To find the angle between a line and a plane it is helpful to draw a right angled triangle using the line that is perpendicular to the plane.

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In order to find the required angle you will first need to find the lengths of the side MB and BE. To find BE draw a right angled triangle ABE using the other information you are given.

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In the triangle ABE you know an angle and the length of the adjacent side. You want to find the length of the opposite side so use the tan ratio.

$\tan (35) = \frac{z}{15}$

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Multiply both sides by 15. Leave as an exact value.

$\begin{array}{rcl} z & = & 15 \times \tan (35) \\ = & 15 \tan (35) \end{array}$

To find MB draw a right angled triangle ABM using the other information you are given.

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As DM : MA = 2 : 3, you know that MA is $\frac{3}{2 + 3}$ of the length of DA.

$M B = \frac{3}{5} \times 15 = 9 cm$

In the triangle ABM you know two lengths and want to find the hypotenuse so you can use Pythagoras' theorem.

$\begin{array}{rcl} M A^{2} + A B^{2} & = & M B^{2} \\ 9^{2} + 15^{2} & = & y^{2} \\ 306 & = & y^{2} \end{array}$

Take the positive square root of both sides. Leave as an exact answer.

$y = \sqrt{306}$

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Now you can use the triangle MBE. In this triangle you want to find an angle and you know the lengths of the sides that are opposite and adjacent to the angle so use the tan ratio. Use the exact values for the lengths or use your calculator's memory function.

$\begin{array}{rcl} \tan (x) & = & \frac{15 \tan (35)}{\sqrt{306}} \\ = & 0.600423 . . . \end{array}$

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Find the angle by using the inverse tan function.

$\begin{array}{rcl} x & = & \tan^{- 1} (\frac{15 \tan (35)}{\sqrt{306}}) \\ = & \tan^{- 1} (0.600423 . . .) \\ = & 30.981574 . . . \end{array}$

Round your answer to 1 decimal place. Remember to include the units.

$Angle between E M and the base = 31.0 ° (1 dp)$ [1]

3D Pythagoras & Trigonometry (CIE IGCSE Maths: Extended)

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