Question : In $\triangle \mathrm{ABC}, \angle \mathrm{A}=5 \mathrm{x}-60^{\circ}, \angle \mathrm{B}=2 \mathrm{x}+40^{\circ}, \angle \mathrm{C}=3 \mathrm{x}-80^{\circ}$. Find $\angle \mathrm{A}$.
Option 1: $75^{\circ}$
Option 2: $90^{\circ}$
Option 3: $80^{\circ}$
Option 4: $60^{\circ}$
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Correct Answer: $80^{\circ}$
Solution :
Given: In $\triangle \mathrm{ABC}, \angle \mathrm{A}=5 \mathrm{x}-60^{\circ}, \angle \mathrm{B}=2 \mathrm{x}+40^{\circ}, \angle \mathrm{C}=3 \mathrm{x}-80^{\circ}$.
We know, the sum of the three angles of a triangle is $180^{\circ}$.
According to the question,
$5x-60^{\circ}+2x+40^{\circ}+3x-80^{\circ}=180^{\circ}$
$⇒10x=180^{\circ}+100^{\circ}$
$⇒x=\frac{280^{\circ}}{10}$
$\therefore x=28^{\circ}$
Now, $\angle \mathrm{A}$= $(5×28^{\circ})-60^{\circ} = 140^{\circ}-60^{\circ} = 80^{\circ}$
Hence, the correct answer is $80^{\circ}$.
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