Rules of trigonometry: mixed practice Problem set B One question at a time

The line \(L_1\) has equation \(5x+2y=30\). The line \(L_2\) has equation \(2x-y=12\).

1. (a) Find the coordinates of the points where \(L_1\) meets the axes.

   • y = 0: \(5x=30\), so x = 6
   • x = 0: \(2y=30\), so y = 15

2. (b) Find the acute angle between \(L_1\) and the x-axis.

   • Gradient of \(L_1\) = \(-\frac{5}{2}\)
   • \(\tan\theta=\frac{5}{2}\) (use the size of the gradient)
   • \(\theta\) = 68.2°

3. (c) Show that \(L_1\) and \(L_2\) meet on the x-axis.

   • \(L_1\) at y = 0: x = 6, the point (6, 0)
   • \(L_2\) at y = 0: \(2x=12\), so x = 6
   • Both lines cross the x-axis at the same point, so they meet on the x-axis

4. (d) Find the acute angle between \(L_1\) and \(L_2\).

   • \(L_2\) has gradient \(\frac{2}{1}\) (positive): angle with the x-axis 63.4°
   • \(L_1\) has gradient negative: angle with the x-axis 68.2°
   • The lines slope opposite ways: 68.2° + 63.4° = 132°
   • This is obtuse. Acute angle = 180° − 132° = 48.4°

In the diagram, angle ABC is a right angle and D lies on BC. BD = 8 cm, angle BAD = 31° and angle DAC = 19°. Find the length x = DC.

• \(AB=\frac{8}{\tan 31^\circ}=13.31\) cm
• \(BC=AB\tan 50^\circ=15.87\) cm (the angle at A in the big triangle is 31° + 19° = 50°)
• \(x=BC-BD=15.87-8=7.87\) cm

In the diagram, angle ABC is a right angle and D lies on BC. BD = 4 cm, DC = 5 cm and angle ACB = 31°. Find the size of angle BAD, marked θ.

• BC = 4 + 5 = 9 cm
• \(AB=BC\tan 31^\circ=5.408\) cm
• \(\tan\theta=\frac{BD}{AB}=\frac{4}{5.408}\)
• \(\theta=36.5^\circ\)

In triangle ABC, AB = 9.6 cm, AC = 7.3 cm and angle BAC = 49°. Find the length of BC.

• \(x^2=7.3^2+9.6^2-2\times7.3\times9.6\times\cos 49^\circ\)
• \(x^2=53.50\)
• \(x=7.31\) cm

In triangle ABC, AB = 6.7 cm, BC = 9.4 cm and CA = 7.9 cm.

1. (a) Find the size of angle ABC, marked θ.

   • \(\cos\theta=\frac{6.7^2+9.4^2-7.9^2}{2\times6.7\times9.4}=0.5624\)
   • \(\theta=55.8^\circ\)

2. (b) Find the area of the triangle.

   • \(\text{Area}=\frac12\times6.7\times9.4\times\sin 55.78^\circ\)
   • = 26.0 cm²

A kite K is flying above level ground. Two cameras, A and B, are 145 m apart on the ground. The ground line AB and the kite are in the same vertical plane. From A the kite is seen at an angle of 63° to the ground line AB, and from B at an angle of 48° to BA.

1. (a) Find the distance y = AK from camera A to the kite.

   • Third angle: 180° − 63° − 48° = 69.0°
   • \(\frac{y}{\sin 48^\circ}=\frac{145}{\sin 69.0^\circ}\)
   • y = 115 m

2. (b) Find the height d of the kite above the ground.

   • The dashed line is perpendicular to AB: \(d=y\sin 63^\circ\)
   • d = 115.4 × sin 63° = 103 m

A triangular orienteering course has checkpoints P, Q and R. PQ = 6.8 km, PR = 5.2 km and angle QPR = 71°. Runners go directly from Q to R.

1. (a) Find the length x of the leg QR.

   • \(x^2=5.2^2+6.8^2-2\times5.2\times6.8\times\cos 71^\circ=50.26\)
   • x = 7.09 km

2. (b) Find the size of angle PQR, marked θ.

