Determine the remaining sides and angles of the triangle ABC

Determine the remaining sides and angles of the triangle ABC. A = 109.49 degree, C = 38.11 degree, c = 120 B = degree a (Do not round until the final answer. Then round to the nearest tenth as needed.) b (Do not round until the final answer. Then round to the nearest tenth as needed.) Use an appropriate area formula to find the area of the triangle with the given side lengths. a = 15 m b = 8 m c = 22 m What is the area of the triangle? m^2 (Round your answer to the nearest tenth.)

Solution

A=109.49 degree, C= 38.11 degree c=120

using sine law

Sin C/c=Sin A/a

sin 38.11/120=sin109.49/a

a= 120 sin 109.49/sin 38.11= 183.3

B=180-(A+C)=32.4 degree

SinB/b=SinC/c

sin 32.4/b= sin38.11/120

b=120 sin 32.4/sin 38.11=104.2

Area of the triangle

A= sqrt(s(s-a)(s-b)(s-c))

s=(a+b+c)/2

In the given question a= 15,b=8,c=22

s= (15+8+22)/2= 22.5

A=sqrt(22.5(22.5-15)(22.5-8)(22.5-22)) = sqrt(22.5*7.5*14.5*.5) = 35 m²