Angles and Lines: 11 Applications

Angles on a Straight Line in a Ratio

propertyAngles on a straight line add to 180°; share it by units

AOB is a straight line. Rays OC and OD split the angle above it so that ∠ AOC : ∠ COD : ∠ DOB = 2 : 3 : 4. Find ∠ COD and ∠ AOD.

Three angles on one side of the straight line AOB, in the ratio 2 : 3 : 4.

AOB is a straight line, so the three angles along it add to 180^°.

step 1 of 6

The three angles lie on one side of a straight line, so they share 180^° in the ratio given. Find one unit, then the parts the question asks for.

All steps

AOB is a straight line, so the three angles along it add to 180^°.
The ratio 2 : 3 : 4 makes 2 + 3 + 4 = 9 units.
9 units = 180^°, so 1 unit = 180^° ÷ 9 = 20^°.
∠ COD = 3 units = 3 × 20^° = 60^°.
∠ AOD is ∠ AOC and ∠ COD together: 2 + 3 = 5 units.
∠ AOD = 5 × 20^° = 100^°.

answer∠ COD = 60^°; ∠ AOD = 100^°

techniqueAngles on a Line · Part-to-Whole Ratios

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

Sharing 360^° among the units instead of 180^°: rays on one side of a straight line make half a turn, not a whole one.
Reading ∠ AOD as the single part ∠ DOB (4 units) instead of the two parts from OA round to OD.

Angles at a Point

propertyAngles round a point add to 360°; remove the known, then share the rest

Four rays OA, OB, OC and OD meet at O. ∠ AOB = 90^° and ∠ BOC = 115^°. ∠ DOA is 15^° more than ∠ COD. Find ∠ COD and ∠ DOA.

Four rays from O. The right angle and the 115° are given; the other two differ by 15°.

The four angles around O make a full turn: they add to 360^°.

step 1 of 6

The four angles make a full turn. Two are known; the other two add to what is left and differ by 15^°, which is a sum-and-difference share.

All steps

The four angles around O make a full turn: they add to 360^°.
The two known angles: ∠ AOB + ∠ BOC = 90^° + 115^° = 205^°.
What is left for ∠ COD and ∠ DOA together: 360^° − 205^° = 155^°.
∠ DOA is 15^° more than ∠ COD. Take that 15^° off first: 155^° − 15^° = 140^° is two equal shares.
∠ COD = 140^° ÷ 2 = 70^°.
∠ DOA = 70^° + 15^° = 85^°. Check: 90 + 115 + 70 + 85 = 360.

answer∠ COD = 70^°; ∠ DOA = 85^°

techniqueAngles at a Point

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

Adding the angles to 180^° as if O sat on a straight line; four rays from one point make a full turn.
Halving 155^° straight away and giving 77.5^°: the two angles are not equal, one is 15^° more.

Vertically Opposite Angles at a Three-Line Crossing

propertyVertically opposite angles are equal; the rest comes off the straight line

Three straight lines AB, CD and EF meet at O. ∠ AOC = 38^° and ∠ COE = 65^°. Find ∠ BOD and ∠ AOF.

Three straight lines through O. Two angles on the upper side of AB are given.

AB, CD and EF are straight lines through O, so each pair of opposite angles at O is equal.

step 1 of 5

Every angle at O has a twin across the point. One asked angle is a twin of a given one; the other is the twin of the third angle on the upper side of AB.

All steps

AB, CD and EF are straight lines through O, so each pair of opposite angles at O is equal.
∠ BOD is vertically opposite ∠ AOC: ∠ BOD = 38^°.
Along the straight line AB, the angles on the upper side add to 180^°: ∠ AOC + ∠ COE + ∠ EOB = 180^°.
∠ EOB = 180^° − 38^° − 65^° = 77^°.
∠ AOF is vertically opposite ∠ EOB: ∠ AOF = 77^°.

answer∠ BOD = 38^°; ∠ AOF = 77^°

techniqueVertically Opposite Angles · Angles on a Line

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

Pairing ∠ BOD with ∠ COE because both sit near D: opposite angles share the vertex and lie across it, on the same two lines.
Taking ∠ AOF = 65^°: OF is opposite OE, so ∠ AOF matches the angle from OE to OB, not the one from OC to OE.

