How do I solve this trigonometry question involving bearings?

Question

Hi, I'm doing a maths assignment and I have no idea how to solve this problem. If you know how to answer these questions please solve and explain. Thanks!!
The problem is:

Points A, B & C are all situated on the same horizontal level with B due north of A and C due west of A. A vertical tower has its base at A. From point C, which is 50 metres from A, the top of the tower has an angle of elevation of 24 degrees and the bearing of B is 050 degrees.

Hence, calculate: the height of the tower (nearest metre)
the distance from B to C (nearest metre)
the distance from B to A (nearest metre)
the angle of elevation of the top of the tower from B
to the nearest degree

Answer 1

First draw out the points A, B and C on a piece of paper as per the description given in the question so that you can understand the solution to the problem better. Mark out the tower position also.



Height of the Tower



let A' be the point at the top of the tower and let h be its height

C is 50m away from A which is the base of the vertical tower

Therefore the tower, the line AC and the line CA' make a right angled triangle with CA' being the hypotenuse.



Since the elevation angle (angle at C) is 24 degrees

h = tan (24) X 50 (tan = opposite / adjacent)

h = 22.26m

or to the nearest metre

h = 22 m



Distance of B to C



From C the bearing of B is 50 degrees

therefore the angle ACB = 90 - 50 = 40 degrees since bearing is always taken from North

Distance from A to C is 50 m

CB is the hypotenuse of the right angled triangle ABC

therefore Cos (40) = 50/BC (cos = adjacent / hypotenuse)

BC = 50 / Cos (40)

BC = 65.27m

or to the nearest metre

BC = 65 m



Distance from B to A



ABC is a right angled triangle

AC = one side = 50 m

BC = hypotenuse = 65 m (from calculation above)

therefore BC^2 = BA^2 + AC^2

65^2 = BA^2 + 50^2

4225 = BA^2 + 2500

4225 - 2500 = 1725 = BA^2

therefore BA = sqrt(1725) = 41.5331

to the nearest metre

BA = 42 m



Angle of Elevation from B



Height of the tower = 22 m

Distance BA = 42 m

therefore Tan (angle of elevation from B) = 22/42

Tan (angle of elevation from B) = 0.5238

Inv Tan 0.5238 = 27.64 degrees

or to the nearest degree

Elevation from B = 28 degrees

Answer 2

ABC is a right angle triangle, with the right angle at A.



Things we know:

Let the top of the tower be P. Angle PCA is 24 degrees.

Angle BCA is 40 degrees.



Height of the tower, AP = 50 tan 24 = 22.26m (22m)



Distance from B to C = 50/cos40 = 65.27m (65m)



Distance from A to B = 42m



Angle of elevation of tower from B = inverse tangent of (22/42) = 28 degrees.

Answer 3

Top of tower: point X

Distance AC = 50m

Angle of elevation to top of tower: 24°

tan (24°) = opp/adj = AX/AC

AX = AC * tan(24°) = 50 tan(24°)

AX = 22m



Distance AC = 50m

From point C, bearing of B is 50°

cos(50°) = adj/hyp = AC/BC

BC = AC / cos(50°) = 50 / cos(50°)

BC = 78m



tan(50°) = opp/adj = AB/AC

AB = AC tan(50°) = 50 tan(50°)

AB = 60m



Distance AB = 60m

Height of tower AX = 22m

Angle of elevation = θ

tan(θ) = AX/AB = 22/60

θ = arctan(22/60)

θ = 20°



EDIT:



OOps, I might have gotten the angle ∠ACB wrong. When you stated that bearing of B from C is 50°, I assumed that facing A, the angle to B was 50° (or 50° as calculated using the unit circle).



But usually a "bearing" is measured either east and west from north or south. In this case, it almost certainly refers to 50° to the east when facing north. Therefore

∠ACB = 90 - 50 = 40

as seen in PremK's answer.



The reasoning in my answers remains the same, but with a different angle:

BC = 50 / cos(40°) = 65

AB = 50 tan(40°) = 42

θ = arctan(22/42) = 28°



Sorry for the mixup, and kudos to PremK for a great answer.