Some Applications of Trigonometry — chapter notes

The line of sight is the line from the observer's eye to the object. The horizontal through the eye is the reference line for every angle in this chapter.

If the object is above the horizontal, the angle between the horizontal and the line of sight is the angle of elevation. It is measured at the observer. As the observer walks towards the object, the angle of elevation increases.

All problems use angles of $30^\circ$, $45^\circ$ and $60^\circ$ and at most two right triangles; exact answers use surds, and decimal answers use $\sqrt3 \approx 1.732$, $\sqrt2 \approx 1.414$.

If the object is below the horizontal through the eye, the angle between the horizontal and the line of sight is the angle of depression. Draw the horizontal through the observer (dashed) and measure down from it — never from the vertical.

The horizontal at the top and the ground are parallel, and the line of sight is a transversal, so

Use this to move the angle into the right triangle at ground level.

Method: draw the figure → mark the angle at the observer and the right angle → label the known side and the unknown → choose the ratio → solve, rationalise, approximate.

• Height and ground distance: $\tan\theta = \dfrac{\text{height}}{\text{distance}}$. A gopuram seen at $60^\circ$ from $40$ m: $h = 40\sqrt3 \approx 69.28$ m.
• Height and slant line: $\sin\theta$. A tether at $30^\circ$ holding a drone $45$ m up is $90$ m long.
• Slant line and ground distance: $\cos\theta$. An $8$ m ladder at $60^\circ$ has its foot $4$ m out.
• Shadows: shadow $= \dfrac{h}{\tan\theta}$; at $45^\circ$ the shadow equals the height; higher sun, shorter shadow.

Angles are measured at the eye. If the observer's eyes are $e$ m above the ground, draw the horizontal from the eye; only the part of the object above it, $h - e$, belongs to the right triangle.

Example: eyes at $1.6$ m, a $21.6$ m flagpole at $45^\circ$: $21.6 - 1.6 = 20$, so the distance is $20$ m. Likewise, an observer on a deck or a balcony splits a cliff or building into a part below eye level (seen at a depression) and a part above (seen at an elevation).

Find the side the two triangles share, write one equation from each, and eliminate.

• Same side (points $C$, $D$ in line with the foot): distances are subtracted. With $30^\circ$ and $60^\circ$ and gap $g$: $h\sqrt3 - \dfrac{h}{\sqrt3} = g$, so $h = \dfrac{\sqrt3\,g}{2}$; e.g. $g = 40$ gives $h = 20\sqrt3$.
• Opposite sides: distances are added: $h\sqrt3 + h = 100$ gives $h = 50(\sqrt3 - 1)$.
• Object on top (tank on a tower, sign on a pole): the triangles share the base. Tower $12$ m at $45^\circ$, top of tank at $60^\circ$: tank $= 12(\sqrt3 - 1)$.
• Two buildings: a horizontal from the top of the shorter one splits the taller into $h$ and $H - h$.

From the top of a tower of height $h$, objects at depressions $\alpha$ and $\beta$ are $\dfrac{h}{\tan\alpha}$ and $\dfrac{h}{\tan\beta}$ from the foot. The larger angle of depression belongs to the nearer object. Same side: subtract; opposite sides: add.

Moving objects: if an object moves (boat, car, drone) and the angle changes, the distance moved is the difference of the two distances; speed $= \dfrac{\text{distance}}{\text{time}}$. A drone at $8$ m/s whose elevation falls from $60^\circ$ to $30^\circ$ in $10$ s flies at $h = 40\sqrt3 \approx 69.28$ m.

Angles do not scale like lengths: halving the distance doubles $\tan\theta$, not $\theta$.