Heights & Distances from Angles of Elevation

Set up every height-and-distance problem the same way:

• Angle of elevation: the angle, measured upward from the horizontal, between your line of sight and the ground, when you look at an object above your eye level.
• Angle of depression: the angle, measured downward from the horizontal, when you look at an object below you (e.g. a boat seen from a lighthouse top).
• In the right triangle with vertical height $h$, horizontal base $d$, and angle of elevation $\theta$ at the observer:

With one right triangle and one given angle $\theta$:

• If the height and distance are the two legs, use $\tan\theta = \dfrac{h}{d}$ to get whichever leg is missing.
• If the line of sight (slant) is given or asked, use $\sin\theta = \dfrac{h}{\ell}$ (height to slant) or $\cos\theta = \dfrac{d}{\ell}$ (base to slant).
• When the angle arrives as $\tan^{-1}(p/q)$, just read $\tan\theta = p/q$ directly — no need to find $\theta$ in degrees.
• A depression angle from a high point equals the elevation angle from the low point; redraw it on the ground triangle.

If $\ell$ is the slant (line-of-sight) distance and $\theta$ the elevation, then the height is

• $\sin\dfrac{A}{2} = \sqrt{\dfrac{1-\cos A}{2}}$, $\quad\cos\dfrac{A}{2} = \sqrt{\dfrac{1+\cos A}{2}}$.
• So $\sin 67.5^\circ = \cos 22.5^\circ = \sqrt{\dfrac{1+\cos 45^\circ}{2}} = \dfrac{1}{2}\sqrt{2+\sqrt{2}}$, and $\sin 22.5^\circ = \dfrac{1}{2}\sqrt{2-\sqrt{2}}$.

Two viewing points stacked vertically (heights $0$ and $p$), same horizontal distance $d$, looking at a target of height $H$:

• From the bottom (elevation $\beta$): $\tan\beta = \dfrac{H}{d}$.
• From the top of the lower object (elevation $\alpha$, at height $p$): $\tan\alpha = \dfrac{H-p}{d}$.
• Eliminate $d$ by dividing or substituting; the target height drops out in terms of $p$ and the two angles.
• How the difference of tangents simplifies — the identity that turns the eliminated equation into the option's closed form: $\tan\beta-\tan\alpha=\dfrac{\sin(\beta-\alpha)}{\cos\alpha\cos\beta}$. Subtracting the two equations gives $p=d(\tan\beta-\tan\alpha)$, so $d=\dfrac{p\cos\alpha\cos\beta}{\sin(\beta-\alpha)}$ and $H=d\tan\beta=\dfrac{p\cos\alpha\sin\beta}{\sin(\beta-\alpha)}$. The distance from the upper observer to the top is then $\sqrt{d^2+(H-p)^2}=\dfrac{d}{\cos\alpha}$, because $d^2(1+\tan^2\alpha)=d^2\sec^2\alpha$ — $\sin^2+\cos^2=1$ collapsing a sum of squares to a single slant length.

A clean special case: if the lower object's height $p$ is itself asked, the same two equations relate $H$, $p$, and the angles.

Tower of height $T$ carrying a flagstaff of height $f$, viewed from distance $d$:

• Elevation of the tower top (flagstaff bottom): $\tan\theta = \dfrac{T}{d}$.
• Elevation of the flagstaff top: $\tan\phi = \dfrac{T+f}{d}$.
• When the two angles are related (e.g. $\phi = 2\theta$), substitute the double-angle identity $\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$ and eliminate $d$. A common clean result is $T = f\cos 2\theta$ for the $\theta,\,2\theta$ case.

A segment between heights $h_1$ (bottom) and $h_2$ (top) subtends angle $\alpha$ at a ground point distance $x$ away. With $\tan\theta_1 = \dfrac{h_1}{x}$ and $\tan\theta_2 = \dfrac{h_2}{x}$, the subtended angle is $\alpha = \theta_2 - \theta_1$, so

A ladder of length $L$ leans so its foot is distance $x$ from a vertical flagstaff and its top reaches height $H - k$ (a point $k$ below the $H$-high top). From the same foot the elevation of the flagstaff top is $\theta$:

• Trig: $\tan\theta = \dfrac{H}{x}$ (so $H = x\tan\theta$).
• Length (Pythagoras): $x^2 + (H-k)^2 = L^2$.

Substitute the first into the second and solve the resulting equation for $x$, then back out $H$. Keep slant length (Pythagoras) and elevation (tangent) as two separate equations.

Tower of height $h$ with foot $N$; three collinear ground points at elevations $\theta_1, \theta_2, \theta_3$:

• Each point's horizontal distance from the foot: $d_i = \dfrac{h}{\tan\theta_i} = h\cot\theta_i$.
• The gap between two points is the difference of their distances: $d_i - d_j = h(\cot\theta_i - \cot\theta_j)$.
• Given one gap, solve for $h$; then any other distance follows. For the classic $30^\circ\!-\!45^\circ\!-\!60^\circ$ trio the cotangents are $\sqrt{3},\,1,\,\tfrac{1}{\sqrt{3}}$.

Tower of height $h$ with foot $O$. Observer $A$ (elevation $x$) and observer $B$ (elevation $y$) lie in perpendicular ground directions from $O$, with $AB = z$:

• Each horizontal distance: $OA = h\cot x$, $OB = h\cot y$.
• If $B$ is due east of $A$ while $A$ is due south of $O$, then $\angle OAB = 90^\circ$, so the ground triangle gives $OB^2 = OA^2 + AB^2$:

Observer at height $p$ above a lake; cloud at height $H$ above the lake, so its image is $H$ below the lake surface. Horizontal distance $d$:

• Elevation of the cloud: $\tan\alpha = \dfrac{H-p}{d}$ (cloud is $H-p$ above the observer's eye).
• Depression of the image: $\tan\beta = \dfrac{H+p}{d}$ (image is $H+p$ below the eye).
• Divide to eliminate $d$: $\dfrac{H-p}{H+p} = \dfrac{\tan\alpha}{\tan\beta}$, then solve for $H$.

A sphere of radius $r$ whose centre is at distance $R$ subtends angle $\alpha$ at the eye (the angle between the two tangent sight-lines). Half of that angle sits in the right triangle formed by the eye, the centre, and the point of tangency: