Right-angled trigonometry and angles between lines Problem set B Worked solutions

Triangle EFG has a right angle at F, EF = 7.2 cm and EG = 11.4 cm. Find angle FEG.

• EF is adjacent to E and EG is the hypotenuse
• \(\cos E=\frac{7.2}{11.4}\)
• From the calculator: E = cos⁻¹(7.2 ÷ 11.4) = 50.8°

Triangle EFG has a right angle at F, angle E = 27° and EG = 14.2 cm. Find the length of FG.

• FG is opposite E and EG is the hypotenuse
• \(\sin 27°=\frac{x}{14.2}\)
• \(x=14.2\times\sin 27°\), so from the calculator x = 6.45 cm

Find the angle between the line y = 0.65x and the positive x-axis.

• Gradient triangle: run 1, rise 0.65
• \(\tan\theta=0.65\)
• From the calculator: θ = tan⁻¹(0.65) = 33.0°

Find the acute angle between the y-axis and each line.

1. (a) y = 6x;

   • Triangle with rise 6 and run 1: the angle with the x-axis has \(\tan\alpha = 6\), α = 80.5°
   • The y-axis is at 90° to the x-axis: θ = 90° − 80.5° = 9.46°

2. (b) y = −0.8x.

   • Triangle with rise 0.8 and run 1: the angle with the x-axis has \(\tan\alpha = 0.8\), α = 38.7°
   • The line slopes down to the right: draw the triangle on the left of the y-axis, the lengths are the same
   • The y-axis is at 90° to the x-axis: θ = 90° − 38.7° = 51.3°

Find the acute angle between the lines y = 7x and y = 2.5x.

• Gradient 7: \(\tan α = 7\), α = 81.9° with the horizontal
• Gradient 2.5: \(\tan β = 2.5\), β = 68.2° with the horizontal
• Same direction: θ = 81.9° − 68.2° = 13.7°

Find the acute angle between the lines y = 2.5x and y = −1.5x.

• Gradient 2.5: \(\tan α = 2.5\), α = 68.2° with the horizontal
• Gradient −1.5: \(\tan β = 1.5\), β = 56.3° with the horizontal
• One line slopes down: α + β = 124.5°
• That is obtuse. The acute angle is θ = 180° − 124.5° = 55.5°

Find the acute angle between the lines y = 5x + 2 and y = −2x + 1.

• Only the gradients matter: the intercepts do not change the angle
• Gradient 5: \(\tan α = 5\), α = 78.7° with the horizontal
• Gradient −2: \(\tan β = 2\), β = 63.4° with the horizontal
• One line slopes down: α + β = 142.1°
• That is obtuse. The acute angle is θ = 180° − 142.1° = 37.9°

A ramp 3.6 m long rises 1.0 m from level ground to a platform. The diagram shows the ramp GT, the height ET and the ground GE.

1. (a) Find the angle θ the ramp makes with the ground.

   • The height 1.0 m is opposite θ and the ramp is the hypotenuse
   • \(\sin\theta=\frac{1.0}{3.6}\)
   • From the calculator: θ = 16.1°

2. (b) A second ramp, 3.2 m long, makes an angle of 12° with the ground. How high does it rise?

   • \(\sin 12°=\frac{h}{3.2}\)
   • \(h=3.2\times\sin 12°\)
   • From the calculator: h = 0.665 m

Two straight cycle tracks on a plan, with distances in metres, are modelled by the lines y = 1.2x + 2 and y = −0.7x + 11. Find the acute angle at which the tracks cross.

• Only the gradients matter
• Gradient 1.2: \(\tan α = 1.2\), α = 50.2° with the horizontal
• Gradient −0.7: \(\tan β = 0.7\), β = 35.0° with the horizontal
• One line slopes down: α + β = 85.19°
• It is acute: θ = 85.2°

A triangle OBC is formed by the x-axis and the lines y = 1.5x and y = −2x + 10.5, where O is the origin, B is on the x-axis and C is where the two lines meet. Find the three angles of the triangle.

• Angle O: \(\tan O = 1.5\), O = 56.3°
• Angle B: the line slopes down with gradient 2, so B = tan⁻¹(2) = 63.4°
• Angle C: angles in a triangle add to 180°: C = 180° − 56.3° − 63.4° = 60.3°