Question

Law of sines and cosines word problem

Answer 1

Answer:

AC = 0.47 mi

BC = 0.51 mi

Step-by-step 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:

$\frac{0.5}{sin(61)} =\frac{AC}{sin(56)}\\AC=\frac{0.5*sin(56)}{sin(61)} \\AC\approx 0.4739\,\,\,mi$

which can be rounded to two decimals as:

AC = 0.47mi

For side BC we use:

$\frac{0.5}{sin(61)} =\frac{BC}{sin(63)}\\BC=\frac{0.5*sin(63)}{sin(61)} \\BC\approx 0.509\,\,\,mi$

which can be rounded to two decimals as:

BC = 0.51 mi