National 5 Maths
Angles in Shapes

Note that octagons have 8 sides and regular means that each interior angle is equal.

There are two ways of thinking about this question.

Method 1: Use the exterior angles

The sum of the exterior angles is 360°. So each exterior angle is 360 ÷ 8 = 45°.

At each vertex, the interior angle and exterior angle add to 180°, so each interior angle = 180 – 45 = 135°.

Method 2: Use the angles in the centre

Draw lines connecting opposite vertices to split the octagon into eight isosceles triangles.

In the centre of the octagon, there are eight equal angles around a point. Each of these is 360 ÷ 8 = 45°.

The other two angles in each triangle are equal, so they add to 180 – 45 = 135°.

Because each interior angle of the octagon comprises two of the angles on the outsides of the triangles, each interior angle is also 135°.

This is quite a difficult 3-mark question.

First, note that \(\angle\)BDF = \(\angle\)BDO + \(\angle\)ODF.

We will find \(\angle\)BDO and \(\angle\)ODF separately, and then add them.

To find \(\angle\)BDO, note that OB is a radius and AB is a tangent, so \(\angle\)OBA \(= 90^\circ\) and \(\angle\)OBD \(= 90-77=13^\circ\small.\)

OB and OD are both radii, so triangle OBD is isosceles and \(\angle\)BDO = \(\angle\)OBD \(=13^\circ\small.\)

Now we need to find \(\angle\)ODF. It is quite easy, because DE is a diameter, so \(\angle\)DFE \(= 90^\circ\) and we can find \(\angle\)ODF just by making sure that the angles in triangle DEF add to \(180^\circ\small.\) So \(\angle\)ODF \(=180-90-64=26^\circ\small.\)

Finally, we add \(\angle\)BDO and \(\angle\)ODF. So \(\angle\)BDF = \(13+26=39^\circ\small.\)

This 3-mark question looks harder than it actually is.

\(\angle\)OAC \(=180-143=37^\circ\) because of the parallel lines.

OA is a radius and AB is a tangent, so \(\angle\)OAB \(=90^\circ\) and \(\angle\)CAB \(=90-37=53^\circ\small.\)

\(\angle\)ACB is also \(53^\circ\) because the triangle is isosceles, so now we can complete the solution using the fact that the three angles in triangle CAB add to \(180^\circ\small.\)

So \(\angle\)ABC \(=180-2\times 53 = 74^\circ\small.\)

This is a trickier 3-mark question, because it involves quite a few steps.

Note, firstly, that both triangles are isosceles, because OB, OC and OD are all radii.

Now note that the angles OBA and OBE are both right-angles, because the radius OB meets a tangent at B. Also, \(\angle\)CBD \(=90^\circ\small,\) because CD is a diameter.

We can now work our way down the angles at the point B:
• Angle OBD \(=90-58=32^\circ\)
• Angle OBC \(=90-32=58^\circ\)
• Angle ABC \(=90-58=32^\circ\)

We have now found one of the angles in triangle ABC, so next we need to find \(\angle\)ACB. We do this by noting that \(\angle\)OCB = \(\angle\)OBC \(=58^\circ\small,\) because triangle OBC is isoseles, so \(\angle\)ACB \(=180-58 = 122^\circ\small.\)

Now we know both of the other angles in triangle ABC, so \(\angle\)CAB \(=180-32-122 = 26^\circ\small.\)

This is a relatively easy 3-mark question.

\(\angle\)COE \(=180-68=112^\circ\)

Triangle OCE is isosceles so \(\angle\)OCE \(=(180\!-\!112)\div 2 = 34^\circ\)

\(\angle\)ACO \(=90^\circ\) because OC is a radius and AC is a tangent.

So \(\angle\)ACE \(=90+34=124^\circ\small.\)

This unfamiliar looking 2-mark question is surprisingly easy.

Each exterior angle of a regular decagon is \(360 \div 10 = 36^\circ\small.\)

The angle inside the H shape is obviously \(90^\circ\small.\)

So the shaded angle is \(90+36=126^\circ\small.\)