Standard +0.3 This is a straightforward application of the parallelogram area formula (A = ab sin θ) to find sin α, then using the Pythagorean identity to find cos α, and finally calculating tan α. While it requires multiple steps and careful attention to which angle is being asked for (the larger angle), the techniques are standard AS-level trigonometry with no novel problem-solving required.
A parallelogram has sides of length 6 cm and 4.5 cm.
The larger interior angles of the parallelogram have size \(\alpha\)
Given that the area of the parallelogram is 24 cm², find the exact value of \(\tan \alpha\)
[4 marks]
Question 6:
6 | Translates given information into an
equation by using the formula for the
area of triangle or parallelogram to
form a correct equation | AO3.1a | M1 | AB × AD × sin = 24
hence 6 × 4.5 × sin = 24
24 8
sin
27 9
Sides of right angled triangle are 8, 9
and 17
8
Hence tan
17
is one of the largest angles and
must be obtuse hence tangent is
negative
8 8 17
tan
17 17
Rearranges ‘their’ equation to obtain a
correct value of sin | AO1.1b | A1F
Uses ‘their’ sin value to identify an
appropriate right-angled triangle
or uses identities
and deduces exact ratio of
tan – positive or negative
Condone only positive ratio seen | AO2.2a | M1
Relates back to mathematical context
of problem and hence chooses
negative ratio – accept any equivalent
exact form
FT ‘their’ tan values for obtuse α | AO3.2a | A1F
Total | 4
Q | Marking Instructions | AO | Marks | Typical Solution
A parallelogram has sides of length 6 cm and 4.5 cm.
The larger interior angles of the parallelogram have size $\alpha$
Given that the area of the parallelogram is 24 cm², find the exact value of $\tan \alpha$
[4 marks]
\hfill \mbox{\textit{AQA AS Paper 1 Q6 [4]}}