Thread: [SOLVED] Urgentt HELP! 3D Trigonometry Problem

1. [SOLVED] Urgentt HELP! 3D Trigonometry Problem

Hii I reallyy need to solve this question. could you guys help me and also solve it? thankss

Two buildings of equal height are 40m apart and at a point on the horizontal line joining their feet the angles of elevation of the top of the building are 47 degrees and 28 degrees.

Show that the height (h) of the building is given by:

*=Degrees

h = 40tan47*tan28*
----------------
tan47*+tan28*

----------=Divider/Over

Show all working.

Thanks guys

2. Hi

$\displaystyle \tan \theta_1 = \frac{h}{d+d'}$ [1]

$\displaystyle \tan \theta_2 = \frac{h}{d'}$ [2]

Replace d' in [1] by its expression in terms of h and $\displaystyle \tan \theta_2$ derived from [2]

EDIT : the point seems to be actually BETWEEN the 2 buildings
That does not change the method anyway

3. Heyy Uhh

Could you explain it a bit more..

Im confused a bit..

and could you also solve it?

also what does it have to do with the h = 40tan47*tan28* thing? im kinda dumb. please help out

Thanks

4. Take the second scheme of my previous post

$\displaystyle d = 40 m$
$\displaystyle \theta_1 = 28 deg$
$\displaystyle \theta_2 = 47 deg$

The tangent is defined as the ratio of the opposite leg to the adjacent leg

$\displaystyle \tan \theta_1 = \frac{h}{d-d'}$ [1]

$\displaystyle \tan \theta_2 = \frac{h}{d'}$ [2]

Replace $\displaystyle d' = \frac{h}{\tan \theta_2}$ in [1] to find the expression of h with respect to $\displaystyle d$, $\displaystyle \tan \theta_1$ and $\displaystyle \tan \theta_2$

5. Heyy

I got up to here..
what to I do after this?

/tan/theta_1 = \frac{h}{d-\frac{h}{/tan/theta_2}}

basically

tan theta 1 = h over d minus h over tan theta 2

6. OK so you are getting

$\displaystyle \tan \theta_1 = \frac{h}{d-\frac{h}{\tan \theta_2}} = \frac{h \:\tan \theta_2}{d \:\tan \theta_2-h}$

$\displaystyle \tan \theta_1\d \:\tan \theta_2-h) = h \:\tan \theta_2$

$\displaystyle d\:\tan \theta_1 \: \tan \theta_2 = h \:\tan \theta_2 + h \:\tan \theta_1$

Finally

$\displaystyle h = \frac{d\:\tan \theta_1 \: \tan \theta_2}{\tan \theta_1 + \tan \theta_2}$

7. Wow..

Thanks alot

I get nearly everything

cept for the second last part.

what happens to the 2 h's? doesnt it get like added up or something?

Thanks

8. $\displaystyle \tan \theta_1 = \frac{h}{d-\frac{h}{\tan \theta_2}}$

Multiplying numerator and denominator by $\displaystyle \tan \theta_2$

$\displaystyle \tan \theta_1 = \frac{h \:\tan \theta_2}{d \:\tan \theta_2-h}$

Multiplying both sides by $\displaystyle d \:\tan \theta_2-h$

$\displaystyle \tan \theta_1\d \:\tan \theta_2-h) = h \:\tan \theta_2$

Expanding

$\displaystyle d\:\tan \theta_1 \: \tan \theta_2 - h \:\tan \theta_1 = h \:\tan \theta_2$

Adding $\displaystyle h \:\tan \theta_1$ to both sides

$\displaystyle d\:\tan \theta_1 \: \tan \theta_2 = h \:\tan \theta_2 + h \:\tan \theta_1$

Factorizing the RHS

$\displaystyle d\:\tan \theta_1 \: \tan \theta_2 = h\\tan \theta_1 + \tan \theta_2)$

Dividing both sides by $\displaystyle \tan \theta_1 + \tan \theta_2$

$\displaystyle h = \frac{d\:\tan \theta_1 \: \tan \theta_2}{\tan \theta_1 + \tan \theta_2}$

9. Ooof.

Wow.

Thanks alot.

I get it now! Geez I was dumb

Thanks,

You Rock xD

SOLVED