Can you help with this question (math whizzes)...?

Question

At a point A on the ground the angle of elevation of a mountain due east is 16.6°, and at B, due south of the mountain, the angle of elevation is 12.3°. If the mountain is 960metres high, find the distance from A to B, correct to one decimal place

(Note: I think I have to draw this, I can't visualise on a flat diagram/scale diagram where point B would be-Is it a 3-D shape?). Could you please help me and explain to me what the diagram. Thankyou.

Answer 1

Due East

tanθ = y / x

tan 16.6 = 960 / x

x tan 16.6 = 960

x tan 16.6 / tan16.6 = 960 / tan 16.6

x = 960 / tan16.6

x = 960 / 0.298112948

x = 3220.255973 meters

x = 3220 meters rounded

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Due south

tanθ = y / x

tan 12.3 = 960 / x

x tan 12.3 / tan12.3 = 960 / tan12.3

x = 960 / tan12.3

x = 960 / 0.21803526

x = 4402.957568 meters

x = 4403 meters rounded

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Caculating the distance A is from B

Pythagorean Theorem

c² = a² + b²

√c² = ± √(3220)² + (4403)²

√c² = ± √10368400 + 19386409

√c² = ± √29754809

x = ± 5454.479688 Meters

x = 5454.5 meters rounded to one decimal place

The distance A from B is 5454.5 meters.

Answer 2

Do a sketch, let the face of a clock be a guide.

The mountain (point M) is in the centre of the clock. Point A is at 9'o clock position (as the mountain is due east). Point B is at position 6 o'clock (as it is due south of the mountain).

Now, the mountain is 960m high, So distance mountain to point A is 960/(tan(16.6°)) = 3220.26m

Distance mountain to point B is 960/(tan(12.3°)) = 4402.96m

Note that angle between AM and MB is 90°, so distance AB can be found by taking square-root of [ (AM X AM) + (BM X BM) ] = 5454.92m (Pythagoras theorem)

Answer 3

Let
x = horizontal distance from A to top of mountain
y = horizontal distance from B to top of mountain
d = distance between A and B

tan(16.6°) = 960/x
tan(12.3°) = 960/y

x = 960 / tan(16.6°) = 960cot(16.6°)
y = 960 / tan(12.3°) = 960cot(12.3°)

A right triangle can be formed by the mountain peak and points A and B with the right angle at the peak and the hypotenuse the segment AB.

d² = x² + y² = [960cot(16.6°)]² + [960cot(12.3°)]²

d² = 960²*[cot(16.6°)² + cot(12.3°)²]

d = √{960²*[cot(16.6°)² + cot(12.3°)²]}

d = 960√[cot(16.6°)² + cot(12.3°)²] ≈ 5454.913 m

Answer 4

Let M be point at foot of mountain.
A is point due east of M
B is point due south of M
Join M, B and A to form triangle BMA, right angled at M.
This triangle is lying in plane of your paper.
Let tower be MT where MT is perpendicular to plane of paper.
Tan 16.6° = MT/MA = 960/MA
MA = 960/tan 16.6° = 3220
Tan 12.3° = MT/MB = 960/MB
MB = 960/tan 12.3° = 4295
AB² = MA² + MB²
AB = √[ (3220)² + (4295)² ]
AB = 5368.0 m

Answer 5

Ok, call the top of the mountain T and the bottom of the mountain M You now have three triangles. ATM, BTM, and ABM

You also know SOHCAHTOA right? Sin x = length of opposite divided by length of hypotenuse, Cos x = length of adjacent side divided by length of hypotenuse, and Tan x = length of opposite divided by length of adjacent.

so length TM = 960m
Therefore length AM= 960m/Tan 16.6 = 3220.25m
Length BM = 960m/Tan 12.3 = 4402.96

If according to Pythagoras theorem AB^2=AM^2+BM^2 then

AB = SQRT 10370010.06 + 19386056.76 = SQRT 29756066.82

Length from A to B therefore equals 5454.9 metres

Answer 6

xx

distance from A {think right-triangles} call DA
(to "base" of mountain)

tan(16.6 deg) = 960 / DA = .29811
DA = 960/.29811 = 3220.25597

tan(12.3 deg) = 960/DB = .218035
DB = 4402.95757

east and south makes a right triangle again so
distance (AB) = sqrt (DB^2 + DA^2)
=sqrt( 10370048.5301 + 19386035.34538)

=sqrt(29756083.8755) = 5454.9137 meters

Answer 7

Peak of mountain, we'll call point X

Rule of sine: 960/sin(16.6) = AX

960/sin(12.3) = BX

Since angle AXB is also a right angle...

AB = sqt((960/sin(16.6))^2 + (960/sin(12.3)^2))

Now get a calculator...