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 $\tan 60=\frac{h}{x}$


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

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

  • 1
    $\begingroup$ At those angles, the hill is not distant at all. $\endgroup$
    – Matt
    Jan 26 at 4:17
  • $\begingroup$ You solved based on the first two equations. Your solution does not satisfy the third equation. $\endgroup$
    – Matt
    Jan 26 at 4:21
  • 1
    $\begingroup$ Did you use $h = tan(60)x = tan(45)(x+1) = tan(30)(x+2)$? $\endgroup$
    – joseville
    Jan 26 at 4:23
  • 2
    $\begingroup$ With three equations for two variables, you should be suspicious that you have not modeled the problem correctly. There is another degree of freedom that you did not consider. There is no reason to assume that the peak of the hill lies on the straight road. In fact, the road is specified to be straight and horizontal, so the peak clearly cannot be on the straight horizontal road. $\endgroup$
    – Matt
    Jan 26 at 4:30
  • 1
    $\begingroup$ @Matt the peak is not directly on the road, but I think the way OP has modeled it is that the peak is $h$ above the road (if the road were imaginarily extended to cut through the mountain/hill). On the other hand, maybe the road, even if extended, does not cross the point below the peak of the hill. Maybe it's a road that is off to the side of the hill/tangent to the hill in some way. $\endgroup$
    – joseville
    Jan 26 at 4:47

1 Answer 1


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


(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 $d > -\frac{1}{2}$.


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.


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  • 3
    $\begingroup$ Great answer. Thank you. $\endgroup$ Jan 26 at 8:31

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