Right-angled trigonometry and angles between lines Problem set A One question at a time

Triangle PQR has a right angle at Q, PQ = 5.6 cm and QR = 9.3 cm. Find angle QPR.

• QR is opposite P and PQ is adjacent to P
• \(\tan P=\frac{9.3}{5.6}\)
• From the calculator: P = tan⁻¹(9.3 ÷ 5.6) = 58.9°

Triangle LMN has a right angle at L, angle M = 38° and LM = 7.5 cm. Find the length of LN.

• LN is opposite M and LM is adjacent to M
• \(\tan 38°=\frac{x}{7.5}\)
• \(x=7.5\times\tan 38°\), so from the calculator x = 5.86 cm

Find the angle between the line y = 1.8x and the positive x-axis.

• Gradient triangle: run 1, rise 1.8
• \(\tan\theta=1.8\)
• From the calculator: θ = tan⁻¹(1.8) = 60.9°

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

1. (a) y = 4x;

   • Triangle with rise 4 and run 1: the angle with the x-axis has \(\tan\alpha = 4\), α = 76.0°
   • The y-axis is at 90° to the x-axis: θ = 90° − 76.0° = 14.0°

2. (b) y = −1.5x.

   • Triangle with rise 1.5 and run 1: the angle with the x-axis has \(\tan\alpha = 1.5\), α = 56.3°
   • 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° − 56.3° = 33.7°

Find the acute angle between the lines y = 4x and y = 1.2x.

• Gradient 4: \(\tan α = 4\), α = 76.0° with the horizontal
• Gradient 1.2: \(\tan β = 1.2\), β = 50.2° with the horizontal
• Same direction: θ = 76.0° − 50.2° = 25.8°

Find the acute angle between the lines y = 4x and y = −0.5x.

• Gradient 4: \(\tan α = 4\), α = 76.0° with the horizontal
• Gradient −0.5: \(\tan β = 0.5\), β = 26.6° with the horizontal
• One line slopes down: α + β = 102.5°
• That is obtuse. The acute angle is θ = 180° − 102.5° = 77.5°

Find the acute angle between the lines y = 3x − 4 and y = −0.4x + 2.

• Only the gradients matter: the intercepts do not change the angle
• Gradient 3: \(\tan α = 3\), α = 71.6° with the horizontal
• Gradient −0.4: \(\tan β = 0.4\), β = 21.8° with the horizontal
• One line slopes down: α + β = 93.37°
• That is obtuse. The acute angle is θ = 180° − 93.37° = 86.6°

A ladder 5.4 m long leans against a vertical wall. Its foot is on level ground 1.9 m from the wall. The diagram shows the ladder FT, the wall WT and the ground WF.

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

   • The foot 1.9 m is adjacent to θ and the ladder is the hypotenuse
   • \(\cos\theta=\frac{1.9}{5.4}\)
   • From the calculator: θ = 69.4°

2. (b) A second ladder, 4.8 m long, makes an angle of 66° with the ground. How high up the wall does it reach?

   • \(\sin 66°=\frac{h}{4.8}\)
   • \(h=4.8\times\sin 66°\)
   • From the calculator: h = 4.39 m

On a map, with distances in km, two straight roads are modelled by the lines y = 0.8x + 1 and y = −1.6x + 7. Find the acute angle at which the roads cross.

• Only the gradients matter
• Gradient 0.8: \(\tan α = 0.8\), α = 38.7° with the horizontal
• Gradient −1.6: \(\tan β = 1.6\), β = 58.0° with the horizontal
• One line slopes down: α + β = 96.65°
• That is obtuse. The acute angle is θ = 180° − 96.65° = 83.3°

A triangle OBC is formed by the x-axis and the lines y = 2.5x and y = −x + 7, 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: \(\tan O = 2.5\), O = 68.2°
• Angle B: the line slopes down with gradient 1, so B = tan⁻¹(1) = 45.0°
• Angle C: angles in a triangle add to 180°: C = 180° − 68.2° − 45.0° = 66.8°