maths ~-00~~唔係好多題

1. Let x m be the distance between the building and Danny,

l m be the vertical distance between Sam and Danny, then (l - 20)m be the vertical distance between Cherry and Danny.

l / x = tan36°

l = xtan36° ───── (1)

(l - 20) / x = tan22° ───── (2)

Sub (1) into (2), we have

(xtan36° - 20) / x = tan22°

xtan36° - 20 = xtan22°

x(tan36° - tan22°) = 20

x = 62 (cor. to the nearest m)

So, the distance between Danny and the building is 62 m.




2. a. ∠PQR = (90° - 38°) + (90° - 52°) = 90°

So, ΔPQR is a right-angled triangle.

PQ = speed X time

PQ = 2 X (2 X 60) = 240 m

QR = speed X time

QR = 1 X (5 X 60) = 300 m

So PR = √(PQ² + QR²)

PR = √[(240)² + (300)²]

PR = √147 600

PR = 384 m (cor. to the nearest m)


b. tan∠PRQ = PQ / QR

tan∠PRQ = 240 / 300 = 4 / 5

∠PRQ = 38.6598°

So, the bearing of P from R

= N[90° - 38.6598° - (90° - 52°)]E

= N13°E (cor. to the nearest degree)

1.)19
2a.)384
2b.)唔明問咩