Geometric Two Triangles Height Building

[CookizREvil]

So I've been given this question in Physics, (I know, not math.) but the question is strictly Geometrical.

It's ridiculous.


"A group of physics scouts decide to measure the height of a distant building from an unknown distance. They take two angle observations 22.25 meters apart from a height of 1.25 meters. At the closer observation point, Dorfo notes that the string is at the 82.5 degree mark. At the second observation point, DorKo sees the string at the 84.7 degree mark."


The question wants the height of the building, or the unknown distance.


I got this one formula, it says, "Height = Delta D[Tan(Angle2) Times Tan(Angle1) OVER Tan(Angle2) Minus Tan(Angle1)] + Height of Meter Stick."


And what I got is always 0, or I'll get the height of the meter stick, 1.25 meters.

I don't understand WHAT I'm doing wrong! My calculator is in Degree mode, and I typed everything correctly.

Please help mee.

 

[skeeter]

I got this one formula, it says, "Height = Delta D[Tan(Angle2) Times Tan(Angle1) OVER Tan(Angle2) Minus Tan(Angle1)] + Height of Meter Stick."

\(\displaystyle \L H = (22.25) \frac{tan(84.7)*tan(82.5)}{tan(84.7) - tan(82.5)} + 1.25\)

hmmm ... my calculator sez H = 573.4456276

this should be your calculator syntax if using a TI ...

22.25(tan(84.7)*tan(82.5))/(tan(84.7) - tan(82.5)) + 1.25

 

[CookizREvil]

I got this one formula, it says, "Height = Delta D[Tan(Angle2) Times Tan(Angle1) OVER Tan(Angle2) Minus Tan(Angle1)] + Height of Meter Stick."

\(\displaystyle \L H = (22.25) \frac{tan(84.7)*tan(82.5)}{tan(84.7) - tan(82.5)} + 1.25\)

hmmm ... my calculator sez H = 573.4456276

this should be your calculator syntax if using a TI ...

22.25(tan(84.7)*tan(82.5))/(tan(84.7) - tan(82.5)) + 1.25

Thanks a lot. I forgot to put the Two tangent quantities in parentheses.