Perpendicular Lines and a Ray Between Them

propertyPerpendicular lines make right angles; the ray splits one of them

Straight lines PQ and RS are perpendicular and cross at O. Ray OT lies between OQ and OS, with ∠ QOT = 27^°. Find ∠ TOS and ∠ TOR.

PQ and RS cross at right angles. OT lies between OQ and OS.

PQ ⊥ RS, so the angles at O between the lines are right angles: ∠ QOS = 90^° and ∠ QOR = 90^°.

step 1 of 5

Perpendicular lines give four right angles at O. One right angle is split by OT; the other asked angle reaches across a second right angle.

All steps

PQ ⊥ RS, so the angles at O between the lines are right angles: ∠ QOS = 90^° and ∠ QOR = 90^°.
OT lies inside ∠ QOS with ∠ QOT = 27^°.
∠ TOS = 90^° − 27^° = 63^°.
∠ TOR runs from OT through OQ to OR: 27^° + 90^° = 117^°.
Check on the straight line RS: ∠ TOS + ∠ TOR = 63^° + 117^° = 180^°.

answer∠ TOS = 63^°; ∠ TOR = 117^°

techniqueParallel and Perpendicular Lines · Complementary Angles

examsPSLE · SAT · GCSE Higher

Common pitfalls

Writing ∠ TOR = 180^° − 27^° = 153^°: that is the angle from OT to OP, on the straight line PQ, not to OR.
Treating 63^° and 27^° as a pair adding to 180^°; they add to one right angle.

Parallel Lines with One Transversal

propertyCorresponding angles are equal (F); alternate angles are equal (Z); co-interior angles add to 180° (C)

AB is parallel to CD. A straight line EF cuts AB at G and CD at H. ∠ EGB = 118^°. Find ∠ GHD and ∠ CHG.

EF crosses the parallels AB and CD at G and H. One angle is given.

AB ∥ CD and EF crosses both, so each angle at G has an equal partner at H.

step 1 of 5

One transversal across two parallels makes eight angles of only two sizes. Name the pair that links the given angle to each asked one.

All steps

AB ∥ CD and EF crosses both, so each angle at G has an equal partner at H.
∠ EGB and ∠ GHD are corresponding angles (the F shape): ∠ GHD = 118^°.
CD is a straight line, so ∠ CHG = 180^° − 118^° = 62^°.
Check with alternate angles (the Z shape): ∠ BGH = 180^° − 118^° = 62^°, and ∠ BGH = ∠ CHG.
Check with co-interior angles (the C shape): ∠ BGH + ∠ GHD = 62^° + 118^° = 180^°.

answer∠ GHD = 118^°; ∠ CHG = 62^°

techniqueAngles in Parallel Lines

examsO-Level · SAT · GCSE Higher

Common pitfalls

Making every angle at the transversal 118^°: only the four obtuse ones are; the other four are 62^°.
Calling ∠ EGB and ∠ CHG alternate angles: they are on the same side of the transversal, and are neither equal nor a Z pair.

Parallel Lines with a Bent Transversal

propertyDraw a third parallel through the bend; the angle splits into two alternate angles

AB is parallel to CD. Point E lies between the two lines, with ∠ ABE = 34^° and ∠ EDC = 51^°. Find ∠ BED.

AB and CD are parallel. The path B-E-D bends at E, between them.

E sits between the parallel lines, so no single Z or F shape reaches from B to D.

step 1 of 6

No Z or F shape reaches from B to D because the path bends at E. A line through E parallel to both cuts the unknown angle into two alternate angles, one with each parallel, both on the far side of the bend from A and C.

All steps

E sits between the parallel lines, so no single Z or F shape reaches from B to D.
Draw a line through E parallel to AB and CD, and mark a point X on it to the right of E.
AB ∥ EX: ∠ BEX is alternate to ∠ ABE, so ∠ BEX = 34^°.
CD ∥ EX: ∠ XED is alternate to ∠ EDC, so ∠ XED = 51^°.
∠ BED is the two parts together: 34^° + 51^° = 85^°.
The line through E is the whole method: it turns one bent transversal into two straight ones.

answer∠ BED = 85^°

techniqueAngles in Parallel Lines

examsO-Level · SAT · GCSE Higher

Common pitfalls

Subtracting, 51^° − 34^° = 17^°: the parallel through E shows the two alternate angles sit side by side, so they add.
Answering 180^° − 85^° = 95^° from an imagined triangle BED; BD is not drawn and nothing says it is.

