Answer:
Perimeter = 32.3
Area = 95.24
Step-by-step explanation:
Given:
△ABC, m∠A=60°
m∠C=45°, AB = 9
To find:
Perimeter of △ABC
Area of △ABC
Solution:
Using angle sum property in a triangle:
m∠A + m∠B + m∠C = 180°
m∠B = 180° - 45° - 60° = 75°
As per Sine Rule:
[tex]\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC}[/tex]
Where
[tex]a[/tex] is the side opposite to [tex]\angle A[/tex]
[tex]b[/tex] is the side opposite to [tex]\angle B[/tex]
[tex]c[/tex] is the side opposite to [tex]\angle C[/tex]
[tex]\dfrac{BC}{sin60^\circ} = \dfrac{AC}{sin 75^\circ} = \dfrac{AB}{sin 45^\circ} \\\Rightarrow \dfrac{BC}{\frac{\sqrt3}{2}} = \dfrac{9}{\frac{1}{\sqrt2}} \\\Rightarrow BC = 12.73\times 0.87 \\\Rightarrow BC = 11.08[/tex]
[tex]\Rightarrow \dfrac{AC}{0.96} = \dfrac{9}{\frac{1}{\sqrt2}} \\\Rightarrow AC = 12.73\times 0.96 \\\Rightarrow AC = 12.22[/tex]
Perimeter of △ABC = AB + BC + AC = 9 + 11.08 + 12.22 = 32.3
Area of a triangle is given as:
[tex]\dfrac{1}2\times ab sin(angle\ between\ a\ and\ b)[/tex]
[tex]\Rightarrow \dfrac{1}{2}\times AB\times AC\times sinA\\\Rightarrow 9\times 12.22\times sin 60\\\Rightarrow \bold{95.24}[/tex]
