Angle Of Depression Problem

[mathisfunny]

Answer given is 11.4 m. I got 55.6 m (3 s.f). Please help

 

[skeeter]

I worked out h = 55.6 m also ...

 

[Dr.Peterson]

I worked out h = 55.6 m also ...

I agree.

So the next thing to do is to check their answer: if the height is as they say, what are the distances and angles in the problem?

And then, what misreading of the problem would lead to their answer?

 

[skeeter]

I calculate the horizontal distance from the base of the lighthouse to the first buoy as d₁ = 60.7 m ; to the second buoy as d₂ = 63.3 m.

With height of the lighthouse h = 55.6 m, the depression angles work out to the given values in the problem statement ...

$\arctan\left(\dfrac{h}{d_1}\right) = 42.5^\circ$

$\arctan\left(\dfrac{h}{d_2}\right) = 41.3^\circ$

 

[The Highlander]

View attachment 31974
Answer given is 11.4 m. I got 55.6 m (3 s.f). Please help

I agree.

So the next thing to do is to check their answer: if the height is as they say, what are the distances and angles in the problem?

And then, what misreading of the problem would lead to their answer?

The "Answer" given cannot possibly be correct (for the question as it is presented)!

If the Lighthouse were indeed 11.4m high then the correct answer to the question would be 11m (not 11.4m) because the question specifically states that the answer must be provided "to the nearest metre"!

However, according to my workings the answer would be 11m.
I get the height of the Lighthouse (ignoring the height of the keeper himself or assuming his eyes to be level with the top of the tower, lol) to be 10.816m (to 3s.f.).

(Maybe the question setter is defining "angle of depression" differently from me; I haven't bothered to check that.)

 

[The Highlander]

I calculate the horizontal distance from the base of the lighthouse to the first buoy as d₁ = 60.7 m ; to the second buoy as d₂ = 63.3 m.

View attachment 31974
Answer given is 11.4 m. I got 55.6 m (3 s.f). Please help

I think you may need to re-read the question (more carefully) and start by drawing a sketch of the situation described.
(It looks to me that your approach may be a bit too simplistic.)

Show us your (labelled) sketch, please, and we can progress form there.

 

[skeeter]

 

[The Highlander]

View attachment 31983

Yes, that's how I would regard the angle of depression but it's your (labelled) sketch of the situation described in the question that we need in order to see how you are are arriving at your calculated results.

 

[skeeter]

"We" ???


$d_1^2 + 18^2 = d_2^2$

$h = d_1\tan(42.5^\circ)$

$h = d_2\tan(41.3^\circ)$

solve the system for height, h

satisfied?

 

[Dr.Peterson]

View attachment 31984
$d_1^2 + 18^2 = d_2^2$

$h = d_1\tan(42.5^\circ)$

$h = d_2\tan(41.3^\circ)$

solve the system for height, h

satisfied?

I did the same.

Checking the claimed answer, if h were 11.4, then I get d₁ = 12.44 and d₂ = 12.98, which is far too small. Using h=10.816 would be even worse.

Of course, we should be having this discussion with the OP. But ...

However, according to my workings the answer would be 11m.
I get the height of the Lighthouse (ignoring the height of the keeper himself or assuming his eyes to be level with the top of the tower, lol) to be 10.816m (to 3s.f.).

(Maybe the question setter is defining "angle of depression" differently from me; I haven't bothered to check that.)

Now can you show your work, since three of us disagree with your answer? I suspect you may be imagining a different situation.

 

[The Highlander]

"We" ???

I agree.

So the next thing to do is to check their answer: if the height is as they say, what are the distances and angles in the problem?

And then, what misreading of the problem would lead to their answer?

Indeed!

I tried to sketch it 'orthogonally' and then (stupidly) took 18m as the hypotenuse (even though I had marked B₁ as a right angle, Doh!) because I allowed my eyes to deceive me.
(Surely you'll forgive me? G does look like a right angle! ?)

So I, erroneously, started with x₁² + x₂² = 18²!

I too now get the Lighthouse height to be 55.624m.

My mistake! Apologies for any (righteous) indignation caused. ?

However, reworking things, I have now noticed that I also made another mistake in evaluating the height (see Spoiler) on my calculator and my original answer should have been 11.415m (not 10.816m as I posted earlier) so it looks as if, perhaps, whoever provided the given "Answer" made the same mistake as me!

Good to know I may not be the only idiot around!

Spoiler: Height Expression (I used)

$\displaystyle 324 × \frac{(tan²42.5° × tan²41.3°)}{(tan²41.3° + tan²42.5°)}$

 

[The Highlander]

I did the same.


Of course, we should be having this discussion with the OP. But ...

Now can you show your work, since three of us disagree with your answer? I suspect you may be imagining a different situation.

See above Dr.P. ??

 

[Dr.Peterson]

so it looks as if, perhaps, whoever provided the given "Answer" made the same mistake as me!

Good to know I may not be the only idiot around!

Idiots can be useful for understanding other idiots. (Not saying that anybody in particular here actually is an idiot ... there's a little of that in all of us.)

In effect, they just made a sign error (as all of us do from time to time). My own solution was $$h=\frac{18\tan 42.5^\circ\tan 41.3^\circ}{\sqrt{\tan^2 41.3^\circ-\tan^2 42.5^\circ}}$$ which differs from yours only in the one sign. I hadn't considered that as a source of their error, but I should have.

 

[The Highlander]

Idiots can be useful for understanding other idiots. (Not saying that anybody in particular here actually is an idiot ... there's a little of that in all of us.)

In effect, they just made a sign error (as all of us do from time to time). My own solution was $$h=\frac{18\tan 42.5^\circ\tan 41.3^\circ}{\sqrt{\tan^2 41.3^\circ-\tan^2 42.5^\circ}}$$ which differs from yours only in the one sign. I hadn't considered that as a source of their error, but I should have.

Agreed, except that: wouldn't your expression give an Imaginary answer (given that tan²42.5° is greater than tan²41.3°)?
Hence $\displaystyle \sqrt{\tan^2 41.3^\circ-\tan^2 42.5^\circ}$ is the square root of a negative number.

Such fun when we are all making basic boo boos. ?

 

[The Highlander]

At least none of us has actually given the (correctly) worked through solution to the OP yet, lol.

Ya think s/he might be able to work it out for her/himself now?

 

[skeeter]

The OP cited a solution of 55.6 m, which is correct for the given problem statement.

 

[The Highlander]

The OP cited a solution of 55.6 m, which is correct for the given problem statement.

Of course, I'm just so used to the OP's not having any answer it slipped my mind that s/he'd posted 55.6m as her/his "solution". However, that's still wrong as the "correct" answer ("for the given problem statement") would be 56m.
(I guess we all make the odd mistake now & then.) ?

 

[Steven G]

I choose not to show my work after seeing how easy it is to make a mistake(s).

 

[The Highlander]

I choose not to show my work after seeing how easy it is to make a mistake(s).

Ooops, you want sin|x|, not |sinx|

That pledge didn't last long! ?

 

[Harry_the_cat]

I worked out h = 55.6 m also ...

To the nearest metre, the correct answer is 56m.