Exact Trig Values

[NeedingWD40]

I can't see my way through this one.... I need the height, I can raise a perpendicular at the RHS of the base, and get 6 up, but can't see how to find the rest of the height, I can raise a perpendicular at the end of the dotted line to get a 60 degree right angle but I have no side lengths so no scale factor...

 

[Deleted member 4993]

View attachment 19809
I can't see my way through this one.... I need the height, I can raise a perpendicular at the RHS of the base, and get 6 up, but can't see how to find the rest of the height, I can raise a perpendicular at the end of the dotted line to get a 60 degree right angle but I have no side lengths so no scale factor...

You have the right idea of dropping a perpendicular from the top vertex. After that, your drawing will look somewhat as:



Now define BD = x & AD = h

Since mACD = 45^o, AD = CD = 6+x

AD/BD = tan(60^o) = \(\displaystyle \sqrt{3} \)

(6+x)/x = \(\displaystyle \sqrt{3} \)

x = ?

CD = AD = ?

Area of ABC = 1/2 * (6) * (AD) = ?

BTW, this method was used to estimate the height of Mt. Everest - before actually scaling it.

 

[NeedingWD40]

Thank you!
so 6 = √3*x – x
6/(√3 - 1) = x
6(√3 + 1)/3 - 1 = x
3√3 + 3 = x

so AD = 6 + 3 + 3√3 = 9 + 3√3

and the triangle has area 1/2 * 6 * (9 + 3√3) = 27 + 9√3

Yes? Nice problem!