CAIE IGCSE Mathematics 6. Topic 6 Question Bank
Practise triangle geometry using Pythagoras, trigonometric ratios, exact values, graphs and non-right-angled triangle methods.
- Syllabus
- 2028–2030
- Course
- Mathematics 0580
- Level
- Extended
Practise triangle geometry using Pythagoras, trigonometric ratios, exact values, graphs and non-right-angled triangle methods.
The quadrilateral ACDE is formed by two right-angled triangles ABE and BCD. AC=17 cm,AE=18 cm and BD=6 cm.
Show that CD=10 cm.
18AB=cos60∘ or AB=18cos60∘
M1
cos60∘=21 and AB=9
A1
Correct use of Pythagoras' theorem, i.e. CD2=62+(17−their AB)2 oe
M1
Correct evaluation for their AB: CD2=36+64 or CD=36+64
M1
100=10
A1
Find the perimeter of the quadrilateral ACDE.
Give your answer in the form p+kq.
39+93
B3 for BE=93 or for answer equivalent to 39+93 but not in required form
OR M2 for 18BE=23 oe or better
or M1 for 18BE=sin60∘ oe or better
M1 for 18+17+10+ their BE-6 oe
M2 for 182−their 92 oe
or M1 for BE2+(their 9)2=182 oe

NOT TO SCALE
The diagram shows a field, ABCD with B north of A.
BD is a path across the field.
AB=85 m,AD=72 m,BD=129 m, angle BDC=39∘ and angle BCD=60∘.
Calculate CD.
147 or 147.1...
M2 for sin60∘129sin81∘ oe
or M1 for CDsin81∘=129sin60∘ oe
Show that angle ABD=31.6∘, correct to 1 decimal place.
[cos=]2×85×129852+1292−722
M2
M1 for 722=852+1292−2×85×129cosABD31.58…
A2
A1 for 0.851 to 0.852 or 109659341 or equivalent fraction
Find the shortest distance from A to BD.
44.5 or 44.51 to 44.54
M2 for implicit correct method, e.g. 85d=sin31.6∘ oe
or M1 for recognition that the line from A is perpendicular to BD
On the axes, sketch the graph of y=sinx for 0∘⩽x⩽360∘.

Correct sketch
2
B1 for correct shape but inaccurate
Describe fully the symmetry of the graph of y=sinx for 0∘⩽x⩽360∘.
Rotational symmetry order 2
Centre (180,0)
B1 for rotational symmetry.
Solve 4sinx−1=2 for 0∘⩽x⩽360∘.
x= and x=
48.6 or 48.59 to 48.60 and 131.4 or 131.40 to 131.41
B2 for one correct value, or M1 for sinx=0.75 or better.
If 0 scored, SC1 for two answers adding to 180.

NOT TO SCALE
The diagram shows triangle ABC on horizontal ground.
AC=15 m,BC=8 m and AB=20 m.
BP and CQ are vertical poles of different heights.
BP=3 m and CQ=4 m.
AQ and PQ are straight wires.
Show that angle ACB=117.5∘, correct to 1 decimal place.
cosACB=2×15×8152+82−202
M2
117.54 to 117.55
A2
A1 for −8037, −240111, or -0.4625
Calculate the area of triangle ABC.
53.2 or 53.19 to 53.23
M1 for 0.5×8×15×sin117.5∘ oe
Calculate the length of AQ.
15.5 or 15.52 to 15.53
M1 for 152+42 oe
Calculate the angle of elevation of Q from P.
7.1 or 7.13 or 7.125 to 7.126
M2 for tanP=84−3 oe or M1 for vertical line =4-3
Another straight wire connects A to the midpoint of PQ.
Calculate the angle between this wire and the horizontal ground.
11.5 nfww or 11.48 to 11.49...
B1 for height 3.5
M2 for 152+42−2×15×4cos117.5∘
M1 for tan=their 17.216...3.5 oe