A triangular parcel of land has 110 meters of frontage, and the other boundaries have lengths of 79 meters and 92 meters. What angles does the frontage make with the two other boundaries?

Question

(Round your answers to one decimal place.)

Answer 1

Use the law of cosines:

1. cos α = [b² + c² – a²]/2bc
2. cos β = [a² + c² – b²]/2ac
3. cos γ = [b² + a² – c²]/2ab

where a, b and c are the lengths of the sides of a given triangle; and α, β, and γ are the angles opposite to each give side.

In our case, we are given the three sides, and asked to find the angles at the ends of the 110-meter side.

Plugging the values for the sides in, we conclude that:

The angle opposite the 92-meter side imeasures approximately 55.37 degrees.

The angle opposite the 79-meter side measures approximately 44.96 degrees.

To check, let's find the approximate measure of the 110-meter side. It turns out to be 79.68 degrees.

Adding these three angle measurements, we obtain 180.01 degrees, which is very close to 180 degrees.

Answer 2

Law of Cosines

c^2 = a^2 +b^2 -2abCosC where C is the angle opposite side c

92^2 = 79^2 + 110^2 -2(79)(110)CosC

CosC = (92^2-79^2-110^2)/(-2(79)(110) =about .5683

C= arcos.5683 = 55.368 degrees= 55^o 22' 6" = ,966 radians

rounding to one decimal place for radians would be

C=1.0 rad or C = .3 pi, A= 0.8 rad or A= .2 pi

rounded with degrees

C=55.4, A= 45.0

unlikely but the problem could even have meant

rounded to one decimal of the minutes or seconds

then use either the Law of Sines or Law of Cosines again

SinA/a = SinC/c

SinA = aSinC/c

A = arcsin(aSinC/c)

or

CosA = (79^2-110^2-92^2)/(-2(110)(92)) =about .708

A = arccos.708= 44.955 degrees= 44^o 57' 19" = .785 radians

B = 180-A-C = about 180-45-55 = 80 degrees

largest angle is opposite the longest side

b>c>a, so B>C>A

it your answers don't come out that way you made a mistake somewhere

or use an online triangle calculator for SSS given the 3 Sides, it calculates the 3 angles for you

at least use it to check your answers

Answer 3

Let θ be the angle formed by the sides of length 110 and 92.

Using the Law of Cosines, we have 79² = 110² + 92² - 2(92)(110)cosθ

20240cosθ = 14323

cosθ = 0.7076581

θ = 45.0°

Let β be the angle formed by the sides of length 110 and 79.

92² = 79² + 110² - 2(79)(110)cosβ

Solve for β.