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6b · E6.4–E6.6

Trig graphs, sine and cosine rules, 3D

Topic 6 Trigonometry · Extended only · Papers 2 and 4

EXTENDED Everything on this page is Extended content only. Core trigonometry stops at right-angled triangles (6a).

🎯What you need to be able to do

  • Recognise, sketch and use the graphs of \( y = \sin x \), \( y = \cos x \) and \( y = \tan x \) for \( 0^{\circ} \le x \le 360^{\circ} \).
  • Solve equations such as \( \sin x = k \) for \( 0^{\circ} \le x \le 360^{\circ} \), finding every solution.
  • Use the sine rule and the cosine rule for sides and angles in any triangle, including obtuse angles and the ambiguous case.
  • Use area \( = \tfrac{1}{2}ab\sin C \).
  • Use Pythagoras and trigonometry in three dimensions, including the angle between a line and a plane.

📚The mathematics

The graphs and their symmetry

\( \sin x \) and \( \cos x \) repeat every \( 360^{\circ} \) and stay between \(-1\) and 1; \( \tan x \) repeats every \( 180^{\circ} \) and has asymptotes at \( 90^{\circ} \) and \( 270^{\circ} \). The calculator gives one solution of \( \sin x = k \); the graph’s symmetry gives the other:

\( \sin x = k \): \( x \) and \( 180^{\circ} - x \)
\( \cos x = k \): \( x \) and \( 360^{\circ} - x \)
\( \tan x = k \): \( x \) and \( x + 180^{\circ} \)
The graph of y = sin x from 0 to 360 degrees with the horizontal line y = 0.75. The line crosses the curve twice, at 48.6 degrees and at 131.4 degrees, which is 180 minus 48.6; the two solutions are symmetric about 90 degrees.
\( \sin x = 0.75 \): the calculator gives \( 48.6^{\circ} \); symmetry about \( 90^{\circ} \) gives \( 180^{\circ} - 48.6^{\circ} = 131.4^{\circ} \).

The sine rule, the cosine rule and the area

\( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \)
\( a^{2} = b^{2} + c^{2} - 2bc\cos A \)
area \( = \tfrac{1}{2}ab\sin C \)

All three are on the List of formulas. Side \(a\) is opposite angle \(A\). Choose the rule by what you know:

  • Cosine rule: two sides and the angle between them (find the third side), or all three sides (find an angle — rearrange to \( \cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc} \)).
  • Sine rule: any other combination with a matching side-and-opposite-angle pair.

The ambiguous case. When the sine rule gives an angle, \( \sin C = k \) also has the obtuse solution \( 180^{\circ} - C \). Both are possible if the angles still add up to less than \( 180^{\circ} \); the cosine rule never has this problem, because \( \cos \) is negative for obtuse angles.

Triangle ABC with AB = 9 cm, AC = 7 cm and the angle at A equal to 52 degrees. BC, found by the cosine rule, is 7.24 cm, angle B is 49.6 degrees, and the area is 24.8 square centimetres.
The worked example: two sides and the included angle, so the cosine rule first, then the sine rule.

Three dimensions

Find a right-angled triangle that lies in a flat plane inside the solid, draw it separately, and solve it. The angle between a line and a plane is the angle between the line and its projection onto the plane (the “shadow” directly beneath it): drop a perpendicular from the top of the line to the plane to make the right angle.

A cuboid 8 cm long, 6 cm wide and 5 cm high. The space diagonal AG is drawn, with its projection AC on the base. AC is 10 cm, CG is 5 cm vertical, AG is 11.2 cm, and the angle GAC between the diagonal and the base is 26.6 degrees.
The angle between \(AG\) and the base is angle \(GAC\), in the vertical right-angled triangle \(ACG\).

