Question

In Fig. 16.196, if ∠AOB = 80° and ∠ABC=30°, then find ∠CAO.

Answer 1

The value of ∠CAO =60°

Step-by-step explanation:

• Given data

        ∠AOB = 80°

        ∠ABC=30°

• Let

       ∠CAO = x

• Consider triangle ΔOAB , It is isosceles triangle because its side are equal.

        OA =OB       (radius of circle)

        ∠OAB =∠OBA      (opposite angle are equal to equal side)

• Use triangle angle property in ΔOAB

        ∠OAB +∠OBA +∠AOB = 180°

        ∠OAB +∠OAB + 80° = 180°

        So

        ∠OAB = 50°        ...1)

• Where

        $\mathbf{\angle ACB =\frac{\angle AOB}{2}}$    (because both angle made on same segment)

        $\mathbf{\angle ACB =\frac{ 80}{2}}$

        ∠ACB = 40°        ...2)

• Now consider triangle ΔABC and use triangle property

       ∠ACB +∠CBA +∠BAC = 180°

       It can also be written as

       ∠ACB +∠CBA +∠OAB +∠CAO= 180°

       40° +30° + 50° + x = 180°

       On solving, we get

      x = 60° = ∠CAO        Answer

Answer 2

Given :  ∠AOB = 80° and ∠ABC = 30°.

 

To find : The value of  ∠CAO.

 

Solution :

In ∆ AOB,  

OA = OB (radius)

∠OAB  =  ∠OBA

[Angles opposite to equal sides of a triangle are equal]

∠OAB  +   ∠OBA + ∠AOB = 180°

[Sum of the angles of a triangle is 180° ]

∠OAB + ∠OAB + ∠AOB = 180°

2 ∠OAB  + 80° = 180°

2 ∠OAB   = 180° - 80°  

2∠OAB = 100°

∠OAB = 100°/2

∠OAB = 50°

∴ ∠OAB = ∠OBA = 50°

Since, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∠AOB = 2 ∠BCA

80° = 2 ∠BCA

∠BCA = 80°/2

∠BCA = 40°

Now,

In ∆ABC,  

∠A + ∠B + ∠C = 180°

[Sum of the angles of a triangle is 180° ]

∠A + 30° + 40° = 180°

∠A + 70°  = 180°

∠A  = 180° - 70°

∠A = 110°

∠BAC = 110°

∠BAC =   ∠CAO + ∠OAB

110° =   ∠CAO + 50°

 ∠CAO = 110° - 50°

 ∠CAO  = 60°

Hence the value of  ∠CAO is 60°.

HOPE THIS ANSWER WILL HELP YOU…..

 

Some questions of this chapter :

In Fig. 16.198, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD:∠ABE

https://brainly.in/question/15910499

 

In Fig. 16.199, AB is a diameter of the circle such that ∠A=35° and ∠Q=25°, find ∠PBR.

https://brainly.in/question/15910508