Three Parallel Lines with Two Transversals

propertyWork one transversal at a time; at the shared point, add the two parts

AB, CD and EF are three parallel lines. G is on AB, H on CD and K on EF; GH and HK are straight. ∠ AGH = 112^° and ∠ HKE = 47^°. Find ∠ GHK.

Three parallels. The path G-H-K bends on the middle line.

AB ∥ CD ∥ EF. GH crosses the top two lines; HK crosses the bottom two. At H the path bends.

step 1 of 6

The path G-H-K bends on the middle line. Each half is a plain transversal across two parallels; the answer is the two angles at H, one from each half, added.

All steps

AB ∥ CD ∥ EF. GH crosses the top two lines; HK crosses the bottom two. At H the path bends.
On the transversal GH: ∠ AGH and ∠ GHC are co-interior, so ∠ GHC = 180^° − 112^° = 68^°.
CD is a straight line: ∠ GHD = 180^° − 68^° = 112^°.
On the transversal HK: ∠ DHK and ∠ HKE are alternate angles, so ∠ DHK = 47^°.
∠ GHK runs from HG through HD to HK: 112^° + 47^° = 159^°.
Check: the angle on the other side of the bend is 360^° − 159^° = 201^°, a reflex angle, so 159^° is the one inside the bend.

answer∠ GHK = 159^°

techniqueAngles in Parallel Lines · Angle Chases

examsO-Level · SAT · GCSE Higher

Common pitfalls

Using ∠ GHC = 68^° on the wrong side of H and giving 68^° + 47^° = 115^°: HC and HK are on opposite sides of the transversal GH.
Reading ∠ HKE as corresponding to ∠ GHD: they belong to different transversals, so no F, Z or C shape joins them.

Exterior Angle of a Triangle

propertyAn exterior angle equals the sum of the two interior angles opposite it

In triangle ABC, side BC is extended to D. ∠ ACD = 121^° and ∠ ABC = 67^°. Find ∠ BAC.

BC is extended to D, so ∠ ACD sits outside the triangle.

BC is extended to D, so ∠ ACD is an exterior angle of triangle ABC.

step 1 of 5

The exterior angle at C is made of the two interior angles that are not at C. One of them is given, so the other is a subtraction; the triangle's 180^° checks it.

All steps

BC is extended to D, so ∠ ACD is an exterior angle of triangle ABC.
An exterior angle equals the sum of the two interior angles opposite it: ∠ ACD = ∠ BAC + ∠ ABC.
∠ BAC = 121^° − 67^° = 54^°.
Check inside the triangle: ∠ ACB = 180^° − 121^° = 59^°, from the straight line BCD.
54^° + 67^° + 59^° = 180^°.

answer∠ BAC = 54^°

techniqueAngles in a Triangle · Angles on a Line

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

Writing ∠ BAC = 180^° − 121^° − 67^°: that mixes an exterior angle into the triangle's own 180^°.
Taking the exterior angle as 180^° minus the opposite angle ∠ BAC; it is 180^° minus the adjacent one, ∠ ACB.

Bisectors of Two Angles on a Straight Line

propertyHalf of each angle on a straight line adds to half of 180°

AOB is a straight line and ∠ AOC = 70^°. OD bisects ∠ COB and OE bisects ∠ AOC. Find ∠ DOE.

On the straight line AOB, ∠ COB = 180° − 70° = 110°.

AOB is a straight line, so ∠ AOC + ∠ COB = 180^° and ∠ COB = 180^° − 70^° = 110^°.

step 1 of 6

Each bisector halves its angle, and the two halves next to OC make ∠ DOE. Because the two whole angles add to 180^°, the halves add to 90^° whatever ∠ AOC is.

All steps

AOB is a straight line, so ∠ AOC + ∠ COB = 180^° and ∠ COB = 180^° − 70^° = 110^°.
OD bisects ∠ COB: ∠ COD = 110^° ÷ 2 = 55^°.
OE bisects ∠ AOC: ∠ EOC = 70^° ÷ 2 = 35^°.
∠ DOE is the two halves next to OC: ∠ EOC + ∠ COD = 35^° + 55^° = 90^°.
In general: half of ∠ AOC plus half of ∠ COB is half of 180^°.
So the bisectors of two angles on a straight line are always perpendicular, whatever ∠ AOC is.

answer∠ DOE = 90^°

techniqueAngles on a Line · Constructing an Angle Bisector

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

Halving 70^° and stopping: ∠ DOE needs the half of ∠ COB as well.
Drawing OD inside ∠ AOC: OD bisects ∠ COB, on the other side of OC from OE.