✏️Worked example (calculator)

(a) Solve \( 4\sin x = 3 \) for \( 0^{\circ} \le x \le 360^{\circ} \). [2] (b) In triangle \(ABC\), \( AB = 9 \) cm, \( AC = 7 \) cm and angle \( BAC = 52^{\circ} \). Find \(BC\), angle \(ABC\) and the area. [6] (c) A cuboid is 8 cm by 6 cm by 5 cm high. Find the length of a space diagonal and the angle it makes with the base. [4]

(a) \( \sin x = 0.75 \Rightarrow x = 48.6^{\circ} \) or \( 180 - 48.6 = 131.4^{\circ} \).

(b) \( BC^{2} = 9^{2} + 7^{2} - 2(9)(7)\cos 52^{\circ} = 52.43\ldots \), so \( BC = 7.24 \) cm. Sine rule: \( \sin B = \dfrac{7\sin 52^{\circ}}{7.241\ldots} = 0.7618 \), so \( B = 49.6^{\circ} \) (it must be acute, because it is opposite the shortest side). Area \( = \tfrac{1}{2}(9)(7)\sin 52^{\circ} = 24.8 \) cm².

(c) Base diagonal \( \sqrt{8^{2} + 6^{2}} = 10 \); space diagonal \( \sqrt{10^{2} + 5^{2}} = 11.2 \) cm. Angle with the base: \( \tan^{-1}\tfrac{5}{10} = 26.6^{\circ} \).

Check it. In (b), angle \( C = 180 - 52 - 49.6 = 78.4^{\circ} \), and the largest angle should face the longest side, \( AB = 9 \) ✓.
Keeping the rounded side. Using \( BC = 7.24 \) instead of the full value in the sine rule can shift the angle in the first decimal place. Keep \( 7.2410\ldots \) in the calculator.

📝Practise

All Extended, calculator style unless stated.

1. Solve \( \tan x = 2.5 \) for \( 0^{\circ} \le x \le 360^{\circ} \). [2]
\( x = 68.2^{\circ} \) or \( 68.2 + 180 = 248.2^{\circ} \).
2. Solve \( \cos x = -0.4 \) for \( 0^{\circ} \le x \le 360^{\circ} \). [2]
\( x = 113.6^{\circ} \) or \( 360 - 113.6 = 246.4^{\circ} \).
3. A triangle has sides 5 cm, 7 cm and 9 cm. Find its largest angle. [3]
Opposite the 9 cm side: \( \cos\theta = \dfrac{25 + 49 - 81}{2(5)(7)} = -0.1 \), so \( \theta = 95.7^{\circ} \). The negative cosine shows at once that it is obtuse.
4. In triangle \(ABC\), angle \( A = 40^{\circ} \), angle \( B = 65^{\circ} \) and \( a = 12 \) cm. Find \(b\). [2]
\( b = \dfrac{12\sin 65^{\circ}}{\sin 40^{\circ}} = 16.9 \) cm.
5. Find the area of a triangle with sides 8 cm and 11 cm and an included angle of \( 120^{\circ} \). [2]
\( \tfrac{1}{2}(8)(11)\sin 120^{\circ} = 38.1 \) cm².
6. In triangle \(ABC\), \( AB = 10 \) cm, \( BC = 7 \) cm and angle \( A = 35^{\circ} \). Find the two possible values of angle \(C\). [3]
\( \sin C = \dfrac{10\sin 35^{\circ}}{7} = 0.8194 \), so \( C = 55.0^{\circ} \) or \( 180 - 55.0 = 125.0^{\circ} \). Both work, since \( 35 + 125 < 180 \): this is the ambiguous case.
7. A cone has base radius 5 cm and slant height 13 cm. Find its vertical height, and the angle between the slant edge and the base. [3]
The vertical height, the radius and the slant edge form a right-angled triangle: \( h = \sqrt{13^{2} - 5^{2}} = 12 \) cm. The angle with the base: \( \cos\theta = \tfrac{5}{13} \Rightarrow \theta = 67.4^{\circ} \).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — graph \( y = \sin x \) in degree mode with a horizontal line to see both solutions
  • Corbettmaths — sine rule, cosine rule and 3D trigonometry