Question

Fire tower A is x = 27 kilometers due west of tower B. A fire is spotted from the towers, and the bearings from A and B are θ = N 70° E and φ = N 54° W, respectively (see figure). Find the distance d of the fire from the line segment AB. (Round your answer to two decimal places.)

Answer 1

Hey

Let the distance d meet AB at D. 

Using trig: 
AD = d*cot(θ) = d*cot(13 deg) 

BD = d*cot(ϕ) = d*cot(31 deg) 

So: 
AB = AD + BD 
30 km = d*cot(13 deg) + d*cot(31 deg) 
30 km = d*[cot(13 deg) + cot(31 deg)] 
d = (30 km)/[cot(13 deg) + cot(31 deg)] = 5.0035 km (round as needed).

Hoped That Helped You

Answer 2

The distance between the fire towers and the fire is an illustration of bearing and distance

The distance d of the fire from segment AB is 24.76 kilometers

The given parameters are:

$AB = 27$

$\theta = 70^o$

$\omega = 54^o$

Calculate distance d (see attachment) using tangent ratio.

Considering triangle BCD, we have:

$\tan(54) = \frac{d}{27 - x}$

Make d the subject

$d = (27 - x)\times \tan(54)$

Considering triangle ACD, we have:

$\tan(70) = \frac{d}{x}$

Make d the subject

$d =x \times \tan(70)$

Substitute $d =x \times \tan(70)$ in $d = (27 - x)\times \tan(54)$

$x \times \tan(70) = (27 - x) \times \tan(54)$

$x \times 2.7475 = (27 - x) \times 1.3764$

$2.7475x = (27 - x) \times 1.3764$

Open brackets

$2.7475x = 37.1628 - 1.3764x$

Collect like terms

$2.7475x +1.3764x= 37.1628$

$4.1239x= 37.1628$

Divide both sides by 4.1239

$x= 9.0112$

Substitute $x= 9.0112$ in $d =x \times \tan(70)$

$d = 9.0112 \times \tan(70)$

$d = 24.76$

Hence, the distance d of the fire from segment AB is 24.76 kilometers

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