In ΔGHI, g = 150 inches, h = 590 inches and i=640 inches. Find the area of ΔGHI to the nearest 10th of an square inch.

Answer 1

Using the Law of Cosines, i² = g² + h² - 2ghcosI

640² = 150² + 590² - 2(150)(590)cosI

39000 = -177000cosI

cosI = -0.220339

So, ∠I = Cos^-1(-0.220338983) = 102.7289439°

Sides g and h are the sides of ∠I.

So, Area = (1/2)ghsinI = (1/2)(150)(590)sin102.7289439º = 43,162.5 in²

Answer 2

find an angle between 2 of the sides

use the law of cosines, generalized version of the Pythagorean Theorem for non-right triangles

c^2 = a^2+b^2 -2abCosC, C is opposite side c and between a and b

640^2 =150^2 +590^2 - 2(150)(590)CosC

CosC = (640^2 -150^2 -590^2)/(-2(150)(590)=-.2203389

C =arccos(-220339)=about 102.73 degrees

sinC =sin102.73 = .97542

Area = .5abSinC = .5(150)(590)(.97542)= about 43,162.5 square inches rounded to 1 decimal

Answer 3

Area of a triangle formula when all sides are given to you:

A=sqrt(s(s-g)(s-h)(s-i)) where s is semi perimeter, g, h and i are the given sides.

s=(g+h+i)/2, you should get 690 inches

Now plug everything in the above area formula for A

You should get 9000*sqrt(23) approximately 43,162.5 sq inches

Answer 4

See the videa.