Triangle with angles 30⁰,45⁰,105⁰, find area

Question

The length side across $105⁰$ is $4$ cm.

What is the area?

$\sin 45⁰/x = \sin 30⁰/y = \sin 105⁰/4 \text { cm}$

No calculator allowed.

$\sin 30⁰=t/x$

$\sin 45⁰=t/y$

What is the simplest, brilliant way to solve it?

Answer 1

Cut it into two triangles.

$30,60, 90$s and $45,45,90$ are well understood.

So draw the perpencular from the vertex of the $105$ degree angle down to the side that is $4$ cm. Call that height $h$.

Then the $30, 60, 90$ triangle will have sides $h$, $h\sqrt 3 $ and hypotenuse $2h$.

And the $45,45,90$ will have sides $h$, $h$, and hypotenuse $h\sqrt 2$.

And the side across the $105$ degree vertex is $h\sqrt3 $ (from the $30,60,90$ triangle) $+h$ (from the $45,45,90$) and so $h + h\sqrt 3 = 4$

So $h = \frac 4{1 + \sqrt 3}$ and the area is $\frac 12*4*\frac 4{1+\sqrt 3}=\frac 8{1+\sqrt 3}$.

Answer 2

Drop the altitude onto the length-4 side, and you break the triangle up into a 30-60-90 and a 45-45-90 triangle.

Write $4$ as $x+y$ where $x$ is the base of the 30-60-90 triangle and $y$ is the base of the 45-45-90 triangle. You can solve for the height (along which we joined the triangles) in two ways, which gives you $x$ and $y$, and at that point you have all the information you need.

Answer 3

The Sine theorem you used will also work: $$\frac{x}{\sin 45^\circ}=\frac{4}{\sin 105^\circ} \Rightarrow\\ x=\frac{4\cdot \sin 45^\circ}{\sin (60^\circ+45^\circ)}=\frac{4\cdot \frac{1}{\sqrt2}}{\sin 60^\circ\cdot \cos 45^\circ+\cos 60^\circ\cdot \sin 45^\circ}=\\ \frac{4\cdot \frac{1}{\sqrt2}}{\frac{\sqrt3}{2}\cdot \frac1{\sqrt2}+\frac12\cdot \frac1{\sqrt2}}=\frac{8}{\sqrt3+1}\\ S_{\Delta}=\frac12\cdot 4x\cdot \sin 30^\circ=\frac12\cdot 4\cdot \frac{8}{\sqrt3+1}\cdot \frac12=\frac{8}{\sqrt3+1}.$$