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Tagged: Cos Rule, Mixed Trig, Sine Rule

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Use the diagram to find:
a. The length of XY.
b. The length of YZ.
a. $$\frac{YZ}{\sin40^{\circ}}=\frac{120}{\sin18^{\circ}}$$ $$YZ=\frac{120\sin40^{\circ}}{\sin18^{\circ}}$$
YZ = 249.6m
b. $$\cos58^{\circ}=\frac{WY}{249.6}$$
WY =249.6 × cos 58°
WY = 132.3m
XW = XY + WY
= 120 + 132.3
= 252.3

Which trigonometric formula would be the most useful in calculating the length of side YZ?
a. A = ½absin C
b. c^{2} = a^{2} + b^{2} – 2ab cos C
c. $$\cos C=\frac{a^2+b^2c^2}{2ab}$$
d. $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$
Option A will find the area, which is not what we are after
Option B is the cos rule looking for a side, which requires two sides and the included angle, and for you to be looking for the side opposite the angle, which we do not have here
Option C is the cos rule looking for an angle, and since we are looking for a side, this is no help
Option D is the correct rule, as it is the sin rule, which requires pairs of angles/sides. the 19° is opposite YZ and the 131° is opposite the 15m
∴ D

Use the diagram to find:
a. The length of BD, correct to 1 decimal place,
b. The size of ∠DBC, correct to the nearest degree.a. BD^{2} = 18^{2} + 35.7^{2} – 2 × 18 × 35.7 × cos 100°
BD^{2} = 1821.66
BD = 42.7b. $$\frac{\sin B}{41.5}=\frac{\sin 80^{\circ}}{42.7}$$
$$\sin B=\frac{41.5\sin 80^{\circ}}{42.7}$$
B = 73°

ABCD is a quadrilateral in which AB = 14cm, BC = 10cm, AD = 17.5cm, ∠A = 50° and ∠C = 55°.
a. Find BD.
b. Find ∠BDC.a. BD^{2} = 14^{2} + 17.5^{2} – 2 × 17.5 × 14 × cos 50º
BD^{2} = 187.2840713
BD = 13.69cmb. $$\frac{\sin D}{10} = \frac{\sin 55}{13.69}$$
$$\sin D = 10\times\frac{\sin 55}{13.69}$$
sin D = 0.598
D = 37°

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