Three Dimensional Trigonometry


One of the most effective ways to solve Three Dimensional Trigonometry questions is to list all of the trigonometric ratios appeared in the questions. Then connect each relevant ratios to the question.

Worked Examples of Three Dimensional Trigonometry

Question 1

\(X\) and \(Y\) are two points on the same bank of a straight river and \(B\) is the bottom of a vertical pole on the other bank directly opposite \(X\). \(P\) is the mid-point of \(XY\). \(\theta_1, \theta_2, \theta_3 \) are the angles of elevation of the top \(T\) of the pole from \(X\), \(Y\), and \(P\) respectively. Prove that \( 4 \cot^2 \theta_3 = 3 \cot^2 \theta_1 + \cot^2 \theta_2 \), given that \( \angle BXY = 90^{\circ}\).


\( \begin{aligned} \displaystyle \require{color}
\text{Let } TB &= h \\
\cot \theta_1 &= \frac{BX}{h} \\
BX &= h \cot \theta_1 \color{red} \cdots (1) \\
\cot \theta_2 &= \frac{BY}{h} \\
BY &= h \cot \theta_2 \color{red} \cdots (2) \\
\cot \theta_3 &= \frac{BP}{h} \\
BP &= h \cot \theta_3 \color{red} \cdots (3) \\
BX^2 + XY^2 &= BY^2 &\color{red} \triangle BXY \\
XY^2 &= BY^2 – BX^2 \color{red} \cdots (4) \\
BX^2 + XP^2 &= BP^2 &\color{red} \triangle BXP \\
XP^2 &= BP^2 – BX^2 \color{red} \cdots (5) \\
XY &= 2 XP \\
XY^2 &= 4 XP^2 \color{red} \cdots (6) \\
BY^2 – BX^2 &= 4 \big(BP^2 – BX^2\big) &\color{red} \text{put } (4), (5) \text{ into } (6) \\
&= 4 BP^2 – 4 BX^2 \\
4 BP^2 &= 3 BX^2 + BY^2 \color{red} \cdots (7) \\
4 h^2 \cot^2 \theta_3 &= 3 h^2 \cot^2 \theta_1 + h^2 \cot^2 \theta_2 &\color{red} \text{put } (1), (2), (3) \text{ into } (7) \\
\therefore 4 \cot^2 \theta_3 &= 3 \cot^2 \theta_1 + \cot^2 \theta_2 \\
\end{aligned} \\ \)

Question 2

A bushwalker on a horizontal straight road that runs north-east observes from a point \(P\) that a hill bears north-west and its peak \(R\) has an angle of elevation of \(15^{\circ} \). On walking 200 m farther along this road to a point \(Q\), the angle of elevation of \(R\) is now \(12^{\circ} \). Find the height \(h\) of the hill to the nearest metre.


\( \begin{aligned} \displaystyle \require{color}
\tan 15^{\circ} &= \frac{h}{NP} \\
NP &= h \cot 15^{\circ} \color{red} \cdots (1) \\
\tan 12^{\circ} &= \frac{h}{NQ} \\
NQ &= h \cot 12^{\circ} \color{red} \cdots (2) \\
NP^2 + PQ^2 &= NQ^2 &\color{red} \triangle NPQ \\
h^2 \cot^2 15^{\circ} + 200^2 &= h^2 \cot^2 12^{\circ} &\color{red} \text{by } (1) \text{ and } (2) \\
h^2 \cot^2 12^{\circ} – h^2 \cot^2 15^{\circ} &= 200^2 \\
h^2 \big( \cot^2 12^{\circ} – \cot^2 15^{\circ} \big) &= 200^2 \\
h^2 &= \frac{200^2}{\cot^2 12^{\circ} – \cot^2 15^{\circ}} \\
\therefore h&\approx 70 \\
\end{aligned} \\ \)

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