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:
[tex]\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[/tex]
#Bisect angle C, and use sine rule again to find the lengths of bisected AB:
[tex]\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[/tex]
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.
