Question

In the adjoining figure, OC = OD.

Find :

1. ∠CBD
2. ∠CAD
3. ∠BCA

Answer 1

We are given the figure:

∠ABC = 25°

∠DAO = 90°

∠ADO = 30°

OC = OD

⇒ ΔCOD is isosceles

O lies on the diagonal CD

We are to find:

1. ∠CBD
2. ∠CAD
3. ∠BCA

i. Find ∠CBD

From the figure:

∠ABC = 25°

Triangle ΔCBD include angle ∠CBD, which is adjacent to ∠ABC

Since ∠ABC = 25° and triangle ΔCBD is isosceles as OC = OD and O lies on the diagonal, the base angles ∠CBD = ∠CDB

Let’s calculate ∠C + ∠D in triangle CBD:

∠CBD = 50°

ii. Find ∠CAD

In triangle ΔAOD:

∠DAO = 90°

∠ADO = 30°

So, ∠CAD

= 180° – (90° + 30°)

= 60°

But we must include the full angle at A across triangle CAD, not just in triangle AOD. 

From symmetry and the figure kite-shaped, angle ∠CAB on left is same as angle ∠DAB and they are supplementary. 

So, ∠CAD

= 60° + 60° 

= 120°

iii. Find ∠BCA

In triangle ΔABC, use angle sum:

∠ABC = 25°

∠CAB = 60°

So, ∠BCA

= 180° – (25° + 60°) 

= 95°