Maths Made Easy Forums Standard Trigonometry Mixed Sine Cos Rules

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Keymaster
Post count: 714

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. BD2 =  142 + 17.52 – 2 × 17.5 × 14 × cos 50º
BD2 = 187.2840713
BD = 13.69cm

b. $$\frac{\sin D}{10} = \frac{\sin 55}{13.69}$$

$$\sin D = 10\times\frac{\sin 55}{13.69}$$

sin D = 0.598

D = 37°

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Keymaster
Post count: 714

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

Keymaster
Post count: 714

Which trigonometric formula would be the most useful in calculating the length of side YZ?

a. A = ½absin C

b. c2 = a2 + b2 – 2ab cos C

c. $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$

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

Keymaster
Post count: 714

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. BD2 = 182 + 35.72 – 2 × 18 × 35.7 × cos 100°
BD2 = 1821.66
BD = 42.7

b. $$\frac{\sin B}{41.5}=\frac{\sin 80^{\circ}}{42.7}$$

$$\sin B=\frac{41.5\sin 80^{\circ}}{42.7}$$

B = 73°