[MOVED] using angle of depression, elevation to find height

[ilovemath3]

You need part (a) in order to complete part (b):

a) From point A that is 25 feet above level ground, the angle of elevation of a building is 31 degrees 20 minutes, and the angle of depression is 12 degrees 50 minutes. Approximate the height of the building

Could someone work this and tell me what they get? I would like to compare your answer to mine:

. . .h = 25 + R = 25 + 66.8128 = 91.8128

Here's the part where I need help:

b) Generalize the problem to the case where A is d feet above the ground and the angles of elevation and depression are alpha and beta, respectively. Express h of the building in terms of d, alpha, and beta.

Note: The answer is supposed to be

. . .h = d[1+cot(beta) tan(alpha)]

I would greatly appreciate any tips or advice. Thank you!

 

[Denis]

a = alpha, b = beta

Horizontal distance from A to building: d * COT(b)

Vertical distance from A to top of building: [d * COT(b)] * TAN(a)

h = d + [d * COT(b)] * TAN(a)

h = d[1 + COT(b) TAN(a)]

 

[soroban]

Re: [MOVED] using angle of depression, elevation to find hei

Hello, ilovemath3!

Here's the first part.
Denis did a great job with the second part.

a) From point \(\displaystyle A\) that is 25 feet above level ground,
the angle of elevation of a building is 31°20'
and the angle of depression is 12°50'.
Approximate the height of the building.

                              * B
                          *   |
                      *       |
                  *           |y
              *               |
          * 31°20'   x        |
    A * - - - - - - - - - - - + C
      |     *  12°50'         |
    25|           *           |25
      |                 *     |
      * - - - - - - - - - - - * D

The building is \(\displaystyle BD.\)
Let \(\displaystyle AC\,=\,x,\;BC\,=\,y\)
The height of the building is: \(\displaystyle \,h\:=\:y\,+\,25\)

In right triangle \(\displaystyle ACD:\;\tan12^o50'\:=\:\frac{25}{x}\;\;\Rightarrow\;\;x\:=\:\frac{25}{\tan12^o50'}\:\approx\:109.74\)

In right triangle \(\displaystyle ACB:\;\tan31^o20'\:=\:\frac{y}{x}\;\;\Rightarrow\;\;y\:=\:x\cdot\tan31^o20'\)

Hence: \(\displaystyle \,y\:=\109.74)(\tan31^o20')\:\approx\:66.8118\)

Therefore: \(\displaystyle \:h\:=\:66.818\,+\,25\:=\:91.8118\) feet.

 

[ilovemath3]

thank you so much for the help

i dont understand how denis is getting his answer, could you please clarify how he got it? thanks

edit

i got down to y + d = h

therefore h = d + [d * COT(b)] * TAN(a) but i dont udnerstand how he gets

h = d[1 + COT(b) TAN(a)]

thanks !!!

 

[stapel]

i got down to y + d = h, therefore h = d + [d * COT(b)] * TAN(a) but i dont udnerstand how he gets "h = d[1 + COT(b) TAN(a)]"

I believe he factored:

. . . . .h = d + d cot(b) tan(a)

. . . . .h = d [1] + d [cot(b) tan(a)]

. . . . .h = d [1 + cot(b) tan(a)]

That's all there is to it, I think.

Eliz.