Question
✓
Answer
- 60º
Solution:
Let $ \angle\text{A}=(3\text{x})^\circ,\ \angle\text{B}=(2\text{x})^\circ$ and $\angle\text{C}=\text{x}^\circ$
Then,
3x + 2x + x = 180º [Sum of the angles of a triangle]
⇒ 6x = 180º
⇒ x = 30º
Hence, the angles are
$\angle\text{A}=3×30^\circ=90^\circ,\ \angle\text{B}=2×30^\circ=60^\circ$ and $\angle\text{C}=30^\circ$
Side BC of triangle ABC is produced to E.
$\therefore\ \angle\text{ACE}=\angle\text{A}+\angle\text{B}$
$⇒\angle\text{ACD}+\angle\text{ECD}=90^\circ+60^\circ$
$⇒90^\circ+\angle\text{ECD}=150^\circ$
$⇒\angle\text{ECD}=60^\circ$
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