# pls help my problem

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• Sep 24th 2012, 09:56 PM
sevenfold99
pls help my problem
give me the solution from the top of a building the angle of depression of a point on the same horizontal plane with the base of the building is observed to be 60 degree what will be the angle of depression of the same point when viewed from a position one third of the way up the building
• Sep 24th 2012, 10:14 PM
MarkFL
Re: pls help my problem
If you draw a sketch, look at some ratios, you will find a nice relationship between the two angles of depression.
• Sep 24th 2012, 10:21 PM
sevenfold99
Re: pls help my problem
can u give a formula for my problem
• Sep 24th 2012, 10:36 PM
MarkFL
Re: pls help my problem
Did you draw a sketch of the resulting triangles? Trigonometry frequently involves drawing diagrams and looking for similar triangles, etc.
• Sep 24th 2012, 10:43 PM
sevenfold99
Re: pls help my problem
can u give me the answer
for me
• Sep 24th 2012, 10:52 PM
MarkFL
Re: pls help my problem
No, I'm not going to simply give you the answer. I will help you figure out how to do the problem...if you are willing to show some effort.

Put pen/pencil to paper, and draw a sketch. Tell us what you find, and we will be glad to help.
• Sep 24th 2012, 10:58 PM
sevenfold99
Re: pls help my problem
ok sir the triangle are two how can i find the formula can u help me pls
• Sep 24th 2012, 11:07 PM
MarkFL
Re: pls help my problem
Okay, I assume you have labeled the distances and angles...

Using trigonometry, what is the ratio of the horizontal distance to the point to the height of the building?

Hint: use the tangent function. Do you see why?

edit: Or use the special angles within the larger triangle.
• Sep 24th 2012, 11:11 PM
sevenfold99
Re: pls help my problem
y can i use tangent funtion?
• Sep 24th 2012, 11:28 PM
sevenfold99
Re: pls help my problem
tan 60 degree is 10.23856
• Sep 24th 2012, 11:38 PM
MarkFL
Re: pls help my problem
No, $\displaystyle \tan(60^{\circ})=\sqrt{3}$.

The reason the tangent function is useful here, is that it relates an angle in a right triangle to the ratio of the opposite to the adjacent side.

$\displaystyle \tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

If $\displaystyle \theta=60^{\circ}$, what do the opposite and adjacent sides of the larger triangle represent?
• Sep 25th 2012, 12:00 AM
sevenfold99
Re: pls help my problem
0.57735
• Sep 25th 2012, 12:06 AM
MarkFL
Re: pls help my problem
I have no idea where you got that number from, but there is no need for decimal equivalents.

In your diagram, what is the side that is opposite the 60° angle?