Question

Can anybody please help with this? Pretty Urgent!!!

Answer 1

In AOAB, ∠OAB + ∠ABO + ∠BOA = 180°

∠OAB + ∠OAB + 90° = 180° ⇒ 2∠OAB = 180°- 90°

[angles opposite to equal sides are equal] [angle sum property of a triangle] [from Eq. (i)]

⇒ ∠OAB = 90°/2 = 45° …(i)

In ΔACB, ∠ACB + ∠CBA + ∠CAB = 180°

∴ 45°+ 30°+ ∠CAB = 180°

⇒ ∠CAB = 180° – 75° = 105°

∠CAO+ ∠OAB = 105°

∠CAO + 45° = 105°

∠CAO = 105° – 45° = 60°

I'm really sorry for the wrong answer in the comment section!!

Answer 2

Answer:

Step-by-step explanation:

In AOAB, ∠OAB + ∠ABO + ∠BOA = 180°

∠OAB + ∠OAB + 90° = 180° ⇒ 2∠OAB = 180°- 90°

[angles opposite to equal sides are equal] [angle sum property of a triangle] [from Eq. (i)]

⇒ ∠OAB = 90°/2 = 45° …(i)

In ΔACB, ∠ACB + ∠CBA + ∠CAB = 180°

∴ 45°+ 30°+ ∠CAB = 180°

⇒ ∠CAB = 180° – 75° = 105°

∠CAO+ ∠OAB = 105°

∠CAO + 45° = 105°

∠CAO = 105° – 45° = 60°