Trigonometry in isoceles trisected triangles

Question

Hello,

A $\triangle ABC$ has a point $D$ on $BC$ such that $\angle BAD = 30$ and $\angle CAD = 45$. Both $AB$ and $AC$ are of the length $\sqrt{2}$. Find the length of $AD$.

This is what I have done:-

Taking $AD$ as $x$ and $DC$ as $y$, and then using law of cosines for $y$, we get $y^2 = 2 + x^2 - 2x$.

Taking $BD$ as $k$, and applying law of cosines on $k$, we get $k^2 = x^2 + 2 - x*root6$

Taking law of cosines on $k + y$ we get, $(k + y)^2 = 5 - root3 ( cos 75 = root3 - 1/ 2*root2)$

And now I'm stuck at this step. I have no idea on how to go forward after this step.

ANY HELP IS APPRECIATED(if possible in trigonometry plz).

This is not a part of the question but can someone give me a link for the quick reference of mathjax editing?

Answer 1

Since $|AB|=|AC|$,

\begin{align} \angle ACD&=\angle ABD=52.5^\circ ,\\ \angle DAB&=97.5^\circ ,\\ \angle ADC&=82.5^\circ . \end{align}

Using the standard notation for $\triangle ABC$,
the area $[ABC]$ in terms of known side $a$ and the angles $\alpha$, $\beta$, $\gamma$ is \begin{align} [ABC]&=\frac{a^2\sin\beta\sin\gamma}{2\sin\alpha} .\\ \text{Then}\quad[ADC] \text{ in terms of } |AC|:\quad [ADC]&=\frac{(\sqrt2)^2\sin52.5^\circ\sin45^\circ}{2\sin82.5^\circ} =\frac{\sin52.5^\circ}{\sin82.5^\circ} ,\\ \text{and}\quad[ADC] \text{ in terms of } x:\quad [ADC]&=\frac{x^2\sin82.5^\circ\sin45^\circ}{2\sin52.5^\circ} = \frac{x^2\sin82.5^\circ}{2\sin52.5^\circ} ,\\ x&=\frac{\sqrt2\sin52.5^\circ}{\sin82.5^\circ} . \end{align}

Edit:

\begin{align} x&=\frac{\sqrt2\sin52.5^\circ}{\sin82.5^\circ} \\ &=\sqrt2\cdot\frac{\sin(82.5^\circ-30^\circ)}{\sin82.5^\circ} \\ &=\sqrt2\left(\cos30^\circ -\sin30^\circ\cot82.5^\circ \right) \\ &=\sqrt2\left(\tfrac{\sqrt3}2 -\tfrac12\tan(\tfrac{30^\circ}4) \right) \\ &=\sqrt2\left(\tfrac{\sqrt3}2 -\tfrac12(\sqrt2\sqrt3-\sqrt3+\sqrt2-2) \right) \\ &=(\sqrt2-1)(1+\sqrt3) . \end{align}

Answer 2

Answer to a changed problem.

The easiest way to solve the problem is with introduction of the coordinate system. Let $A=(0,0)$ and $D=(0,d)$. We have to find $d$. Then $AC$ is a line $y=x$ and $AB$ is a line $y=-{\sqrt{3}\over 3}x$. Since $B,C$ lies on the circle $x^2+y^2=2$ we can easily calculate coordinate of $B$ and $C$:

$$ B = ({\sqrt{6}\over 2},-{\sqrt{2}\over 2})\;\;\;\;{\rm and}\;\;\;\; C = (1,1)$$ Since $B,C,D$ are colinear we get $$d= (\sqrt{3}+1)(\sqrt{2}-1)$$