Elementary problem in geometry

Question

The problem asks to find the angle at $C$. The distance between $A$ and $B$ is $12 \space m$ and the distance between $B$ and $C$ is $8\space m$.

Anyone got an idea?

Answer 1

Let $h$ be the height of the blue wall, $x$ the length of the adjacent side for the triangle for $C$. Then we have $$ h = \tan (40^o) (12 + 8 + x) = \tan(45^o) (8 + x) = \tan (c^o)x $$ You can use the middle two to solve for $x$, then use the last two to solve for $c$ (as you already know $x$). Then as a bonus you can even extract $h$.

Answer 2

Let's assume the P is the point of intersection of all three lines of elevation & then drop an altitude $PQ=h$ to the (horizontal) plane. Now, we have

In right $\Delta PQA$ $$\tan 40^o=\frac{PQ}{AQ}=\frac{h}{AQ}$$ $$\implies \color{blue}{AQ=h\cot 40^o}$$ Similarly, in right $\Delta PQB$ $$\tan 45^o=\frac{PQ}{BQ}=\frac{h}{BQ}$$ $$\implies \color{blue}{BQ=h\cot 45^o=h}$$ $$\implies AB=AQ-BQ \implies 12=h\cot 40^o-h \implies 12\tan 40^o=h-h\tan 40^o$$ $$\color{red}{h=\frac{12\tan 40^o}{1-\tan 40^o}}\tag 1$$ If $\alpha$ is angle at $C$ then

In right $\Delta PQC$ $$\tan \alpha=\frac{PQ}{CQ}=\frac{h}{CQ}$$ $$\implies \color{blue}{CQ=h\cot \alpha}$$ $$\implies BC=BQ-CQ \implies 8=h-h\cot \alpha$$ $$\implies \cot \alpha =\frac{h-8}{h}$$ $$\implies \tan\alpha=\frac{h}{h-8}$$ Now, substituting the value of $h$ from eq(1) in the above expression, we get $$\implies \tan\alpha=\frac{\frac{12\tan 40^o}{1-\tan 40^o}}{\frac{12\tan 40^o}{1-\tan 40^o}-8}$$ $$=\frac{12\tan 40^o}{20\tan 40^o-8}=\frac{3\tan 40^o}{5\tan 40^o-2}$$
$$\implies \color{blue}{\alpha=\tan^{-1}\left(\frac{3\tan 40^o}{5\tan 40^o-2}\right)\approx 48.91^o}$$