急! F5 Trigonometry 11q9

24.
(a)(i)
In ΔABD :
DB = √(9² + 12²) m = 15 m (Pythagorean theorem)

In ΔPDK :
tanφ = DK/PD = (15 - 5)/10
φ = 45°

(a)(ii)
In ΔCBD :
cos∠CBD= CB/BD = 9/15 = 0.6

In ΔQCK :
CK² = 5² + 9² - 2 x 5 x 9 x 0.6 m² (cosine law)
CK = √52 m

In ΔDCK :
tanθ = CK/QC = (√52)/10
θ = 35.8°

(b)
In ΔPDK :
PK = √[10² + (15 - 10)²] m = √125 m

In ΔPAD :
PA = √(10² + 9²) m = √181 m

cos∠ADK = cos∠CBD = 0.6

In ΔADK :
AK² = 9² + (15 - 5)² - 2 x 9 x (15 - 5) x 0.6 m² (cosine law)
AK = √73

In ΔPAK :
cos∠APK = [(√181)² + (√125)² - (√73)²] / [2 x √181 x√125] (cosine law)
Angle that spotlight P needed to rotate = ∠APK = 39.2°

In ΔQCK :
QK = √[10² + (√52)²] m = √152 m

In ΔQCB :
QB = √(10² + 9²) m = √181 m

In ΔKQB :
cos∠KQB = [(√152)² + (√181)² - 5²] / (2 x √152 x √181)
Angle that spotlight Q needed to rotate = ∠KQB = 21.8°


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15.
(a)
In ΔPQR :
(c m) / sin[180° - (α + β)] = PR / sinβ (sine law)
(c m) / sin(α + β) = PR / sinβ
Hence, PR = c sinβ / sin(α + β) m

In ΔPBR :
tanθ = h / [c sinβ / sin(α + β)]
Hence, h = tanθ x [c sinβ / sin(α + β)]
h = c tanθ sinβ / sin(α + β)

(b)(i)
h = c tanθ sinβ / sin(α + β)
h/c = tan40° sin46° / sin(54° + 46°) = 0.613

(b)(ii)
In ΔPQR :
(c m) / sin[180° - (54° + 46°)] = QR / sin54° (sine law)
QR = c sin54° / sin80° m

In ΔQBR :
tan∠BQR = h / (c sin54° / sin80°) = 0.613 x (sin80°/ sin54°)
Angleof elevation of B from Q = ∠BQR = 36.7°

(b)(iii)
From (a) :
PR = c sinβ / sin(α + β) m
PR = c sin46° / sin(54° + 46°) m = 0.730c m

In ΔPMR :
RM² = PM² + PR² - 2 x PM x PR x cosα
RM² = (0.5c)² + (0.730c)² - 2 x 0.5c x 0.730c x cos54°
RM = 0.595c m

In ΔBMR :
tan∠BMR = h / 0.595c = 0.613/0.595
Angleof elevation of B from M = ∠BMR = 45.9°

In ΔPMR :
0.595c / sin54° = 0.730c / sin∠PMR (sinelaw)
Bearing of B from M = N ∠PMR E = N 83.0° E