Geometry Challenge

[Albert Chan]

(01-11-2024 09:38 PM)Johnh Wrote:  Those trig equivalencies are like old-school mathe-magic!

If I were doing this problem in a exam, I would use vectors, instead of trigonometry.
Solving trig problem may need addition/removal of lines in sketch, which take time.

With vectors, we could do geometric tricks with algebra. (swapping lines is same as swapping terms)

Also, identities may be derived geometrically.
Isosceles triangle (2 sides = 1, exterior angle = 2θ), we have:

1 + cis(2θ) = 2*cos(θ) * cis(θ)

Getting this with trig identities is a bit more messy.

1 + cis(2θ) = (1+cos(2θ), sin(2θ)) = (2*cos(θ)^2, 2*sin(θ)*cos(θ)) = 2*cos(θ) * cis(θ)

BTW, your trig setup is equivalent to Thomas Kleem's vector setup.

(01-09-2024 03:09 PM)Thomas Klemm Wrote:  

Consider \(B\) the origin of \(\mathbb{C}\).

Then:

\(
\begin{align}
A &= (1 \measuredangle 70^{\circ}) \\
\\
C &= 1 \\
\\
D
&= C + (1 \measuredangle 10^{\circ}) \\
&= 1 + (1 \measuredangle 10^{\circ}) \\
\\
\overline{AD}
&= D - A \\
&= 1 + (1 \measuredangle 10^{\circ}) - (1 \measuredangle 70^{\circ}) \\
\end{align}\)

Solve with vectors (angles in degree)

cis(10) - cis(70) = cis(10) * (1 - cis(60) = 1 + cis(-120) = cis(-60)) = cis(10 - 60) = cis(-50)

vec(AD) = 1 + cis(-50) = 2*cos(-25) * cis(-25)

|AD| = 2*cos(25) ≈ 1.8126
∠BAD = 180 - 70 - 25 = 85