# ans pls Questions. Dear Student,
We have to find in each figure.

(i) It is given that   As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Now ,$x=\frac{1}{2}\angle COB=22{\frac{1}{2}}^{°}$

Hence (ii) As we know that = x                 [Angles in the same segment]

line is diameter passing through centre,

So,  (iii) It is given that   So And  Then Hence (iv)  (Linear pair) 12236919 And
x =  Hence, (v) It is given that  is an isosceles triangle. Therefore And,

$\angle AOB=2\left(\mathrm{Reflex}\angle ACB\right)\phantom{\rule{0ex}{0ex}}$ Hence, (vi) It is given that  And

$\angle COA+\angle AOB=180°\phantom{\rule{0ex}{0ex}}⇒\angle COA=180°-60°\phantom{\rule{0ex}{0ex}}⇒\angle COA=120°$

So Hence, 12236919

(vii) (Angle in the same segment) In we have Hence (viii) As (Radius of circle)

Therefore, is an isosceles triangle. So (Vertically opposite angles)

Hence, (ix) It is given that   …… (1)          (Angle in the same segment)

$\angle ADB=\angle ACB=32°$  ......(2)           (Angle in the same segment)

Because and are on the same segment of the circle.

Now from equation (1) and (2) we have Hence, (x) It is given that  $\angle BAC=\angle BDC=35°$                (Angle in the same segment)

Now in $∆BDC$ we have

$\angle BDC+\angle DCB+\angle CBD=180°\phantom{\rule{0ex}{0ex}}⇒35°+65°+\angle CBD=180°\phantom{\rule{0ex}{0ex}}⇒\angle CBD=180°-100°=80°$

Hence, (xi)  (Angle in the same segment)

In we have Hence (xii)  (Angle in the same segment) is an isosceles triangle

So, (Radius of the same circle)

Then Hence Regards

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