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In Figure 2 OAB is a sector of a circle, radius 5 m - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 2
Question 6
In Figure 2 OAB is a sector of a circle, radius 5 m. The chord AB is 6 m long. (a) Show that cos AOB = \( \frac{7}{25} \) (b) Hence find the angle AOB in radians, ... show full transcript
Worked Solution & Example Answer:In Figure 2 OAB is a sector of a circle, radius 5 m - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 2
Step 1
Show that cos AOB = \( \frac{7}{25} \)
Answer
To find ( \cos AOB ), we can use the cosine rule in triangle OAB:
[ \cos AOB = \frac{OA^2 + OB^2 - AB^2}{2 \cdot OA \cdot OB} ]
Substituting the known values:
- ( OA = OB = 5 , m )
- ( AB = 6 , m )
[ \cos AOB = \frac{5^2 + 5^2 - 6^2}{2 \cdot 5 \cdot 5} = \frac{25 + 25 - 36}{50} = \frac{14}{50} = \frac{7}{25} ]
Step 2
Step 3
Calculate the area of the sector OAB.
Answer
The area of a sector is given by:
[ \text{Area} = \frac{1}{2} r^2 \theta ]
For sector OAB:
- Radius, ( r = 5 , m )
- Angle, ( \theta \approx 1.287 , radians )
Now substituting the values:
[ \text{Area} = \frac{1}{2} \times 5^2 \times 1.287 \approx 16.087 , m^2 ]
Step 4
Hence calculate the shaded area.
Answer
The shaded area is the sector's area minus the area of triangle OAB.
-
Calculate the area of triangle OAB:
The area is given by: [ \text{Area}_{triangle} = \frac{1}{2} \times OA \times OB \times \sin AOB ]
Substituting the values:
- ( OA = OB = 5 , m )
- ( AOB \approx 1.287 , radians )
[ \text{Area}_{triangle} = \frac{1}{2} \times 5 \times 5 \times \sin(1.287) \ .
]
- Substituting into the area formula gives: [ \text{Segment Area} = \text{Sector Area} - \text{Triangle Area} ]
Finally, subtract the area of the triangle from the area of the sector to find the shaded area.