Question

Law of sines and cosines word problem

Answer 1

AC = 0.47 mi

BC = 0.51 mi

Explanation:

Notice that we are in the case of an acute triangle for which we know two angles ( < A = 63 and < B = 56) and one side (AB = 0.5).

We can find the measure of the third angle using the property of addition of three internal angles of a triangle:

< A + < B + < C = 180

63 + 56 + < C = 180 degrees

< C = 180 - 63 - 56 = 61 degrees.

Now we use the law of sines to find the length of sides AC and BC:

`(0.5)/(sin(61)) =(AC)/(sin(56))\\AC=(0.5*sin(56))/(sin(61)) \\AC\approx 0.4739\,\,\,mi`

which can be rounded to two decimals as:

AC = 0.47mi

For side BC we use:

`(0.5)/(sin(61)) =(BC)/(sin(63))\\BC=(0.5*sin(63))/(sin(61)) \\BC\approx 0.509\,\,\,mi`

which can be rounded to two decimals as:

BC = 0.51 mi