From the top of a cliff 160m high, two buoys are observed. Calculate the distance between the buoys.

Question

From the top of a cliff 160m high, two buoys are observed. Their bearings are 337° and 308°. Their respective angle of depression are 3° and 5°. Calculate the distance between the buoys.

I got this question in my Year 11 Maths worksheet, I tried making a diagram for this question but I have not been able to do so. One thing for the question I figured that io could use the trig ratios to calculate the distance from the cliff to each of the buoys, therefore, using the cosine rule to get the distance b/w the buoys. But I am not able to do this without the diagram so I wanted some help in that regard.

Answer 1

General nudge: Drawing all of this in a single diagram will be very difficult and confusing. Think about how this problem can be broken into two parts and then diagram each separately.

More specific: You need to know the distance from the cliff base to each of the buoys before you can determine the distance between the buoys. Start by drawing a side-on diagram to solve for that distance and then draw a second bird's eye view diagram to determine the distance between the buoys.

Answer 2

The figure above shows the situation at hand. The blue dot is the observation point, and the two buoys are the two red dot labelled $A$ and $B$.

If the origin of the Cartesian coordinate system is at the observation point, then the coordinates of the two buoys is

$A = ( 160 \cot \theta_1 \sin \phi_1 , 160 \cot \theta_1 \cos \phi_1, -160) $

$B = (160 \cot \theta_2 \sin \phi_2, 160 \cot \theta_2 \cos \phi_2, -160 )$

where $\theta$ is the depression angle, and $\phi$ is the bearing. Hence,

$\overline{AB} = 160 \sqrt{ (\cot \theta_1 \sin \phi_1 - \cot \theta_2 \sin \phi_2 )^2 + (\cot \theta_1 \cos \phi_1 - \cot \theta_2 \cos \phi_2)^2 }$

And this reduces to

$\overline{AB} = 160 \sqrt{ \cot^2 \theta_1 + \cot^2 \theta_2 - 2 \cot \theta_1 \cot \theta_2 \cos(\phi_2 - \phi_1) }$

Using the values given,

$\overline{AB} = 160 \sqrt{ \cot^2(3^\circ) + \cot^2(5^\circ) - 2 \cot(3^\circ) \cot (5^\circ) \cos(337^\circ - 308^\circ) } $

And this evaluates to

$\overline{AB} \approx 1702.55 $

Answer 3

Just in case someone is interested in the non-flat-earth solution, assuming the surface of the earth is approximately a sphere:

The base of the cliff and the two buoys form a spherical triangle. Let $\alpha_i,\;i\in\{1,2\},$ represent the "side length of the spherical triangle" between the base of the cliff and buoy $i$ measured as the angle at the center of the earth. Let $\theta_i$ denote the depression angles and $\phi_i$ denote the bearings. Let $\gamma$ represent the "side length of the spherical triangle" between the two buoys measured as the angle at the center of the earth. $R$ is the radius of the earth, $h$ is the height of the cliff.

By means of the law of sines, applied to the triangle formed by the top of the cliff, the buoy and the center of the earth, we find $$ \alpha_i = \arcsin\left(\frac{R+h}{R}\;\sin(90^{\circ}-\theta_i)\right)-(90^{\circ}-\theta_i) $$ You have to be careful when you apply the law of sines here, because there is an obtuse angle at the buoy.

The first spherical law of cosines gives us $\gamma.$ $$ \cos\gamma = \cos\alpha_1 \cos\alpha_2 + \sin\alpha_1 \sin\alpha_2 \cos(\phi_1 - \phi_2) $$ Then the distance between the two buoys (length of the geodesic) is $$ d = R \; \gamma \; \frac{\pi}{180^{\circ}}\approx 1713.08\;\mathrm{m} $$