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.)

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

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chibuikeumeh chibuikeumeh · Answer 2 · 2021-10-21T20:26:30+02:00

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:

[tex]AB = 27[/tex]

[tex]\theta = 70^o[/tex]

[tex]\omega = 54^o[/tex]

Calculate distance d (see attachment) using tangent ratio.

Considering triangle BCD, we have:

[tex]\tan(54) = \frac{d}{27 - x}[/tex]

Make d the subject

[tex]d = (27 - x)\times \tan(54)[/tex]

Considering triangle ACD, we have:

[tex]\tan(70) = \frac{d}{x}[/tex]

Make d the subject

[tex]d =x \times \tan(70)[/tex]

Substitute [tex]d =x \times \tan(70)[/tex] in [tex]d = (27 - x)\times \tan(54)[/tex]

[tex]x \times \tan(70) = (27 - x) \times \tan(54)[/tex]

[tex]x \times 2.7475 = (27 - x) \times 1.3764[/tex]

[tex]2.7475x = (27 - x) \times 1.3764[/tex]

Open brackets

[tex]2.7475x = 37.1628 - 1.3764x[/tex]

Collect like terms

[tex]2.7475x +1.3764x= 37.1628[/tex]

[tex]4.1239x= 37.1628[/tex]

Divide both sides by 4.1239

[tex]x= 9.0112[/tex]

Substitute [tex]x= 9.0112[/tex] in [tex]d =x \times \tan(70)[/tex]

[tex]d = 9.0112 \times \tan(70)[/tex]

[tex]d = 24.76[/tex]

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

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