Reflected Path off a Flat Wall

propertyThe path leaves the wall at the angle it arrived; the wall is a straight line

A ball rolls from Q toward a straight wall XY, hits it at P and rolls away to R. It leaves the wall at the same angle it arrived. The path QP makes an angle of 34^° with the wall. Find ∠ QPR, the angle between the path in and the path out.

The ball comes in from Q at 34° to the wall and leaves toward R.

A ray bounces off a flat wall at the same angle it arrived: the angle between the incoming path and the wall equals the angle between the outgoing path and the wall.

step 1 of 5

A bounce is a reflection: the angle with the wall is the same on both sides of P. The wall is a straight line through P, so the angle between the two paths is what is left of 180^°.

All steps

A ray bounces off a flat wall at the same angle it arrived: the angle between the incoming path and the wall equals the angle between the outgoing path and the wall.
The incoming path makes 34^° with the wall at P, so the outgoing path makes 34^° with the wall on the other side of P.
The wall XY is a straight line through P: the three angles at P on the ball's side add to 180^°.
∠ QPR = 180^° − 34^° − 34^° = 112^°.
Measured from the perpendicular to the wall instead, each path is 90^° − 34^° = 56^° from it, and 56^° + 56^° = 112^° again.

answer∠ QPR = 112^°

techniqueReflection · Angles on a Line

examsGCSE Higher

Common pitfalls

Doubling 34^° and answering 68^°: that is what the two paths make with the wall together, not the angle between the paths.
Answering 56^°, the angle from the perpendicular, when the question asks for the angle between the two paths.

A Pointer Turned Both Ways, Several Times Round

propertyA whole turn is 360° and leaves the direction unchanged; the net clockwise rotation is the clockwise turning less the counter-clockwise turning

A surveying pointer starts facing North. It makes 3 complete turns clockwise, then turns 225^° clockwise, and finally turns counter-clockwise through 112 right angles. (a) In which compass direction does the pointer face at the end? (b) The net clockwise rotation is the angle turned clockwise minus the angle turned counter-clockwise. Find the pointer's net clockwise rotation in degrees. (c) The pointer is then fixed to an arm that turns clockwise at 0.5^° per second from 09:15 to 10:55. Find the total angle the arm turns through, in right angles.

Three complete turns: 3 × 360° = 1080°, and the pointer faces North again.

Three complete turns are 3 × 360^° = 1080^° clockwise, and after each whole turn the pointer faces North again.

step 1 of 6

A complete turn is 360^° and brings the pointer back to where it started, so only the part turns change the direction. The net rotation keeps count of everything: the clockwise turning counts forward and the counter-clockwise turning counts back.

All steps

Three complete turns are 3 × 360^° = 1080^° clockwise, and after each whole turn the pointer faces North again.
Turning 225^° clockwise from North passes East at 90^° and South at 180^°, then goes 45^° further: the pointer faces South-West.
112 right angles are 112 × 90^° = 135^°. Turning 135^° counter-clockwise from South-West takes 45^° back to South and 90^° more to East. (a) The pointer faces East.
(b) Clockwise the pointer turned 1080^° + 225^° = 1305^°, so the net clockwise rotation is 1305^° − 135^° = 1170^°. Check: 1170^° = 3 × 360^° + 90^°, a quarter turn clockwise past North, which is East.
From 09:15 to 10:55 is 1 h 40 min, which is 100 min, or 100 × 60 = 6000 s.
(c) The arm turns 6000 × 0.5^° = 3000^°, and 3000 ÷ 90 = 3313, so it turns through 3313 right angles.

answer(a) East (b) 1170^° (c) 3313 right angles

techniqueAngles at a Point · Angles · Time Durations

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

Adding the counter-clockwise turn to the clockwise ones and answering 1080^° + 225^° + 135^° = 1440^°: a turn the other way undoes part of the clockwise turning, so it is taken away.
Leaving out the three complete turns in (b) and answering 90^°: they do not change the direction, but the pointer still turned through them, and the net rotation counts every degree.