IBDP Maths SL 3.3 Applications of trigonometry AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)
Find the perimeter of the shaded sector \( POR \).
The perimeter of the sector includes the arc \( PQR \) and the radii \( OP \) and \( OR \).
Given: Arc length \( PQR = 10 \, \text{cm} \), radius \( OP = OR = 4 \, \text{cm} \).
Perimeter = Arc \( PQR + OP + OR \):
\[ 10 + 4 + 4 = 18 \, \text{cm} \]
Answer: The perimeter of the shaded sector is \( 18 \, \text{cm} \).
Part (b)
Find \( \theta \).
Use the arc length formula: \( \text{Arc length} = r \theta \).
Given: Arc \( PQR = 10 \, \text{cm} \), radius \( r = 4 \, \text{cm} \).
\[ 10 = 4 \theta \]
\[ \theta = \frac{10}{4} = 2.5 \, \text{radians} \]
Answer: \( \theta = 2.5 \, \text{radians} \)
Part (c)
Find the area of the shaded sector \( POR \).
Use the sector area formula: \( \text{Area} = \frac{1}{2} r^2 \theta \).
Given: \( r = 4 \, \text{cm} \), \( \theta = 2.5 \, \text{radians} \) (from part b).
\[ \text{Area} = \frac{1}{2} \cdot 4^2 \cdot 2.5 = \frac{1}{2} \cdot 16 \cdot 2.5 = 8 \cdot 2.5 = 20 \, \text{cm}^2 \]
Answer: The area of the shaded sector is \( 20 \, \text{cm}^2 \).
▶️ Answer/Explanation
Explanation:
Step 1: Find diagonal AC
Using Pythagoras’ theorem in right-angled ΔCDA:
\[ AC = \sqrt{DA^2 + CD^2} = \sqrt{7.8^2 + 10.4^2} = 13 \]
Step 2: Apply cosine rule
In ΔABC, using the cosine rule for ∠ABC:
\[ \cos B = \frac{AB^2 + BC^2 – AC^2}{2 \times AB \times BC} = \frac{6.5^2 + 9.1^2 – 13^2}{2 \times 6.5 \times 9.1} \] \[ \cos B = \frac{-43.94}{118.3} ≈ -0.3714 \]
Step 3: Calculate angle
\[ ∠ABC = \cos^{-1}(-0.3714) ≈ 111.8° \] Rounded to nearest degree: \[ \boxed{112°} \]
Markscheme:
• Correct use of Pythagoras (M1)
• AC = 13 (A1)
• Correct cosine rule setup (M1)
• ∠ABC = 111.8° (A1)
• 112° A1
[5 marks]