Question

please answer.......... ​

Answer 1

Given :-

• AD is Diameter.
• ∠BCD = 150°

To find :-

• ∠BAD
• ∠ADB

We know that :-

• ABCD is Cyclic Quadrilateral.
• In a cyclic quadrilateral, the sum of a pair of opposite angles is 180°.
• If the sum of two opposite angles are supplementary, then it’s a cyclic quadrilateral.
• Angle subtanded by same arc in a semicircle is a half of the angle subtanded at the center.
• Sum of all angles of the triangle is 180°.

Construction :-

• Join BD

Solution :-

(1) ∠ABD is

As we have known above that Angle subtanded by same arc in a semicircle is a half of the angle subtanded at the center. So,

${\sf{\longmapsto{ \angle ABD = \cfrac{1}{2} \: \: ( \angle AOD)}}}$

${\sf{\longmapsto{ \angle ABD = \cfrac{1}{2} \: \: ( 180 \degree)}}}$

${\sf{\longmapsto{ \angle ABD = \cfrac{180 \degree}{2}}}}$

${{ \underline{ \boxed{{\sf{\longmapsto{ \angle ABD = 90 \degree}}}}}}}$

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(2) ∠BAD is

As also we have above known that In a cyclic quadrilateral, the sum of a pair of opposite angles is 180°.

${ \bf{\longmapsto{ \angle BAD + \angle BCD = 180 \degree}}}$

${ \bf{\longmapsto{ \angle BAD + 150 \degree = 180 \degree}}}$

${ \bf{\longmapsto{ \angle BAD = 180 \degree - 150 \degree }}}$

${ \large{ \underline{ \boxed{ \red { \bf{\longmapsto{ \angle BAD = 30 \degree }}}}}}}$

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(3) ∠ADB is

As we have known that Sum of all angles of the triangle is 180°. By construction we got the Triangle ABD in which Angle ADB is :-

${\bf{\implies{ \angle ABD + \angle BAD + \angle ADB = 180 \degree}}}$

${\bf{\implies{ 90\degree + 30\degree + \angle ADB = 180\degree }}}$

${\bf{\implies{ 120\degree + \angle ADB = 180\degree }}}$

${\bf{\implies{ \angle ADB = 180\degree - 120\degree }}}$

${ \large{ \underline{ \boxed{ \red{\bf{\implies{ \angle ADB = 60\degree }}}}}}}$

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Final answer

1. ∠BAD is 30°
2. ∠ADB is 60°

Answer 2

I hope it helps u lot .

Solution :-

(i) Join BD

Now,

ABCD is a cyclic quadrilateral

∴ ∠BAD + ∠BCD = 180° (Opposite angles of a cyclic quadrilateral)

=> ∠BAD + 150° = 180°

=> ∠BAD = 180° - 150°

= 30°

(ii) ∠ABD = 90° (Angle in a semi-circle)

Now, In △ABD,

we have ∠ABD + ∠BAD + ∠ADB = 180°

90° + 30° + ∠ADB = 180°

=> ∠ADB = 180° – 120°

= 60°

hope it helps ....