Question

Towers A and B are located 10 miles apart. A ranger spots a fire at a 42-degree angle from tower A. Another fire ranger spots the same fire at a 64-degree angle from tower B. To the nearest tenth of a mile how far from
tower B is the fire?
7.0
7.4
9.4
13.4

Answer 1

Answer:

The distance between tower B and the fire is $\approx 9.4 \,\,miles$  which agrees with the third option listed among the list of  possible answers.

Step-by-step explanation:

In order to understand which are our known elements of the triangle, we represent the quantities given in the diagram (please see attached image).

We understand that we can solve for the third angle in the triangle (the one defined by the vertex where the fire is by using:

$180^o-64^o-42^o=74^o$

Now that we know all angles, we can use a proportion from the law of sines to find side "x" which is the distance between the fire and tower B:

$\frac{10\,mi}{sin(74^o)} =\frac{x}{sin(64^o)} \\\frac{10\,mi\,*\,sin(64^o)}{sin(74^o)} =x\\x\approx 9.4 \,\,miles$

Answer 2

Answer:

7.0

Step-by-step explanation:

$\frac{sinA}{a} =\frac{SinC}{c}$

sinA is known which is sin(42)

sinC is equal to 180-sinB-sinA, which is 74

c is equal to 10, so plugging this in, the equation is:

$\frac{.6691306064}{a} = \frac{.9612616959}{10}$

Cross multiply to get:

.9612616959a=6.691306064

Divide by .9612616959 on both sides to get:

a=6.960961924, which is rounded to 7.0 miles