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Concept 05 / 11Trigonometry

Heights and Distances

Angles of elevation and depression, solving for heights and distances with right triangles, the two-observation formula, problems in three dimensions and on inclined planes, and using the sine rule when right triangles are not available.

Figure 1
The angle of elevation of an object above the observer's eye level is the angle between the horizontal and the line of sight upward; the angle of depression of an object below is the angle between the horizontal and the line of sight downward. Because the horizontal lines at the observer and the object are parallel, the angle of depression from AA to BB equals the angle of elevation from BB to AA (alternate angles). Basic method. Draw a clear figure, mark every right angle, and write one right-triangle relation per unknown, usually with tan⁡\tan: height == horizontal distance ×tan⁡(elevation)\times\tan(\text{elevation}). Keep unknowns symbolic until the end, then eliminate. Two observations in line with a tower. From points AA and BB, a distance dd apart on the same side and in line with the foot of a tower of height hh, the elevations are α\alpha (farther point) and β\beta (nearer point), as in the figure. The horizontal distances are hcot⁡αh\cot\alpha and hcot⁡βh\cot\beta, differing by dd: h(cot⁡α−cot⁡β)=dh(\cot\alpha - \cot\beta) = d, so h=dcot⁡α−cot⁡β=dtan⁡αtan⁡βtan⁡β−tan⁡α=dsin⁡αsin⁡βsin⁡(β−α)h = \frac{d}{\cot\alpha - \cot\beta} = \frac{d\tan\alpha\tan\beta}{\tan\beta - \tan\alpha} = \frac{d\sin\alpha\sin\beta}{\sin(\beta - \alpha)}. The last form comes from the sine rule in the oblique triangle formed by AA, BB and the top. If the points are on opposite sides of the tower, the distances add: h(cot⁡α+cot⁡β)=dh(\cot\alpha + \cot\beta) = d. Useful standard results. If the elevation of the top of a tower from a point is θ\theta and from a point dd closer is 2θ2\theta, then the triangle formed is isosceles (the exterior angle 2θ2\theta equals the sum of two angles θ\theta), so the distance from the nearer point to the top is dd and h=dsin⁡2θh = d\sin 2\theta. If a flagpole of height pp stands on a tower of height hh, and both subtend the same angle at a point, the angle bisector theorem or the tangent of a difference gives the relation between distances. The angle subtended by a vertical segment at a point is greatest when the point lies on the circle through the ends of the segment that touches the ground (useful in 'best viewing distance' problems): the optimal distance is h1h2\sqrt{h_{1}h_{2}} for a segment from height h1h_{1} to h2h_{2}. Three-dimensional problems. Observations from points not in a straight line with the foot (for example, due south and then due east) lead to a right triangle on the ground whose sides are hcot⁡αh\cot\alpha and hcot⁡βh\cot\beta; use Pythagoras on the ground plane: d2=h2(cot⁡2α+cot⁡2β)d^{2} = h^{2}(\cot^{2}\alpha + \cot^{2}\beta). Reflections and depressions: the image of a cloud in a lake appears as far below the water surface as the cloud is above it. With elevation α\alpha of the cloud and depression β\beta of its reflection from a height hh above the lake, the cloud's height is H=h(tan⁡β+tan⁡α)tan⁡β−tan⁡αH = \frac{h(\tan\beta + \tan\alpha)}{\tan\beta - \tan\alpha}. Key Takeaways: - Elevation from AA to BB equals depression from BB to AA. - Height == horizontal distance ×tan⁡(elevation)\times\tan(\text{elevation}). - Same side: h=dcot⁡α−cot⁡β=dsin⁡αsin⁡βsin⁡(β−α)h = \frac{d}{\cot\alpha - \cot\beta} = \frac{d\sin\alpha\sin\beta}{\sin(\beta - \alpha)}; opposite sides: h=dcot⁡α+cot⁡βh = \frac{d}{\cot\alpha + \cot\beta}. - Elevations θ\theta and 2θ2\theta with separation dd: h=dsin⁡2θh = d\sin 2\theta. - 3D: ground distances form a right triangle; d2=h2(cot⁡2α+cot⁡2β)d^{2} = h^{2}(\cot^{2}\alpha + \cot^{2}\beta). - Cloud and reflection: H=htan⁡β+tan⁡αtan⁡β−tan⁡αH = h\frac{\tan\beta + \tan\alpha}{\tan\beta - \tan\alpha}. - Pitfall: measuring elevations from the ground instead of from the observer's eye height when a height is given. Brain-Teaser 1: From a point on the ground, the elevation of a tower's top is 30∘30^{\circ}. Moving 2020 m towards it, the elevation becomes 60∘60^{\circ}. Find the height. Solution: With α=30∘\alpha = 30^{\circ}, β=60∘=2α\beta = 60^{\circ} = 2\alpha: h=dsin⁡2α=20sin⁡60∘=103≈17.3h = d\sin 2\alpha = 20\sin 60^{\circ} = 10\sqrt{3} \approx 17.3 m. Check with the general formula: 20cot⁡30∘−cot⁡60∘=203−13=2032=103\frac{20}{\cot 30^{\circ} - \cot 60^{\circ}} = \frac{20}{\sqrt{3} - \frac{1}{\sqrt{3}}} = \frac{20\sqrt{3}}{2} = 10\sqrt{3}. The isosceles-triangle shortcut and the formula agree. Brain-Teaser 2: A person due south of a tower sees its top at elevation 60∘60^{\circ}; walking 100100 m east, the elevation becomes 30∘30^{\circ}. Find the height. Solution: The distances from the foot are hcot⁡60∘=h3h\cot 60^{\circ} = \frac{h}{\sqrt{3}} and hcot⁡30∘=h3h\cot 30^{\circ} = h\sqrt{3}. These are the hypotenuse and a leg of a right triangle on the ground with the eastward walk as the other leg: 3h2=h23+10023h^{2} = \frac{h^{2}}{3} + 100^{2}. So 8h23=10000\frac{8h^{2}}{3} = 10000 and h=10038=256≈61.2h = 100\sqrt{\frac{3}{8}} = 25\sqrt{6} \approx 61.2 m. The key is realising the ground triangle, not the vertical one, carries the 100100 m. Brain-Teaser 3: From a point on the ground 2020 m from the foot of a building, the angles of elevation of the bottom and the top of a flagstaff standing on the building are 45∘45^{\circ} and 60∘60^{\circ}. Find the heights of the building and of the flagstaff. Solution: The building's height is 20tan⁡45∘=2020\tan 45^{\circ} = 20 m. The top of the flagstaff is at height 20tan⁡60∘=20320\tan 60^{\circ} = 20\sqrt{3} m. So the flagstaff is 203−20=20(3−1)≈14.620\sqrt{3} - 20 = 20(\sqrt{3} - 1) \approx 14.6 m tall. Both elevations are measured from the same point at the same horizontal distance, so each height is simply that distance times the tangent, and the flagstaff is the difference. A common error is to use tan⁡(60∘−45∘)\tan(60^{\circ} - 45^{\circ}), which is the tangent of the angle the flagstaff subtends, not a ratio of lengths.

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