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I was asked a question like this on an exam today and I'm wondering if I got it right or not.

A boat is laying a pipe on the ocean floor using two cables. The angle of depression for one of the cables is $38^\circ$ degrees and the other is $44^\circ$.

The length of the pipe is $118$ meters. What is the depth from the bottom of the boat to the pipe?

Here's a picture to visual the situation (please assume the bottom of the boat is flat)

Visualization

This is what I did:

Using the alternate interior angles rule I was able to assume that the two angles of elevation were also $44^{\circ}$ and $38^{\circ}$. The top angle would be $98^{\circ}$.:

enter image description here

Then using sine, I was able find out the hypotenuse for one of the outside triangles (in this case I used the one on the left).

$$\frac{118}{\sin{98^\circ}} = \frac{x}{\sin{38^\circ}}$$ $$x\sin{98^\circ} = 118 \times\sin{38^\circ}$$ $$\frac{x\sin{98^\circ}}{\sin{98^\circ}} = \frac{118 \times\sin{38^\circ}}{\sin{98^\circ}}$$ $$ x \approx 73.36 $$

Then, I was able to come up with another triangle which has a hypotenuse of $73.36$ meters. The other angle is $46^\circ$ because $90^\circ-44^\circ = 46^\circ$

Right angle triangle

Then using SOHCAHTOA (because its a right angle triangle) I was able to solve $x$ (which represents the depth) using Sine. Like so:

$$\sin{44^\circ} = \frac{x}{73.36}$$ $$x = \sin{44^\circ} \times 73.36$$ $$x \approx 50$$

The answer I got on the test was more like $80$ meters, but that's probably because I forgot what the angles and measurements exactly were on the exam. Is this the right approach or am I totally off course?

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  • $\begingroup$ What you did is correct, the idea at least. I didn't check the calculations though. $\endgroup$ – The very fluffy Panda Apr 17 '14 at 23:50
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I didn't recheck your calculations so I'll assume your approximations are correct. Other than that you did a fine job finding the depth. This problems basically boils down to finding the altitude of a triangle given two angle measures. Your approach is right on the money and exactly the approach I would use had I remembered the law of sines lol.

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  • $\begingroup$ Good to hear! :D $\endgroup$ – ub3rst4r Apr 18 '14 at 0:17

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