7.3 Solve Simple Trigonometric Problems in Three Dimensions

7.3 Solve Simple Trigonometric Problems in Three Dimensions

Example 8

The diagram shows a prism. Triangle $PQR$ is a cross section of the prism, where
$m\overline{PR} = 20$ cm, $m\overline{MP} = 12$ cm,
$m\angle PRQ = 30^\circ$ and $m\angle PQR = 90^\circ$.
Calculate the size of the angle that the line $MR$ makes with the plane $RQLN$.

[Diagram: prism with cross section triangle PQR]

Solution In right angled triangle $PQR$, we have

$\dfrac{m\overline{PQ}}{20} = \sin 30^\circ$

$m\overline{PQ} = 20 \times \dfrac{1}{2}$

$= 10$ cm

In right angled triangle $MPR$, we have

$m\overline{MR} = \sqrt{(m\overline{PR})^2 + (m\overline{MP})^2} = \sqrt{(20)^2 + (12)^2} = \sqrt{400 + 144} = \sqrt{544}$

$= 23.3238$ cm

In right angled triangle $MLR$, we have

$\sin x = \dfrac{\text{opp.}}{\text{hyp.}} = \dfrac{10}{23.3238} = 0.4287$

$\therefore x = \sin^{-1}(0.4287)$

$= 25.4^\circ$

Example 9

The diagram shows a cuboid $ABCDEFGH$ in which $m\overline{AB} = 8$ cm, $m\overline{AF} = 6$ cm and $m\overline{FC} = 16$ cm. Find the length of $\overline{BC}$.

[Diagram: cuboid ABCDEFGH]

Solution In right angled triangle $AFC$, we have

$m\overline{AC} = \sqrt{(m\overline{FC})^2 - (m\overline{AF})^2} = \sqrt{(16)^2 - (6)^2}$

$= \sqrt{256 - 36} = \sqrt{220} = 14.83$ cm

In right angled triangle $ABC$, we have

$m\overline{BC} = \sqrt{(m\overline{AC})^2 - (m\overline{AB})^2} = \sqrt{(14.83)^2 - (8)^2}$

$= \sqrt{220 - 64}$

$= \sqrt{156} = 12.5$ cm