Right-triangle trigonometry in two dimensions

(a) 6.88 cm

1. $QR$ is opposite angle $P$ and $PR$ is the hypotenuse, so use sine.
2. $$QR = 12 \sin 35^\circ = 6.882\ldots$$

• M1 $12 \sin 35^\circ$ (or $\sin 35^\circ = \frac{QR}{12}$).
• A1 6.88 cm (awrt 6.88).

(a) $0.92$ m

1. $$4.8\times \sin(11)$$
2. The rise is opposite the $11^\circ$ angle and the ramp is the hypotenuse.
3. Use sine: rise = 4.8 sin $11^\circ$ = 0.915883... m.
4. To 2 decimal places, the rise is 0.92 m.

• P1 Establishing $4.8\times \sin(11)$ or an equivalent valid method.
• A1 Correct answer: $0.92$ m

(a) $3.39$ m

1. Height is opposite the angle and cable length is the hypotenuse.
   $$7\times \sin(29)$$
2. Therefore $3.39$ m.

• M1 Height is opposite the angle and cable length is the hypotenuse.
• A1 Correct answer: $3.39$ m

(b) $6.12$ m

1. The horizontal distance is adjacent to the angle.
   $$7\times \cos(29)$$
2. Therefore $6.12$ m.

• M1 The horizontal distance is adjacent to the angle.
• A1 Correct answer: $6.12$ m

(a) $16.7$°

1. Use opposite divided by adjacent and the inverse tangent.
2. Therefore $16.7$°.

• M1 Use opposite divided by adjacent and the inverse tangent.
• A1 Correct answer: $16.7$°

(b) $10.44$ m

1. Use the perpendicular rise and run in Pythagoras.
   $$\sqrt{3^{2}+10^{2}}$$
2. Therefore $10.44$ m.

• M1 Use the perpendicular rise and run in Pythagoras.
• A1 Correct answer: $10.44$ m

(a) $6.30$ m

1. The height is opposite and the ground distance adjacent.
   $$9\times \tan(35)$$
2. Therefore $6.30$ m.

• M1 The height is opposite and the ground distance adjacent.
• A1 Correct answer: $6.30$ m

(b) $19.3$°

1. Keep the same height and use twice the horizontal distance.
2. Therefore $19.3$°.

• P1 Keep the same height and use twice the horizontal distance.
• A1 Correct answer: $19.3$°

(a) $44.4$°

1. The height is opposite the angle and the string is the hypotenuse.
   $$\sin(x)=28/40$$
2. Therefore $44.4$°.

• M1 The height is opposite the angle and the string is the hypotenuse.
• A1 Correct answer: $44.4$°

(b) $28.6$ m

1. Use Pythagoras with the string as the hypotenuse.
   $$\sqrt{40^{2}-28^{2}}$$
2. Therefore $28.6$ m.

• M1 Use Pythagoras with the string as the hypotenuse.
• A1 Correct answer: $28.6$ m

(a) No: the angle is $\tan^{-1}(0.09) = 5.14^\circ$, which is more than $5^\circ$.

1. The rise is opposite the angle and the horizontal distance is adjacent, so use tangent.
2. $$\tan \theta = \frac{0.9}{10} = 0.09$$
3. $$\theta = \tan^{-1}(0.09) = 5.14^\circ$$
4. $5.14^\circ$ is more than $5^\circ$, so the ramp does not meet the rule.

• P1 $\tan \theta = \frac{0.9}{10}$.
• A1 $\theta = 5.14^\circ$ (awrt 5.1).
• C1 "No", because $5.14^\circ > 5^\circ$.

(a) $40.2$ m

1. Let BP = x metres and mast height = h metres. Then BQ = x + 35.
2. $$35/(1/\tan(29)-1/\tan(47))$$
3. Let BP = x metres and mast height = h metres. Then BQ = x + 35.
4. Tangent gives h = x tan $47^\circ$ and h = (x + 35) tan $29^\circ.$
5. Equate the heights: x(tan $47^\circ$ $-$ tan $29^\circ$) = 35 tan $29^\circ.$
6. Therefore h = 35 tan $29^\circ$ tan $47^\circ$/(tan $47^\circ$ $-$ tan $29^\circ$) = 40.159137... m.
7. The height is 40.2 m to 1 decimal place.

• P1 Let BP = x metres and mast height = h metres. Then BQ = x + 35.
• P1 Establishing $35/(1/\tan(29)-1/\tan(47))$ or an equivalent valid method.
• A1 Correct answer: $40.2$ m