Question

in the given figure angle AOB=90 and angle ABC=30 find angleCAO​

Answer 1

Answer:

In ∆OAB, ∠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°

Answer 2

Answer:

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°