Question

Given: △ABC, m∠A=60° m∠C=45°, AB=8 Find: Perimeter of △ABC, Area of △ABC . FIRST CORRECT ANSWER GETS POINTS AND BRAINLIEST!!!! THANK YOU SO MUCH!
pls help me!!!!!!!!!!!

Answer 1

Answer:

Given : In △ABC, m∠A=60°, m∠C=45°,and AB=8 unit

Firstly, find the angles B

Sum of measures of the three angles of any triangle equal to the straight angle, and also expressed as 180 degree

∴m∠A+ m∠B+m∠C=180                      ......[1]

Substitute the values of m∠A=60° and m∠C=45° in [1]

$60^{\circ}+ m\angle B+45^{\circ}=180^{\circ}$

$105^{\circ}+ m\angle B=180^{\circ}$

Simplify:

$m\angle B=75^{\circ}$

Now, find the sides of BC

For this, we can use law of sines,

Law of sine rule is an equation relating the lengths of the sides of a triangle  to the sines of its angles.

$\frac{\sin A}{BC} = \frac{\sin C}{AB}$

Substitute the values of ∠A=60°, ∠C=45°,and AB=8 unit to find BC.

$\frac{\sin 60^{\circ}}{BC} =\frac{\sin 45^{\circ}}{8}$

then,

$BC = 8 \cdot \frac{\sin 60^{\circ}}{\sin 45^{\circ}}$

$BC=8 \cdot \frac{0.866025405}{0.707106781} =9.798$ unit

Similarly for  AC:

$\frac{\sin B}{AC} = \frac{\sin C}{AB}$

Substitute the values of ∠B=75°, ∠C=45°,and AB=8 unit to find AC.

$\frac{\sin 75^{\circ}}{AC} =\frac{\sin 45^{\circ}}{8}$

then,

$AC = 8 \cdot \frac{\sin 75^{\circ}}{\sin 45^{\circ}}$

$AC=8 \cdot \frac{0.96592582628}{0.707106781} =10.9283$ unit

To find the perimeter of triangle ABC;

Perimeter = Sum of the sides of a triangle

i,e

Perimeter of △ABC = AB+BC+AC = 8 +9.798+10.9283 = 28.726 unit.

To find the area(A) of triangle ABC ;

Use the formula:

$A = \frac{1}{2} \times AB \times AC \times \sin A$

Substitute the values in above formula to get area;

$A=\frac{1}{2} \times 8 \times 10.9283 \times \sin 60^{\circ}$

$A = 4 \times 10.9283 \times 0.86602540378$

Simplify:

Area of triangle ABC = 37.856 (approx) square unit