Question

In Fig. 16.206, O is the centre of the circle and ∠BDC=42°. The measure of ∠ACB is
A. 42°
B. 48°
C. 58°
D. 52°

Answer 1

Given :  O is the centre of the circle and ∠BDC = 42°.

To Find : The measure of ∠ACB .

 

Proof :

∠BDC = 42°

∠ABC = 90°  

[Angle in a semi-circle is 90°]

 

Since angles in the same segment of a circle are equal :

∴ ∠BAC = ∠BDC = 42°  

In ∆ABC,

Since Sum of the angles of a triangle is 180° :  

∴ ∠BAC + ∠ABC + ∠ACB = 180°

90° + 42° + ∠ACB = 180°

132° + ∠ACB = 180°

∠ACB = 180° - 132°

∠ACB = 48°

Hence, the measure of ∠ACB is  48°.

Option (B) 48°  is correct.

HOPE THIS ANSWER WILL HELP YOU…..

 

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Answer 2

Given:

/_BDC = 42°

To find:

/_ACB = ?

Solution:

/_BDC = /_CAB = 42° [Angle made on the circle by the same chord]

/_ABC = 90° [Angle on the semicircle]

By Angle Sum Property of a triangle:

/_ACB = 180 - /_ABC - /_CAB

/_ACB = 180 - 90 - 42

/_ACB = 48°

Option (B) 48° is right.