Question 79
In $$\triangle ABC, D$$ and $$E$$ are the points on $$AB$$ and $$AC$$ respectively such that $$AD \times AC = AB \times AE.$$ If $$\angle ADE = \angle ACB + 30^\circ$$ and $$\angle ABC = 78^\circ$$, then $$\angle A = ?$$
Solution
$$AD \times AC = AB \times AE$$
$$ \frac{AD}{AE} = \frac{AB}{AC}$$
$$\triangle ABC\text{ is similar to} \triangle$$ ADE so,
$$\angle ADE = \angle ABC$$
$$\angle ADE = 78\degree$$
$$\angle AED = \angle ACB$$
$$\angle ADE = \angle ACB + 30^\circ$$
$$\angle ACB = 78 -30 = 48 $$
In $$\triangle$$ ABC -
$$\angle$$ ABC + $$\angle$$ ACB + $$\angle$$ A = 180$$\degree$$
$$\angle A = 180 - 78 - 48 = 54\degree$$
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