The angles of elevation of the top of a distant hill in the forest as seen from three consecutive km stones on a straight horizontal road are 30, 45 and 60 degrees. Find the height of the hill.\

My try: Let km stone C,B, A are at distances $x,x+1,x+2$ from base of hill $OP$
Then $\tan 60=\frac{h}{x}$

$\tan45=\frac{h}{x+1}$

$\tan 30=\frac{h}{x+2}$

I solved and got $h=\frac{3+\sqrt3}{2}$ but answer is $\sqrt{\frac{3}{2}}$

1 Answer

Following up on @Matt's comment, the answer is indeed $\boxed{h = \sqrt{\frac{3}{2}}}$ when you consider a more general scenario.

Setup

(Diagram at the bottom.)

Let $P$ be the peak of the hill that lies $h > 0$ above the origin $O$, which is a horizontal distance $r \geq 0$ from the closest point on the horizontal straight road $R$. $A$, $B$, and $C$ lie on the road distances $d$, $d+1$, and $d+2$ from $R$, respectively, and distances $0 < a < b < c$ from $O$, respectively. The closer a point is to the origin $O$, the bigger the angle to the peak $P$ (i.e. $a < b$ implies $\measuredangle OAP > \measuredangle OBP$); thus the angles from points on the road $A$, $B$, and $C$ to the peak $P$ must be $60° > 45° > 30°$, respectively, as implied by $a < b < c$.

Negative $d$

Note, $d$ can be negative, but it can't be less than or equal to $-\frac{1}{2}$. Otherwise, we'd not have $a < b < c$. For example, if $d = -\frac{1}{2}$, then $a = b$, which implies $\measuredangle OAP = \measuredangle OBP$, which directly contradicts $60° = \measuredangle OAP > \measuredangle OBP = 45°$. Further, if $d < -\frac{1}{2}$, then $a > b$, which implies $\measuredangle OAP < \measuredangle OBP$, which also directly contradicts $60° = \measuredangle OAP > \measuredangle OBP = 45°$.

In summary $d > -\frac{1}{2}$.

Solution

We have

$$ \begin{aligned} h &= \tan(60°)a = \tan(45°)b = \tan(30°)c\\ &= \sqrt{3} a = b = \frac{1}{\sqrt{3}} c &&[h] \end{aligned} $$

Square $[h]$

$$ \begin{aligned} 3 a^2 &= b^2 = \frac{1}{3} c^2\\ &\implies a^2 = \frac{1}{3} b^2 &&[*a \to b]\\ &\implies c^2 = 3b^2 &&[*c \to b] \end{aligned} $$

We also have

$$ \begin{aligned} a^2 &= d^2 + r^2 &&[*a]\\ b^2 &= (d+1)^2 + r^2 &&[*b]\\ c^2 &= (d+2)^2 + r^2 &&[*c]\\ \end{aligned} $$

These are valid for positive $d$, of course, but, importantly, for $-\frac{1}{2} < d \leq 0$ as well.

We have a system of $6$ linear equations in $6$ unknowns, $h, r, d, a, b, c$. Hurray!

Subtract $[a]$ from $[b]$

$$b^2 - a^2 = 2d + 1 \qquad[*ba] := [b] - [a]$$

Subtract $[b]$ from $[c]$

$$ \begin{aligned} c^2 - b^2 &= 4d + 4 - (2d + 1) &&[c] - [b]\\ &= 2d + 3 &&[*cb] \end{aligned} $$

We have $\operatorname{RHS}[cb] = 2d + 3 = \operatorname{RHS}[ba] + 2$, so

$$ \begin{aligned} \operatorname{LHS}[cb] &= \operatorname{LHS}[ba] + 2\\ c^2 - b^2 &= b^2 - a^2 + 2\\ c^2 &= 2b^2 - a^2 + 2\\ 3b^2 &= 2b^2 - \frac{1}{3} b^2 + 2 &&[* a \to b] \text{ and } [* c \to b]\\ (3 - 2 + \frac{1}{3}) b^2 &= 2\\ \frac{4}{3} b^2 &= 2\\ b^2 &= \frac{3}{2}\\ b &= \sqrt{\frac{3}{2}}\\ &\boxed{h = b = \sqrt{\frac{3}{2}}} &&[h] \end{aligned} $$

It's also possible to find the values of the other $5$ variables.

Diagram