IB Mathematics SL 3.2 Use of sine, cosine and tangent ratios AI SL Paper 1- Exam Style Questions- New Syllabus
Question
The Bermuda Triangle is a region of the Atlantic Ocean with Miami (M), Bermuda (B), and San Juan (S) as vertices, as shown in the diagram.
A triangle with vertices labeled Miami (M), Bermuda (B), and San Juan (S) in the Atlantic Ocean.
The distances between M, B, and S are given in the following table, correct to three significant figures.
| Distance | Value |
|---|---|
| Miami to Bermuda | \( 1670 \, \text{km} \) |
| Bermuda to San Juan | \( 1550 \, \text{km} \) |
| San Juan to Miami | \( 1660 \, \text{km} \) |
(a) Determine the value of θ, the measure of \( \angle M\hat{S}B \):[3]
(b) Determine the area of the Bermuda Triangle. [2]
▶️ Answer/Explanation
Markscheme
(a)
Use the cosine rule to find \( \theta = \angle M\hat{S}B \):
\( \cos \theta = \frac{SM^2 + SB^2 – MB^2}{2 \times SM \times SB} \). M1
Substitute: \( SM = 1660 \), \( SB = 1550 \), \( MB = 1670 \).
Compute: \( SM^2 = 1660^2 = 2755600 \), \( SB^2 = 1550^2 = 2402500 \), \( MB^2 = 1670^2 = 2788900 \).
Numerator: \( 2755600 + 2402500 – 2788900 = 2369200 \).
Denominator: \( 2 \times 1660 \times 1550 = 5146000 \).
Thus: \( \cos \theta = \frac{2369200}{5146000} \approx 0.460435 \). A1
Calculate: \( \theta = \cos^{-1}(0.460435) \approx 62.5873^\circ \).
Rounded: \( \theta \approx 62.6^\circ \) (or 1.09 radians). A1
[3 marks]
Use the cosine rule to find \( \theta = \angle M\hat{S}B \):
\( \cos \theta = \frac{SM^2 + SB^2 – MB^2}{2 \times SM \times SB} \). M1
Substitute: \( SM = 1660 \), \( SB = 1550 \), \( MB = 1670 \).
Compute: \( SM^2 = 1660^2 = 2755600 \), \( SB^2 = 1550^2 = 2402500 \), \( MB^2 = 1670^2 = 2788900 \).
Numerator: \( 2755600 + 2402500 – 2788900 = 2369200 \).
Denominator: \( 2 \times 1660 \times 1550 = 5146000 \).
Thus: \( \cos \theta = \frac{2369200}{5146000} \approx 0.460435 \). A1
Calculate: \( \theta = \cos^{-1}(0.460435) \approx 62.5873^\circ \).
Rounded: \( \theta \approx 62.6^\circ \) (or 1.09 radians). A1
[3 marks]
(b)
Use the area formula: \( A = \frac{1}{2} \times SM \times SB \times \sin(\theta) \). M1
Substitute: \( SM = 1660 \), \( SB = 1550 \), \( \theta = 62.5873^\circ \).
Compute: \( \sin(62.5873^\circ) \approx 0.887406 \).
Area: \( A = \frac{1}{2} \times 1660 \times 1550 \times 0.887406 \approx 1142043.327 \, \text{km}^2 \).
Rounded: \( A \approx 1140000 \, \text{km}^2 \) (or \( 1.14 \times 10^6 \)). A1
Note: Using \( \theta = 63^\circ \), \( \sin(63^\circ) \approx 0.891007 \), \( A \approx 1150000 \, \text{km}^2 \).
[2 marks]
Use the area formula: \( A = \frac{1}{2} \times SM \times SB \times \sin(\theta) \). M1
Substitute: \( SM = 1660 \), \( SB = 1550 \), \( \theta = 62.5873^\circ \).
Compute: \( \sin(62.5873^\circ) \approx 0.887406 \).
Area: \( A = \frac{1}{2} \times 1660 \times 1550 \times 0.887406 \approx 1142043.327 \, \text{km}^2 \).
Rounded: \( A \approx 1140000 \, \text{km}^2 \) (or \( 1.14 \times 10^6 \)). A1
Note: Using \( \theta = 63^\circ \), \( \sin(63^\circ) \approx 0.891007 \), \( A \approx 1150000 \, \text{km}^2 \).
[2 marks]
Total Marks: 5