# Math Help - Location of a fire

1. ## Location of a fire

Two towers are 30 Km apart where tower A is due west of tower B. A fire is spotted from the towers, and the bearings from A to B are E 14 degrees N, and W 34 degrees N respectively. Find the distance from the fire to line segment AB. I can get the answer by using the law of Sines. How else can I do this problem with out using the law of sines? The altitude of the triangle forms a right angle with AB, but it does not bisect AB in the picture. Angle C is split up to be 74 degrees and 56 degrees. How can I use Sine, cosine, or tangent to find another side?

Thank you. d is 5.6 km using the law of sines. I cannot use this method to show my work. I am stuck.

Angle A= 14 degrees, Angle B= 34 degrees, Altitude is d height we want, and AB=30 Km

I have also said that Tan(14)(30-x)=d and Tan(34)x=d. So I can replace d and get Tan(14)(30-x)=Tan(34)(x) solve for x. I keep getting x equals 15 then plug that in and get a different number around 10?

2. Originally Posted by IDontunderstand
Two towers are 30 Km apart where tower A is due west of tower B. A fire is spotted from the towers, and the bearings from A to B are E 14 degrees N, and W 34 degrees N respectively. Find the distance from the fire to line segment AB. I can get the answer by using the law of Sines. How else can I do this problem with out using the law of sines? The altitude of the triangle forms a right angle with AB, but it does not bisect AB in the picture. Angle C is split up to be 74 degrees and 56 degrees. How can I use Sine, cosine, or tangent to find another side?

Thank you. d is 5.6 km using the law of sines. I cannot use this method to show my work. I am stuck.
let d = perpendicular distance from the fire to segment AB

let x = distance from A to the closest point on AB to the fire

$\displaystyle \tan(14) = \frac{d}{x}$

$\displaystyle \tan(34) = \frac{d}{30-x}$

solve the system for $d$ and $x$