Question

Please help!
Given: △ABC, m∠A=60°
m∠C=45°, AB=8
Find: Perimeter of △ABC,
Area of △ABC

Answer 1

  Given

△ABC, m∠A=60°

m∠C=45°, AB=8

Find out the  Perimeter of △ABC,  and Area of △ABC

To proof

As given in the question

In △ABC

m∠A=60°

,m∠C=45°,

AB = 8unit

∠A +∠B +∠C = 180°

( By using angle sum property of a triangle.)

put the value of angle in the equation

we get

60 + 45 + B = 180

B = 180 - 45 -60

B = 75°

By using the sine law

$\frac{sin \alpha} {a} = \frac{sin\beta}{b}$

Where α,β are the angles

a,b represented the side opposite to the angle  α,β .

Now by using the diagram as given below

Thus

As  AB = 8 cm

$\frac{sin60^{\circ}}{BC}=\frac{sin45^{\circ}}{AB}$

now

$sin45^{\degree}= \frac{1}{\sqrt{2}} , \ sin60^{\degree}= \frac{\sqrt{3}}{{2}}$

put all the value in the above equation

we get

 $\frac{\sqrt{3}}{2BC} = \frac{1}{8\sqrt{2}}$

using the value

√3 = 1.732

√2 = 1.414

put in the above equation

we get

BC = 1.732× 1.414× 4

BC = 9.79 unit

$\frac{sin75^{\circ}}{AC}=\frac{sin45^{\circ}}{AB}$

Putting the value

sin75° = 0.96 and sin 45° = 0.71

put all the value in the above equation

we get

$\frac{0.96}{AC} =\frac{0.71}{8}$

$AC = \frac{ 0.96 \times8}{0.71}$

AC = 10.82(approx) unit

Perimeter of a triangle is the sum of length of all the three sides.

Perimeter of a triangle  = AB + AC + BC

                                       =  8 + 10.82 + 9.79

                                       = 28.61 unit

Now find the area of a triangle

Formula

$Area\ of\ a\ triangle = \frac{1}{2} AB.AC sin\theta$

AC = 10.82(approx) unit , AB = 8 unit , sin60° = 0.86

put in the above equation

$= \frac{1}{2} (10.82)(8)(0.86)$

= 37.22 unit ( approx)

Hence proved