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The diagram shows triangle ABC - OCR - GCSE Maths - Question 20 - 2023 - Paper 4 Question 20
View full question The diagram shows triangle ABC.
AB = 10.6 cm, BC = 8.2 cm and AC = 12.5 cm.
(a) Show that angle BAC = 40.5°, correct to 1 decimal place.
(b) Work out the area of ... show full transcript
View marking scheme Worked Solution & Example Answer:The diagram shows triangle ABC - OCR - GCSE Maths - Question 20 - 2023 - Paper 4
Show that angle BAC = 40.5°, correct to 1 decimal place. Only available for registered users.
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To find angle BAC, we can use the Law of Cosines:
e x t c o s ( C ) = a 2 + b 2 − c 2 2 a b ext{cos}(C) = \frac{a^2 + b^2 - c^2}{2ab} e x t cos ( C ) = 2 ab a 2 + b 2 − c 2 Where:
a = 10.6 cm (side AB)
b = 12.5 cm (side AC)
c = 8.2 cm (side BC)
Substituting these values into the formula:
e x t c o s ( C ) = 10. 6 2 + 12. 5 2 − 8. 2 2 2 × 10.6 × 12.5 ext{cos}(C) = \frac{10.6^2 + 12.5^2 - 8.2^2}{2 \times 10.6 \times 12.5} e x t cos ( C ) = 2 × 10.6 × 12.5 10. 6 2 + 12. 5 2 − 8. 2 2 Calculating each term:
10. 6 2 = 112.36 10.6^2 = 112.36 10. 6 2 = 112.36 12. 5 2 = 156.25 12.5^2 = 156.25 12. 5 2 = 156.25 8. 2 2 = 67.24 8.2^2 = 67.24 8. 2 2 = 67.24
Thus, the equation becomes:
e x t c o s ( C ) = 112.36 + 156.25 − 67.24 2 × 10.6 × 12.5 = 201.37 265 ≈ 0.760 ext{cos}(C) = \frac{112.36 + 156.25 - 67.24}{2 \times 10.6 \times 12.5} = \frac{201.37}{265} \approx 0.760 e x t cos ( C ) = 2 × 10.6 × 12.5 112.36 + 156.25 − 67.24 = 265 201.37 ≈ 0.760 Now, we find angle BAC:
C = cos − 1 ( 0.760 ) ≈ 40.5 ° C = \text{cos}^{-1}(0.760) \approx 40.5° C = cos − 1 ( 0.760 ) ≈ 40.5°
This confirms that angle BAC is approximately 40.5°, correct to 1 decimal place.
Work out the area of triangle ABC. Only available for registered users.
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To calculate the area of triangle ABC, we can use Heron's formula:
Find the semi-perimeter (s):
s = A B + B C + A C 2 = 10.6 + 8.2 + 12.5 2 = 15.6 s = \frac{AB + BC + AC}{2} = \frac{10.6 + 8.2 + 12.5}{2} = 15.6 s = 2 A B + BC + A C = 2 10.6 + 8.2 + 12.5 = 15.6
Use Heron’s formula for the area (A):
A = s ( s − A B ) ( s − B C ) ( s − A C ) A = \sqrt{s(s - AB)(s - BC)(s - AC)} A = s ( s − A B ) ( s − BC ) ( s − A C ) A = 15.6 ( 15.6 − 10.6 ) ( 15.6 − 8.2 ) ( 15.6 − 12.5 ) A = \sqrt{15.6(15.6 - 10.6)(15.6 - 8.2)(15.6 - 12.5)} A = 15.6 ( 15.6 − 10.6 ) ( 15.6 − 8.2 ) ( 15.6 − 12.5 ) A = 15.6 ( 5 ) ( 7.4 ) ( 3.1 ) A = \sqrt{15.6(5)(7.4)(3.1)} A = 15.6 ( 5 ) ( 7.4 ) ( 3.1 )
Calculate the area:
A = 15.6 × 5 × 7.4 × 3.1 ≈ 49.0 cm 2 A = \sqrt{15.6 \times 5 \times 7.4 \times 3.1} \approx 49.0 \text{ cm}^2 A = 15.6 × 5 × 7.4 × 3.1 ≈ 49.0 cm 2
Thus, the area of triangle ABC is approximately 49.0 cm².
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