CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If AOQ = 58º, find ATQ.                        [CBSE 2015]


Solution

It is given that AOQ = 58º.

We know that the angle subtended by an arc at the centre is twice the angle subtended by it any point on the remaining part of the circle.

ABQ=12AOQ=12×58°=29°

Now, AT is the tangent and OA is the radius of the circle through the point of contact A.

OAT=90°    (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

In ∆ABT,

BAT+ABT+ATB=180°       (Angle sum property)

90°+29°+ATB=180°              BAT=OAT and ABT=ABQATB=180°-119°=61° ATQ=ATB=61°

Mathematics
RD Sharma (2020, 2021)
All

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image