82
In a $\triangle ABC$, $\angle A = x^\circ$, $\angle B = (3x-2)^\circ$ and $\angle C = y^\circ$. Also, $\angle C - \angle B = 9^\circ$. Determine the three angles of the triangle.
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Sol. $\angle A + \angle B + \angle C = 180^\circ$
$\therefore x + (3x - 2) + y = 180$
$\Rightarrow 4x + y = 182$ -----(1) (1 Mark)
Given, $\angle C - \angle B = 9^\circ$
$\therefore y - (3x - 2) = 9$
$\Rightarrow y-3x = 7$ -----(2) (1/2 Mark)
Solving (1) and (2), we get
$x = 25$ and $y = 82$ (1 Mark)
Hence, $\angle A = 25^\circ$, $\angle B = (3 \times 25 - 2)^\circ = 73^\circ$ and $\angle C = 82^\circ$ (1/2 Mark)
$\therefore x + (3x - 2) + y = 180$
$\Rightarrow 4x + y = 182$ -----(1) (1 Mark)
Given, $\angle C - \angle B = 9^\circ$
$\therefore y - (3x - 2) = 9$
$\Rightarrow y-3x = 7$ -----(2) (1/2 Mark)
Solving (1) and (2), we get
$x = 25$ and $y = 82$ (1 Mark)
Hence, $\angle A = 25^\circ$, $\angle B = (3 \times 25 - 2)^\circ = 73^\circ$ and $\angle C = 82^\circ$ (1/2 Mark)