Height of a mountain

**qleeq** · October 5th 2010, 10:09 AM

Hi,
I got bad luck this term and got stuck with an instructor that doesn't even use the book to teach, so I'm really lost on this one problem.

I have to figure out the height of Mt Kilimanjaro based on two angles of elevation(they are right triangles).

One angle is 13.7 degrees, and another angle, 10.4 degrees is 5 miles behind the first. Approximate the height of mt kilimanjaro to the nearest 10th of a foot.

I have this formula so far:

$tan(13.7)=\frac{h}{d}$

$dtan(13.7) = h$

$d = \frac{h}{tan(13.7)}$

and

$tan(10.4)=\frac{h}{d+5}$

$(d+5)tan(10.4) = h$

$d+5 = \frac{h}{tan(10.4)}$

$d = \frac{h}{tan(10.4)}-5$

After this I'm stuck. h = height of the mountain, or the opposite side, and d = distance from the mountain; adjacent side.

I have to solve for h.
If you could explain this to me in a way that doesn't seem esoteric I'd really appreciate it..I have the solution manual with steps but it really doesn't help at all.

**Unknown008** · October 5th 2010, 10:16 AM

You distributed improperly;

$(d+5)tan(10.4) = h$

$dtan(10.4) + 5tan(10.4) = h$

This should be okay now

**qleeq** · October 5th 2010, 10:33 AM

Originally Posted by Unknown008

You distributed improperly;

$(d+5)tan(10.4) = h$

$dtan(10.4) + 5tan(10.4) = h$

This should be okay now

oops, what i meant was:
$d = \frac{h}{tan(10.4)}-5$

**Unknown008** · October 5th 2010, 10:36 AM

Right.

Equate:

$\dfrac{h}{tan(10.4)}-5 = \dfrac{h}{tan(13.7)}$

Multiply by tan(10.4)tan(13.7)

$htan(13.7)-5tan(10.4)tan(13.7) = htan(10.4)$

$htan(13.7) - htan(10.4) =5tan(10.4)tan(13.7)$

$h(tan(13.7)- tan(10.4)) =5tan(10.4)tan(13.7)$

Finish it now!

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