Question

Three ships, A, B, and C, are anchored in the atlantic ocean. The distance from A to B is 36.318 miles, from B to C is 37.674 miles, and from C to A is 11.164 miles. Find the angle measurements of the triangle formed by the three ships.

Answer 1

Let us draw the triangle

here the sides a= 37.674 miles

b= 11.164 miles

c= 36.318 miles

Lets use the cosine rule to solve for the angles

( we cannot use the sine law since we do not have the measure of any of the angles)

The cosine law

$c^{2} = a^{2} +b^{2} -2ab cos C$

Let us plug in the values

$(36.318)^{2} = (37.674)^{2} + (11.164)^{2} - 2(37.674)(11.164). Cos C$

$1318.997 = 1419.33 + 124.634 - 841.184. Cos C$

$1318.997 = 1543.964 - 841.184. Cos C$

$841.184. Cos C = 1543.964-1318.997$

$841.184. Cos C = 224.967$

$Cos C = \frac{224.967}{841.184}$

$cos C = 0.267$

C = 74.48°

We can use the sine law to calculate the value of angle A

$\frac{a}{SinA} = \frac{c}{Sin C}$

$\frac{37.674}{Sin A} = \frac{36.318}{Sin 74.48}$

$Sin A = \frac{37.674. sin 74.48}{36.318}$

$Sin A = \frac{37.674 X 0.963}{36.318}$

$Sin A = 0.998$

A= $sin^{-1} (0.998)$

A = 87.38°

Now we can easily find the third angle B by subtracting angle A and C from 180°

$B = 180-(74.48 + 87.38)$

B = 180-161.86

B = 18.14°

Hence we have all the three angles ( attached figure)