Question

Two fire trucks on the ground on either side of a burning building are 1.3 miles apart. They each measure the angle of elevation to the fire, which are 58 degrees and 52 degrees. How far is each truck from the fire. Step by step explanation.. Please..Thanks..

Answer 1

Answer:

A-0.5777miles

B-0.7223miles

Step-by-step explanation:

Let A be truck with 58°, B be truck with 52° elevation and C be the position of the fire. We know that the AB is 1.3 miles apart.

-Using sine rule, we find the length of AB and BC as:

$\angle C=180-(58+52)=70\textdegree\\\\\frac{a}{sin \ A}=\frac{b}{sin \ B}=\frac{c}{sin \ C}\\\\\frac{1.3}{sin \ 70}=\frac{b}{sin \ 52}\\\\b=\frac{1.3 sin \ 52}{sin 70}=1.0902mi\\\\\\c=\frac{1.3 sin \ 58}{sin 70}=1.1732$

#Bisect angle C, and use sine rule again to find the lengths of bisected AB:

$\frac{a}{sin \ A}=\frac{b}{sin \ B}=\frac{c}{sin \ C}\\\\\frac{1.0902}{sin \ 90}=\frac{x}{sin \ (90-58)}\\\\x=\frac{1.0902\ sin \ 32}{sin \ 90}=0.5777mi$

Truck B's position is calculated as 1.3mi-0.5777miles=0.7223miles

Hence, truck one is 0.5777miles and truck two is 0.7223 miles from the fire.