The Law of Sines & Cosines — Challenge
Two rules extend trigonometry to every triangle, not just the right-angled ones.
What these problems cover
These questions use 6 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Ratio — A comparison of two quantities, like 3 cups flour to 2 cups water.
- Area — Space inside, measured in square units.
- Included angle — The angle between two named sides. The A in SAS must be included.
- Sine — sin θ = oppositehypotenuse.
- Law of sines — asin A = bsin B = csin C.
- Law of cosines — c2 = a2 + b2 - 2abcos C.
Problem 1
Which equation states the law of sines for a triangle with sides a, b and opposite angles A, B?
- asin A=bsin B
- a2+b2=c2
- a+b=c
- sin A+sin B=1
- asin A=bsin B
Worked answer
asin A=bsin B
The law of sines says each side divided by the sine of its opposite angle gives the same value for all three pairs.
So asin A = bsin B, and both equal csin C.
It applies to any triangle, unlike Pythagoras' theorem, which needs a right angle.
Problem 2
A triangle has sides 13, 14 and 15. What is its area?
- 84
- 91
- 78
- 90
- 96
Worked answer
84
Use Heron’s formula. The semi-perimeter is s = 13+14+152 = 21. Area = √s(s-a)(s-b)(s-c) = √21 · 8 · 7 · 6. 21 · 8 · 7 · 6 = 7056, and √7056 = 84.
Problem 3
Two sides of length 10 meet at an angle of 120°. How long is the third side?
- 10√3
- 10√2
- 20
- 10
- 15
Worked answer
10√3
By the law of cosines, c2 = 100 + 100 - 2(10)(10)cos 120°. cos 120° = -12, so the correction term adds: c2 = 200 + 100 = 300. So c = √300 = 10√3 ≈ 17.3 — longer than either side, as an obtuse angle demands.
Problem 4
In triangle ABC, a = 7, b = 8 and c = 9. Which angle is the largest?
- C
- A
- B
- They are equal
- Cannot be determined
Worked answer
C
In any triangle the largest angle faces the longest side. The longest side is c = 9, and the angle opposite it is C. No calculation is needed — the labelling convention answers it.
Problem 5
A ship sails 20 km on a bearing of 040°, then turns and sails 30 km on a bearing of 130°. How far is it from its starting point?
- 10√13 km
- 10√5 km
- 50 km
- 25 km
- 10√2 km
Worked answer
10√13 km
The two bearings differ by 130° - 40° = 90°, so the second leg is perpendicular to the first. That makes the displacement the hypotenuse of a right triangle with legs 20 and 30. √400+900 = √1300 = 10√13 ≈ 36.1 km.
Problem 6
With A = 30°, a = 6 and b = 10, how many triangles are possible?
- 2
- 1
- 0
- 3
- Infinitely many
Worked answer
2
By the law of sines, sin B = 10 sin 30°6 = 56 ≈ 0.833. That gives B ≈ 56° or B ≈ 124°, since sine takes the same value in quadrants I and II. Both leave room: 30 + 56 = 86 and 30 + 124 = 154, each under 180°. So two different triangles fit the data.
Problem 7
Find the area of a triangle with sides 5 and 8 meeting at 150°.
- 10
- 20
- 17.3
- 5
- 34.6
Worked answer
10
Area = 12 ab sin C. sin 150° = 12, the same as sin 30° — the reference angle is what matters and sine is positive in quadrant II. Area = 12(5)(8)(12) = 10.
Problem 8
In triangle ABC, cos A = 35 with b = 10 and c = 15. Find a.
- √145
- √505
- 5
- 13
- 12.5
Worked answer
√145
The law of cosines with the angle A between sides b and c: a2 = b2 + c2 - 2bccos A. a2 = 100 + 225 - 2(10)(15)(35) = 325 - 180. a2 = 145, so a = √145 ≈ 12.04.