Question

AB is a diameter of circle with center O and AT is tangent. Of angle AOB is 58° find angle ATQ. Plsss answer quickly

Answer 1

GIVEN:

∠AOQ= 58°

∠ABQ= ½ ∠AOQ

∠ABQ= ½(58°)

∠ABQ= 29°

∠BAT= 90°
(The tangent at any point of a circle is perpendicular to the radius through the point of contact)

∠BAT+ ∠ABQ (∠ABT) +∠ ATQ = 180°

∠ATQ =∠BAT -∠ABQ

∠ATQ= 90° - 29° = 61°

Hence, ∠ATQ= 61°

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Hope this will help you....

Answer 2

Answer:

Step-by-step explanation:

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