Mathematics · Oct / Nov 2025 · Paper 11 · Question 3
YMA01
As printed in the paper
The question as text
3. The sides of a triangle are in the ratio 2 : 4 : 5
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[2]
2
2
2
5
=
2
+
4
−××
2 2 4cos'' ''
θ
or e.g.
(
)
(
)
(
)
2
2
2
5
x
=
2
x
+
4
x
−×
2 2
x
×
4 cos'' ''
x
θ
or
(
)
(
)
(
)
2
2
2
2
x
+
4
x
−
5
x
2
+
2
−
2
2
4
5
cos'' ''
θ
=
or e.g.
cos'' ''
θ
=
(
) (
)
××
2 2 4
2
×
2
x
×
4
x
θ
=
108.2
°
Alternative using the cosine rule for a different angle and then the sine rule
2
2
2
4
+
5
−
2
2
2
2
=
+
−××
=
e.g.
2
4
5
2 4 5cos"
A
"
cos"
A
"
2 4 5
××
=
°
22.33...
2
5
=
θ
sin22.33...
sin" "
5sin22.33...
θ
=
θ
=
°
sin" "
108.2
2
(a) Notes
Note that some candidates may use equivalent correct ratios
5
5
x or 2
11 11 11 or 2
4
5
4
5
e.g.
1: 2 : 2 or
x
: 2 : 2
x
:
:
x
:
x
:
x
11
11
11
M1: Attempts to use the cosine rule with the sides 2, 4 and 5 or e.g. 2x, 4x and 5x
or e.g. x, 2x and 2.5x in the correct positions for the largest angle.
2x rather than (
)
2
2
Condone missing brackets e.g.
2x
as long as the intention is clear.
If they start with a rearranged rule, the values or expressions must be in the correct positions.
θ=
° (Degrees symbol not required)
A1:
awrt 108.2
2
+
2
−
2
2
x
4
x
5
x
Allow missing brackets to be recovered e.g.
cos''
θ
°=
''
θ
=
108.2
×
×
2 2
x
4
x
Note that attempts to find any of the other angles, correct or incorrect, can be ignored if the
largest angle is found correctly.
Alternative:
M1:
Attempts to use the cosine rule with the sides 2, 4 and 5 or e.g. 2x, 4x and 5x in the correct
positions for their angle and then attempts to use the sine rule with their angle and the largest
angle in the correct positions.
Note that they may also find the other smaller angle using the sine rule (or cosine rule) and then
subtract the 2 smaller angles from 180°
θ=
° (Degrees symbol not required)
A1:
awrt 108.2
Note that this method is likely to lead to the acute angle (71.79…°) and scores A0
Allow missing brackets to be recovered as above.
3(b)[3]
1
1
=
=
×
°
A
ab
sin
C
140
2
x
4 sin''108.2 ''
x
2
2
or e.g.
1
1
5
=
=
×
°
A
ab
sin
C
140
2
a
a
sin''22.33... ''
2
2
2
280
2
=
=
=
x
36.8...
x
...
8sin(''108.2 '')
°
×
=
Shortest side = 2
36.8...
12.1
(cm)
(b) Notes
1
B1ft: Sets up a correct area equation in one variable e.g. x. using
A
=
ab
sin
C
with A = 140 and
2
with the sides and angle in the correct positions. Follow through on their 108.2°and they must
have a value i.e. not just “C”.
1
It is for e.g.
140
=
×
2
x
×
4 sin''108.2 ''
x
° or with some letter other than “x” but note that
2
1
=
××
°
=
140
x
2 sin''108.2 ''
x
x
...
is also acceptable as the correct ratio is being used and leads
2
directly to the length of the shortest side.
or if they didn’t find the largest angle in part (a), it is for setting up the correct corresponding
equation for their angle.
=
2
M1: Full use of
A
...
x
sin
C
with A = 140 leading to a value for x (or any other letter) that
involves taking the square root of their variable.
231
Note some candidates may use the exact value for sin C e.g.
16
You do not need to be concerned about the order of operations as long as it involves taking the
square root to find “x”.
Condone errors in sides and/or angles used.
A1:
awrt 12.1 (cm). Units are not required but if they are given they must be correct.
Q3 values for reference
80 231
x =
=
=
NB scale factor
" "
36.845...
6.0700...
33
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