Maths Q 快! 40分

(a) Consider triangle PAB
AB^2 = 248^2 + 200^2 – (2)(248)(200) cos 70 deg
AB^2 = 61504 + 40000 – (2)(248)(200)(0.342)
AB = 259.95 m

(b) Consider triangle ABC
Using Heron’s formula
Semi-perimeter s = (a + b + c)/2 = (200+300 +259.95)/2 = 379.95
Area of triangle = [s(s-a)(s-b)(s-c)]^0.5
= [379.95 (379.95-200)( 379.95- 300379.95– 259.95)]^0.5
= 5748.365 m^2

(c) Consider triangle PAB ,
Sin angle PBA/200 m = Sin angle 70 deg/259.95 m
Sin angle PBA = 200 m x Sin angle 70/259.95 m = 0.723
angle PBA = 46.3 deg

Consider triangle ABC
b^2 = a^2 + c^2 – 2ac cos B
300^2 = 200^2 + 259.95^2 – (2)(200)(259.95) cos B
cos B = 0.1690
B = 80.27 deg (angle ABC)

Angle CBQ = 180- 46.3- 80.27= 53.43 deg
Angle BQC = 32 + 58 = 90 deg
Cos angle CBQ = BQ/BC
Cos 53.43 deg = BQ/200
BQ = 200 Cos 53.43 deg = 119.16 m
PQ = PB + BQ = 248 m + 119.6 m = 367.16 m

Consider triangle PRQ
Cos P = PQ/PR
Cos 70 deg = 367.16 m /PR
PR = 367.16 m/0.342= 1073 m
The distance he walked is 1073 m

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