Question

In △ABC, m∠A=72

Answer 1

Answer:

135.86≈ 136

Step-by-step explanation:

As Given in figure 1:

In ΔABC, m∠A = 72°, m∠B = 16° and c = 61 , where a, b, and c are lengths of side of ΔABC.

To find: Perimeter of triangle = ?

Sol:  In ΔABC,

m∠A + m∠B + m∠C = 180° (sum of angles of a triangle)

m∠C = 180° - (72° + 16°)

m∠C = 92°

Now Using Sine Rule:

$\frac{a}{SinA} = \frac{b}{SInB} = \frac{c}{SInC}$

$\frac{a}{Sin 72^{\circ}} = \frac{b}{Sin 16^{\circ}} = \frac{61}{SIn92^{\circ}}$

Now, $\frac{a}{sin72^{\circ}} = \frac{61}{Sin92^{\circ}}$

∴ $a = \frac{61 \times Sin 72^{\circ}}{Sin92^{\circ}} = \frac{61 \times 0.951}{0.999} = \frac{58.011}{0.999} = 58.07$

In the same way, $\frac{b}{sin 16^{\circ}} = \frac{61}{Sin92^{\circ}}$

∴ $b = \frac{61 \times Sin 16^{\circ}}{Sin92^{\circ}} = \frac{61 \times 0.275}{0.999} = \frac{16.775}{0.999} = 16.79$

Therefore, a = 58.07 ≈ 58, b = 16.79 ≈ 17 and c = 61

Now, Perimeter of ΔABC = a + b + c = 58 + 17 + 61 = 136