Question

ABCD is a cyclic quadrilateral. AB is the diameter of the circle, AD = CD and angle ADC = 130° (a). What is the measure of angle ACB? (b) Find angle DCB (c) What is the measure of angle BAD? anser it quickly please ​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given: In figure ABCD is a cyclic quadrilateral. AB is the diameter of the circle, AD = CD and angle ADC = 130°

To find:

(a) What is the measure of angle ACB?

(b) Find angle DCB

(c) What is the measure of angle BAD?

Solution:

(a) What is the measure of angle ACB?

Ans: Diameter of a circle subtends right angle at segment of the circle.

Here,

AB is diameter, it subtends $\angle ACB$ on segment/semicircle.

Thus,

$\bf \green{\angle ACB=90^{\circ}}$

(b) Find angle DCB.

Ans: To find $\angle DCB$, first find the $\angle DCA$

Because, ∆ADC is an isosceles triangle as it is given that AD=CD.

Thus, angles opposite to equal sides are equal.

Therefore, $\angle DCA= \angle CAD$.

As total sum of interior angles of triangle is equal to 180°.

Apply this property;

$\angle DCA+ \angle CAD+130^{\circ}=180^{\circ}$

$2\angle DCA=180^{\circ}-130^{\circ}$

$2\angle DCA=50^{\circ}$

$\angle DCA=25^{\circ}$

As, it is clear that

$\angle DCB= \angle DCA+\angle ACB$

$\angle DCB= 25^{\circ}+90^{\circ}$

$\bf \orange{\angle DCB=115^{\circ}}$

(c) What is the measure of angle BAD?

Ans: According to the property of cyclic quadrilateral; i.e. sum of opposite angles are 180°.

So,

$\angle DCB+ \angle BAD=180^{\circ}$

put the known value

$115^{\circ}+ \angle BAD=180^{\circ}$

$\angle BAD=180^{\circ}-115^{\circ}$

$\bf \blue{\angle BAD=65^{\circ}}$

Final answer:

a) $\bf \angle ACB=90^{\circ}$

b) $\bf\angle DCB=115^{\circ}$

c) $\bf \angle BAD=65^{\circ}$

Hope it will help you.

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