CAIE P1 2018 November — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks7
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Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypeTrigonometric identities with triangles
DifficultyStandard +0.3 This is a slightly-below-average A-level question. Part (i) requires expressing BD in two ways using basic trigonometry (tan in right triangles), then equating and simplifying—a standard technique. Part (ii) involves solving a quadratic in cos θ, which is routine. The setup is clear with a right-angled triangle, and the question guides students through the method explicitly.
Spec1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
\includegraphics{figure_6} The diagram shows a triangle \(ABC\) in which \(BC = 20\) cm and angle \(ABC = 90°\). The perpendicular from \(B\) to \(AC\) meets \(AC\) at \(D\) and \(AD = 9\) cm. Angle \(BCA = \theta°\).
  1. By expressing the length of \(BD\) in terms of \(\theta\) in each of the triangles \(ABD\) and \(DBC\), show that \(20\sin^2 \theta = 9\cos \theta\). [4]
  2. Hence, showing all necessary working, calculate \(\theta\). [3]
Question 6:
AnswerMarks
6(i)9 9
In ∆ ABD, tanθ = → BD = or 9tan(90 – θ) or 9 cotθ
BD tanθ
( )
or ( 20 tanθ )2 −92 (Pythag) or 9sin 90−θ (Sine rule)
AnswerMarks Guidance
  sinθB1 Both marks can be gained for correct equated expressions.
BD
In ∆ DBC, sinθ= → BD = 20sinθ
AnswerMarks
20B1
9
20sinθ =
AnswerMarks Guidance
tanθM1 Equates their expressions for BD and uses sinθ/cosθ = tanθ
or cosθ/sinθ = cot θ if necessary.
AnswerMarks Guidance
→ 20sin²θ = 9cosθ AGA1 Correct manipulation of their expression to arrive at given
answer.
SC:
BD
In ∆ DBC, sinθ = → BD = 20sinθ B1
20
9 BD
In ∆ ABD, BA = and cosθ =
sinθ BA
20sinθ 20sin²θ
cosθ = → cosθ = M1
9/sinθ 9
→ 20sin²θ = 9cosθ A1 Scores 3/4
4
AnswerMarks Guidance
6(ii)Uses s² + c² = 1 → 20cos²θ + 9cosθ – 20 (= 0) M1
→ cosθ = 0.8A1 www
→ θ = 36.9º awrtA1 www. Allow 0.644c awrt. Ignore 323.1º or 2.50c.
Note: correct answer without working scores 0/3.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(i) ---
6(i) | 9 9
In ∆ ABD, tanθ = → BD = or 9tan(90 – θ) or 9 cotθ
BD tanθ
( )
or ( 20 tanθ )2 −92 (Pythag) or 9sin 90−θ (Sine rule)
  sinθ | B1 | Both marks can be gained for correct equated expressions.
BD
In ∆ DBC, sinθ= → BD = 20sinθ
20 | B1
9
20sinθ =
tanθ | M1 | Equates their expressions for BD and uses sinθ/cosθ = tanθ
or cosθ/sinθ = cot θ if necessary.
→ 20sin²θ = 9cosθ AG | A1 | Correct manipulation of their expression to arrive at given
answer.
SC:
BD
In ∆ DBC, sinθ = → BD = 20sinθ B1
20
9 BD
In ∆ ABD, BA = and cosθ =
sinθ BA
20sinθ 20sin²θ
cosθ = → cosθ = M1
9/sinθ 9
→ 20sin²θ = 9cosθ A1 Scores 3/4
4
--- 6(ii) ---
6(ii) | Uses s² + c² = 1 → 20cos²θ + 9cosθ – 20 (= 0) | M1 | Uses s² + c² = 1 to form a three term quadratic in cosθ
→ cosθ = 0.8 | A1 | www
→ θ = 36.9º awrt | A1 | www. Allow 0.644c awrt. Ignore 323.1º or 2.50c.
Note: correct answer without working scores 0/3.
3
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

The diagram shows a triangle $ABC$ in which $BC = 20$ cm and angle $ABC = 90°$. The perpendicular from $B$ to $AC$ meets $AC$ at $D$ and $AD = 9$ cm. Angle $BCA = \theta°$.

\begin{enumerate}[label=(\roman*)]
\item By expressing the length of $BD$ in terms of $\theta$ in each of the triangles $ABD$ and $DBC$, show that $20\sin^2 \theta = 9\cos \theta$. [4]
\item Hence, showing all necessary working, calculate $\theta$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2018 Q6 [7]}}
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