Question

Find all the missing parts to the triangle below

Answer 1

Use Law of Cosines g^2 = f^2 + h^2 -2fhCosG f^2 = g^2 + h^2 -2ghCosF h^2 = f^2 + g^2 -2fgCosH

f^2 = 28^2 + 15^2 -2*28*15Cos87 28^2 = 31^2 + 15^2 -2*31*15CosG
f^2 = 784 + 225 - 43.96 784 = 961+225 - 930CosG
f^2 = 965.0378 784 - 1186 = -930CosG
f = 31 -402 = -930CosG Divide by -930
.432258 = CosG
Cos^-1(.432258) = G
G = 64 degrees

Angle H = 180 - 64 - 87 = 29 degrees

Side f = 31 Angle F = 87 degrees
Side g = 28 Angle G = 64 degrees
Side h = 15 Angle H =29 degrees

Answer 2

We have:

• f=31 units

• m∠G=64°

• m∠H=29°

Explanation:

We are given angle F as:

m∠GF=87°

Now, g=28 and h=15 we are asked to find f.

Using the law of cosines we have:

`f^2=g^2+h^2-2gh\cos F\\\\f^2=(28)^2+(15)^2-2* 28* 15\cos (87)\\\\\\f^2=965.037796\\\\f=31.065`

which is approximately equal to 31 units

Hence, f=31 units

Also,

`g^2=f^2+h^2-2fh\cos G\\\\\\28^2=15^2+31^2-2* 15* 31* \cos G\\\\\\\cos G=0.4322\\\\G=\arccos 0.4322\\\\\\G=64.392`

Hence, to the nearest degree we get:

m∠G=64°

Also, we know that the sum of all the angles of a triangle is 180°

Hence,

m∠F+m∠G+m∠H=180°

i.e.

87°+64+m∠H=180°

m∠H=29°