An isosceles triangle ABC has AC = BC. CD bisects AB at D and $∠$ CAB = 55^o. Find $∠ A C B$ .
70

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Solution

The correct option is A 70
Angles opposite to equal sides are also equal.
$∠ C A B = ∠ C B A = 55^{\circ}$
$∠ C A B + ∠ C B A + ∠ A C B = 180$
$2 \times 55^{\circ} + ∠ A C B = 180$
$∠ A C B = (180 - 110)^{\circ}$
$∠ A C B = 70^{\circ}$

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An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠ CAB = 55o. Find ∠ ACB.70

The correct option is A 70Angles opposite to equal sides are also equal. ∠CAB=∠CBA=55∘ ∠CAB+∠CBA+∠ACB=180 2×55∘+∠ACB=180 ∠ACB=(180−110)∘ ∠ACB=70∘

An isosceles triangle ABC has AC = BC. CD bisects AB at D and $∠$ CAB = 55^o. Find $∠ A C B$ .
70

The correct option is A 70
Angles opposite to equal sides are also equal.
$∠ C A B = ∠ C B A = 55^{\circ}$
$∠ C A B + ∠ C B A + ∠ A C B = 180$
$2 \times 55^{\circ} + ∠ A C B = 180$
$∠ A C B = (180 - 110)^{\circ}$
$∠ A C B = 70^{\circ}$