in the figure, m angle AOB = 90° and m angle ABC = 30°, then find the measure of angle CAO
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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°.
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