What's the measure of the $\angle BAC$ in the triangle below?

Question

For reference: In the right triangle $ABC$, right at $B$, the corner $AF$ is drawn such that $AB = FC$ and $\angle ACB = 2 \angle BAF$. Calculate $\angle BAC$.

My progress:

$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\ AF^2 = x^2+BF^2\\ \triangle AFC: Law ~of~ cosines:\\ AF^2 = x^2+AC^2-2.x.AC.cosC\\ \triangle ABC:\\ cos C = \frac{BC}{AC} =\frac{x+BC}{AC}\\ x^2+(x+BF)^2 = AC^2\\ Th.Stewart \triangle ABC:\\ AC^2.BF+x^3=AF^2BC+BC.x.BF$

...??

Answer 1

Here is a construction that makes things simple. Extend $CB$ such that $BE = BF = y$

Now $ \displaystyle \angle BAE = \frac{\angle C}{2} \implies \angle CAE = 90^0 - \frac{\angle C}{2}$

And we notice that $\triangle ACE$ is isosceles so $AC = x + 2y$

Applying Pythagoras in $\triangle ABC$,

$(x+2y)^2 = x^2 + (x+y)^2$

$4y^2 = x^2 + y^2 - 2xy = (x-y)^2$

That leads to $x = 3y$ and sides of $\triangle ABC$ are in the ratio $3:4:5$

Answer 2

If $y=BF$ and $t=\tan\frac{C}{2}=\frac{y}{x}$, then $\tan C=\frac{x}{y+x}=\frac{1}{1+t}$

Also $\tan{C}=\frac{2t}{1-t^2}$ from the double angle formula. It follows that:

$$\frac{2t}{1-t^2}=\frac{1}{1+t}$$

And you cand find $t=\frac{1}{3}$ by solving this equation. It follows that $\tan C=\frac{3}{4}$, $\tan\angle BAC=\frac{4}{3}$, and finally $\angle BAC=\arctan\frac{4}{3}$

Answer 3

Just a note: This is a method for reference, it approximates your angle answer:

We know that:

$\tan(\frac{c}{2}) = \frac{BF}{x}$

$\tan(90-c) = \frac{x+BF}{x} = 1 + \frac{BF}{x} = 1+\tan(\frac{c}{2})$

Using the relation between sine, cosine and tangent:

$\frac{\cos(c)}{\sin(c)}$ = $1+\frac{\sin(\frac{c}{2})}{\cos(\frac{c}{2})}$

Then using the half-angle formulae:

$\frac{\cos(c)}{\sin(c)}$ = $1\pm\frac{\sqrt{\frac{1-\cos(c)}{2}}}{\sqrt{\frac{1+\cos(c)}{2}}}$

$\frac{\cos(c)}{\sin(c)}$ = $1+\frac{\sqrt{\frac{1-\cos^2(c)}{4}}}{\frac{1+\cos(c)}{2}}$

Using Pythagorean-Trig identity:

$\frac{\cos(c)}{\sin(c)}$ = 1+$\frac{\sin(c)}{\cos(c)+1}$

$\cos^2(c)+\cos(c)$=$\sin^2(c) +\sin(c)\cos(c)+\sin(c)$

$\cos(2c)+\cos(c)-\sin(c)\cos(c)-\sin(c)=0$

$2\cos(2c)+2\cos(c)-\sin(2c)-2\sin(c)=0$

Because $c$ is real:

$(\cos(\frac{c}{2}))(-\sin(c)+2\cos(c)-1)=0$

$\sin(c)=2\cos(c)-1$

$\sqrt{1-\cos^2(c)} = 2\cos(c)-1$

$4\cos^2(c)-4\cos(c)+1=1-\cos^2(c)$

$(\cos(c))(5\cos(c)-4) = 0$

--> $\cos(c)=\frac{4}{5}$, given that $0<c<90$

$c \approx 37^o$

$180^o - 37^o - 90^o = 53^o$