Applications in Trigonometry

a ii) cos ∠KBC = BC/BD = 3/5

Then KC2 = 52 + 92 - 2 x 5 x 9 cos ∠KBC

= 52

KC = √52

Thus tan θ = √52/10

θ = 35.8

b) tan ∠BPD = BD/DP = 3/2

∠BPD = 56.3

With Φ = 45, P has to turn through an angle of 56.3 - 45 = 11.3 QK2 = QC2 + CK2 = 152QB2 = QC2 + CB2 = 181Then in triangle QKB:KB2 = QB2 + QK2 - 2(QB)(QK) cos ∠KQBcos ∠KQB = 0.928cos ∠KQB = 21.8