急! F5 Trigonometry 11q8

22.
(a)
In ΔABQ :
BQ² = AQ² + AB² (Pythagorean theorem)
BQ = √(20² + 8²) m = √464 m = 21.5 m

In ΔCDQ :
QC² = CD² + DQ² (Pythagorean theorem)
QC = √[20² + (12 - 8)²] m = √416 m = 20.4m


(b)
Produce PQ and meet AN at H.

In ΔAQH :
sin∠QAH = QH/AQ
sin30° = QH/(8 m)
QH = 4 m

In ΔBQH :
sin∠QBH = QN/BQ
sin∠QBH = 4/√464
Inclination of BQ = ∠QBH = 10.7°

In ΔBCM :
sin∠CBM = CM/BC
sin30° = CM/(12 m)
CM = 6 m

Draw a horizontal line QK to cut CM at K.

In ΔQCK :
sin∠CQK = CK/QC = (6 - 4)/√416
Inclination of QC = ∠CQK = 5.6°


(c)
In ΔAQH :
cos∠QAH = AH/AQ
cos30° = AH/(8 m)
AH = 4√3 m

In ΔPAH :
tan∠PAH = PH/AH = (6 + 4)/4√3
Angle of elevation of P from A = ∠PAH = 55.3°


(d)
Draw a horizontal line CX to cut PQ at X.

In ΔQCX :
QC² = CX² + QX² (Pythagorean theorem)
CX = √[(√416)² - (6 - 4)²] m = √412 m

In ΔPCX :
tan∠PCX = PX/CX = (6 + 4 - 6)/√412
Angle of elevation of P from C = 11.1°


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23.
(a)
In ΔBAD :
BD / sin∠BAD = AB / sin∠ADB (sine law)
BD / sin(180° - 60°) = (500 m) / sin(60° - 25°)
BD = 500 sin120° / sin35° m = 754.9 m

In ΔABC :
BC / sin∠BAC = AB / sin∠ACB(sine law)
BC / sin(180° - 120°) = (500 m) / sin(120° - 80°)
BC = 500 sin60° / sin40° m = 673.6 m

In ΔBDC :
CD² = BD² + BC² - 2 x BD x BC x cos∠CBD (cosine law)
CD² = 754.9² + 673.6² - 2 x 754.9 x 673.6 x cos(80° - 25°) m²
CD = 663.5 m

(b)
In ΔBDC :
CD / sin∠DBC = BD / sin∠BCD(sine law)
663.5 / sin(80° - 25°) = 754.9 / sin∠BCD
sin∠BCD = 754.9 sin55° / 663.5
∠BCD = 68.7°

Bearing of D from C = 180° + 80° + 68.7° = 328.7°