Thread: [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

Hi



[1]

[2]

Replace d' in [1] by its expression in terms of h and derived from [2]

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

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

Take the second scheme of my previous post





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

[1]

[2]

Replace in [1] to find the expression of h with respect to , and

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

OK so you are getting



d \:\tan \theta_2-h) = h \:\tan \theta_2" alt="\tan \theta_1\d \:\tan \theta_2-h) = h \:\tan \theta_2" />



Finally

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?

explain please.

Thanks

Multiplying numerator and denominator by



Multiplying both sides by

d \:\tan \theta_2-h) = h \:\tan \theta_2" alt="\tan \theta_1\d \:\tan \theta_2-h) = h \:\tan \theta_2" />

Expanding



Adding to both sides



Factorizing the RHS

\tan \theta_1 + \tan \theta_2)" alt="d\:\tan \theta_1 \: \tan \theta_2 = h\\tan \theta_1 + \tan \theta_2)" />

Dividing both sides by

Ooof.

Wow.

Thanks alot.

I get it now! Geez I was dumb

Thanks,

You Rock xD

SOLVED