ = 180 - 58 = 122°

∠OQT is exterior to ΔOQB and is equal to ∠OBQ + ∠BOQ = 122 + 59 = 151°

In quadrilateral AOQT,

∠OAT + ∠ATQ + ∠OQT + ∠AOQ = 360°

90° + ∠ATQ + 151° + 58° = 360°          

[∠OAT=90° since it is a line drawn from the center of the circle to tangent]

∴∠ATQ = 61°

Hope this helps!