The following triangle helps you to see the angle that is required.

We can work out the lengths CF and CA

Use the triangle CBF to work out CF

\(\large \tan30°=\frac{CF}{12} \qquad \qquad \tan30°=\frac{\sqrt{3}}{3}\\ \large \frac{\sqrt{3}}{3}=\frac{CF}{12}\\ \large \frac{12\sqrt{3}}{3}=CF\\ \large CF=4\sqrt{3}\)

Use Pythagoras' Theorem to work out CA

CA² = 12² + 16²

CA² = 400

CA = 20

Now put this information in the diagram

\(\large \tan\theta=\frac{4\sqrt{3}}{20}\\ \large \tan\theta=\frac{\sqrt{3}}{5}\)

Here is the triangle that contains the angle required

We shoud calculate the length AG using Pythagoras' Theorem.

G is the midpoint of BC. Therefore BG is 9cm.

AG² = 9² + 12²

AG² = 225

AG = 15

Here is the angle that we are required to find.

It is a right-angled triangle.

\(\large \tan\theta =\frac{8}{15}\\ \large\theta=\tan^{-1}(\frac{8}{15})\\ \large\theta\approx24.0°\)

For parts b) and c), the challenge is to visualise the triangles required to find the angles. The question aims to show you that the angles in part b) and c) are not the same!

b)

c)

a) WE can find the length EC using Pythagoras' Theorem

EC² = 10² + 6²

EC² = 136

\(\large EC = \sqrt{136}\)

b) The length AC is the same as EC

We have an isoceles triangle. Use just the top half of the triangle and right-angled trigonometry to find the angle ACE (or you can use the whole triangle and the cosine rule).

\(\large \cos\theta=\frac{3}{\sqrt{136}}\\ \large \theta=\cos^{-1}(\frac{3}{\sqrt{136}})\\ \large \theta\approx75.1°\)

c) The angle between EC and the plane ABCD can be visualised in this diagram.

We know the length EC =\(\large \sqrt{136}\)

We need to find another legnth in this right-anghled triangle. Find the height of the triangle:

\(\large h^2=6^2-3^2\\ h=\sqrt{27}\)

Let \(\large \alpha\) be the angle between EC and the plane ABCD

\(\large\sin\alpha=\frac{\sqrt{27}}{\sqrt{136}}\\ \large\alpha\approx26.5°\)