   • \(\frac{\sin\theta}{5.2}=\frac{\sin 71^\circ}{7.089}\)
   • \(\sin\theta=0.6936\)
   • θ = 43.9° (θ is opposite the shorter side 5.2, so it is acute)

A triangular solar panel ABC has area 3.2 m². The edge AB is 2.4 m long and the angle at A is 58°. The edge AC has length x and the edge BC has length y.

1. (a) Find x.

   • \(3.2=\frac12\times2.4\times x\times\sin 58^\circ\)
   • \(x=\frac{2\times3.2}{2.4\sin 58^\circ}=3.14\) m

2. (b) Find y.

   • \(y^2=2.4^2+3.144^2-2\times2.4\times3.144\times\cos 58^\circ=7.649\)
   • y = 2.77 m

In triangle ABC, D lies on BC. AB = 9 cm, AC = 12 cm, AD = 7 cm and angle ADB = 72°. Find the length of BC.

• Triangle ABD: \(9^2=7^2+BD^2-2\times7\times BD\times\cos 72^\circ\), so \(BD^2-4.326BD-32.00=0\) and BD = 8.219 cm
• Angle ADC = 180° − 72° = 108°
• Triangle ADC: \(12^2=7^2+DC^2-2\times7\times DC\times\cos 108^\circ\), so \(DC^2+4.326DC-95.00=0\) and DC = 7.821 cm
• BC = 8.219 + 7.821 = 16.0 cm

The area of triangle PQR is 22 cm². PQ = 8.2 cm, PR = 6.5 cm and angle QPR, marked θ, is acute.

1. (a) Find the size of θ.

   • \(22=\frac12\times8.2\times6.5\times\sin\theta\)
   • \(\sin\theta=0.8255\)
   • \(\theta=55.6^\circ\) (acute)

2. (b) Find the length of QR.

   • \(QR^2=8.2^2+6.5^2-2\times8.2\times6.5\times\cos 55.64^\circ=49.33\)
   • QR = 7.02 cm

In the diagram, D lies on BC. AB = 7.5 cm, angle ABD = 47°, angle ADB = 76° and angle ACD = 34°. Find the value of x = AC.

• Triangle ABD: \(\frac{AD}{\sin 47^\circ}=\frac{7.5}{\sin 76^\circ}\), so AD = 5.653 cm
• Angle ADC = 180° − 76° = 104°
• Triangle ADC: \(\frac{x}{\sin 104^\circ}=\frac{5.653}{\sin 34^\circ}\)
• \(x=9.81\) cm

An aerial BT is fixed to the roof of a tall building. The foot of the building is O, with OB = h metres and BT = d metres. From a point P on level ground, the angle of elevation of the base of the aerial, B, is 28° and the angle of elevation of the top, T, is 31°.

1. (a) Show that \(h=\dfrac{d\tan 28^\circ}{\tan 31^\circ-\tan 28^\circ}\).

   • \(\tan 28^\circ=\frac{h}{PO}\), so \(PO=\frac{h}{\tan 28^\circ}\)
   • \(\tan 31^\circ=\frac{h+d}{PO}\), so \(PO=\frac{h+d}{\tan 31^\circ}\)
   • \(\frac{h}{\tan 28^\circ}=\frac{h+d}{\tan 31^\circ}\), so \(h\tan 31^\circ=(h+d)\tan 28^\circ\)
   • \(h(\tan 31^\circ-\tan 28^\circ)=d\tan 28^\circ\), so \(h=\frac{d\tan 28^\circ}{\tan 31^\circ-\tan 28^\circ}\)

2. (b) The aerial is 4 m long. Find the height of the building.

   • \(h=\frac{4\tan 28^\circ}{\tan 31^\circ-\tan 28^\circ}=30.8\) m

A vertical tree OT of height h metres stands between two walkers at P and Q on level ground. P, O and Q are in a straight line and PQ = 90 m. The angle of elevation of the top of the tree T from P is 35° and from Q is 56°.

1. (a) Write down PO and OQ in terms of h.

   • In triangle POT: \(PO=\frac{h}{\tan 35^\circ}\)
   • In triangle QOT: \(OQ=\frac{h}{\tan 56^\circ}\)

2. (b) Hence find the value of h.

   • PO + OQ = 90, so \(\frac{h}{\tan 35^\circ}+\frac{h}{\tan 56^\circ}=90\)
   • Solve for h: h = 42.8 m