Question

ABCD is a quadrilateral inscribed in a circle with center O, $\mathrm{\angle }ADC=130{}^{\circ }$ and $AD=DC$. Calculate:

A) reflex $\mathrm{\angle }AOC$
B) $\mathrm{\angle }ABC$
C) $\mathrm{\angle }AOD$

Answer 1

Hint: In part (i), use the theorem, the angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle’s circumference to find the value of the reflex $\mathrm{\angle }AOC$. In part (ii), use the theorem, the sum of opposite angles of a cyclic quadrilateral is $180{}^{\circ }$ to find the value of $$\mathrm{\angle }ABC$$. In part (iii), draw the line AD. After that, apply the kite property, the two angles are equal where the unequal sides meet which will give the value of $\mathrm{\angle }OAD$. After that apply the isosceles triangle property to find the value of $\mathrm{\angle }ODA$. Then apply the sum rule of the triangle to find the value of $\mathrm{\angle }AOD$.

Complete step-by-step answer:
Given: - $\mathrm{\angle }ADC=130{}^{\circ }$ and $AD=DC$
A) As we know that, the angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle’s circumference.
Then for the arc AC,
reflex $\mathrm{\angle }AOC=2\mathrm{\angle }ADC$
Substitute the value of $\mathrm{\angle }ADC$ in the above equation,
reflex $\mathrm{\angle }AOC=2\times 130{}^{\circ }$
Multiply the term on the right side,
reflex $\mathrm{\angle }AOC=260{}^{\circ }$
Hence, the value of reflex $\mathrm{\angle }AOC$ is $260{}^{\circ }$.

B) As we know that the sum of the opposite angles of a cyclic quadrilateral is $180{}^{\circ }$.
Then,
$\mathrm{\angle }ABC+\mathrm{\angle }ADC=180{}^{\circ }$
Substitute the value of $\mathrm{\angle }ADC$ in the above equation,
$130{}^{\circ }+\mathrm{\angle }ABC=180{}^{\circ }$
Move $130{}^{\circ }$ on the other side and subtract from $180{}^{\circ }$.
$\mathrm{\angle }ABC=50{}^{\circ }$
Hence, the value of $\mathrm{\angle }ABC$ is $50{}^{\circ }$.

C) Draw a line from O to D.

In quadrilateral AOCD,
$AD=DC$ (given)
$OA=OC$ (radius)
Then by kite property,
$\mathrm{\angle }OAD=\mathrm{\angle }OCD$
Also, the sum of the angles of a quadrilateral is equal to $360{}^{\circ }$.
$\mathrm{\angle }OAD+\mathrm{\angle }AOD+\mathrm{\angle }OCD+\mathrm{\angle }ACD=360{}^{\circ }$
Substitute the values,
$\mathrm{\angle }OAD+100{}^{\circ }+\mathrm{\angle }OAD+130{}^{\circ }=360{}^{\circ }$
Add the like terms,
$2\mathrm{\angle }OAD+230{}^{\circ }=360{}^{\circ }$
Move $230{}^{\circ }$ to the other side and subtract from $360{}^{\circ }$.
$2\mathrm{\angle }OAD=130{}^{\circ }$
Divide both sides by 2,
$\mathrm{\angle }OAD=65{}^{\circ }$
Now in triangle OAD,
$OA=OD$ (radius)
As we know that the angles opposite to the equal sides of the triangles are equal.
$\mathrm{\angle }OAD=\mathrm{\angle }ODA$
Substitute the value of $\mathrm{\angle }OAD$,
$\mathrm{\angle }ODA=65{}^{\circ }$.
As we know that the sum of angles of a triangle is equal to $180{}^{\circ }$,
$\mathrm{\angle }OAD+\mathrm{\angle }ODA+\mathrm{\angle }AOD=180{}^{\circ }$
Substitute the values,
$65{}^{\circ }+65{}^{\circ }+\mathrm{\angle }AOD=180{}^{\circ }$
Add the terms on the left side,
$$130{}^{\circ }+\mathrm{\angle }AOD=180{}^{\circ }$$
Move $130{}^{\circ }$ to the other side and subtract from $180{}^{\circ }$,
$\mathrm{\angle }AOD=50{}^{\circ }$
Hence, the value of $\mathrm{\angle }AOD$ is $50{}^{\circ }$.

Note: Part B can be done in another way.
Step by step answer: -
Given: - $\mathrm{\angle }ADC=130{}^{\circ }$ and $AD=DC$
As we know that the sum of an angle and its reflex is $360{}^{\circ }$.
Then,
reflex $\mathrm{\angle }AOC+\mathrm{\angle }AOC=360{}^{\circ }$
Substitute the value of reflex $\mathrm{\angle }AOC$ in the above equation,
$260{}^{\circ }+\mathrm{\angle }AOC=360{}^{\circ }$
Move $260{}^{\circ }$ on the other side and subtract from $360{}^{\circ }$.
$\mathrm{\angle }AOC=100{}^{\circ }$ ….. (1)
As we know that, the angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle’s circumference.
Then for the arc AC,
$\mathrm{\angle }AOC=2\mathrm{\angle }ABC$
Substitute the value of $\mathrm{\angle }AOC$ from the equation (1),
$100{}^{\circ }=2\mathrm{\angle }ABC$
Divide both sides by 2,
$\mathrm{\angle }ABC=50{}^{\circ }$
Hence, the value of $\mathrm{\angle }ABC$ is $50{}^{\circ }$.