Trigonometry for triangles with no right angle. The hard part is choosing which law.
Choose the law
A triangle has sides 8 and 11 with an angle of 40° between them. Which law finds the third side?
Two sides and the angle between them is the law of cosines: c² = a² + b² − 2ab cos C. A fails because the law of sines needs an angle paired with its opposite side, and here the only known angle sits between the two known sides. B needs a right angle, and 40° is not one — though notice the law of cosines becomes Pythagoras exactly when C = 90°, since cos 90° = 0.
| You know | Use |
|---|---|
| Two angles and any side (AAS, ASA) | law of sines |
| Two sides and a non-included angle (SSA) | law of sines — watch for two answers |
| Two sides and the included angle (SAS) | law of cosines |
| All three sides (SSS) | law of cosines |
The rule of thumb: if you have a matched pair — an angle and the side opposite it — use sines. If not, use cosines.
asin A = bsin B = csin C
A = 50°, B = 60°, a = 12
find b
12sin 50° = bsin 60°
a matched pair, plus one more
b = 12 sin 60°sin 50°
≈ 13.6
Sensible: B is the larger angle, so b should be the longer side — and 13.6 > 12 ✓
c² = a² + b² − 2ab(cos C), where C is the angle opposite c
a = 8, b = 11, C = 40°
the opening problem
c² = 64 + 121 − 2(8)(11)cos 40°
cos 40° ≈ 0.766
= 185 − 176(0.766) ≈ 50.2
subtract
c ≈ 7.08
take the root
The formula gives c², so the square root is a separate final step. And the whole term 2ab cos C is subtracted — compute it fully before taking it off a² + b².
When C = 90°, cos C = 0, so the term 2ab cos C disappears and the law becomes c² = a² + b² — the Pythagorean theorem is the right-angle case of the law of cosines.
a = 5, b = 7, c = 9
find the largest angle, C
cos C = a² + b² − c²2ab
rearranged
= 25 + 49 − 8170 = −770 = −0.1
negative
C ≈ 95.7°
obtuse, as the sign predicted
A negative cosine means an obtuse angle. That is a free check: the largest angle faces the longest side, and here 9 is longest, so C being the biggest angle is exactly right.
Area = (12)ab sin C, where C is the angle between sides a and b
A lot has two sides of 9 m and 14 m meeting at an angle of 150°. Find its area.
Area = (12)(9)(14) sin 150°
the angle is between the two sides
= 63 × 0.5
sin 150° = 0.5
= 31.5 m²
no height needed
Why it works: the height onto side a is b sin C, so (12) × base × height becomes (12)ab sin C. With C = 90°, sin C = 1 and it is the familiar (12) × base × height.
Dropping the (12) gives 63 m², double the true area. Using an angle that is not between the two sides also fails — the formula needs the included angle, exactly like SAS in the law of cosines.
The same two laws; each case adds one twist.
No matched pair yet: A = 35°, C = 75°, b = 20. Find a.
B = 180° − 35° − 75° = 70°
now b and B are a pair
a = 20 sin 35°sin 70°
law of sines
= 20(0.5736) ÷ 0.9397 ≈ 12.2
smallest angle, shortest side ✓
Different: the given side faces the missing angle, so the angle sum comes first.
An obtuse included angle: a = 4, b = 6, C = 120°. Find c.
c² = 16 + 36 − 2(4)(6)cos 120°
cos 120° = −0.5
= 52 − 48(−0.5) = 52 + 24 = 76
minus a negative adds
c = √76 ≈ 8.72
longer than 4 and 6 ✓
Different: an obtuse angle makes the third side longer than Pythagoras would give.
Two possible triangles: a = 7, b = 10, A = 40°. Find B.
sin B = 10 sin 40°7 ≈ 0.9183
law of sines, SSA
B ≈ 66.7° or B ≈ 180° − 66.7° = 113.3°
sine is positive in both
both work: 40° + 113.3° < 180°
two triangles fit the data
Different: a sine gives two angles. Keep the obtuse one only if the angle sum leaves room for it.
A tower's top has elevation 32° from 50 m away
a right triangle
tan 32° = h50
opposite over adjacent
h = 50 tan 32° ≈ 31.2 m
no special law needed
Expect "which law" items, one missing side or angle to compute, an area from two sides and the included angle, and a distance or height word problem. The fastest route is to look for a matched angle-and-opposite-side pair: if there is one, use sines; if not, cosines. The 10-second check: the longest side must face the largest angle.
Choose the law
A triangle has all three sides known: 6, 9 and 12. Which law finds its largest angle?
With three sides and no angles there is no matched pair, so the law of cosines is the only option: cos C = 36 + 81 − 144108 = −0.25, giving C ≈ 104.5°. B needs an angle to start from, and there is none. C requires a right angle — and since 36 + 81 = 117 ≠ 144, this triangle does not have one.
Hint: ask whether you have an angle paired with the side opposite it. Without one, the law of sines has nothing to work from.
Every lesson in the Grade 7-12 Math Review has one of these. 30 questions, marked as you answer, with the working shown for each.
Question 1 of 30
What does the Law of Sines state?
Each side divided by the sine of its opposite angle gives the same value.
Question 2 of 30
What does the Law of Cosines state for side c?
It is the Pythagorean theorem with a correction term that vanishes when C is 90°, since cos 90° = 0.
Question 3 of 30
Two sides and the angle between them are known. Which law should be used?
The Law of Sines needs a side with its opposite angle. With SAS, only the Law of Cosines will start.
Question 4 of 30
Two angles and one side are known. Which law should be used?
Knowing two angles gives the third, and then any side pairs with its opposite angle.
Question 5 of 30
In a triangle, a = 8, A = 30° and B = 45°. What is b, to one decimal place?
From b/sin B = a/sin A, b = 8 sin 45° ÷ sin 30° = 8(0.7071) ÷ 0.5 ≈ 11.3.
Question 6 of 30
In a triangle, a = 5, b = 7 and the angle between them is 60°. What is the third side, to two decimal places?
c² = 25 + 49 − 2(5)(7)(0.5) = 74 − 35 = 39, so c = √(39) ≈ 6.24.
Question 7 of 30
A triangle has sides 7, 8 and 9. What is the cosine of the angle opposite the side of 9?
From 81 = 49 + 64 − 2(7)(8)cos C, we get 112 cos C = 32, so cos C = 32/112 = 2/7.
Question 8 of 30
A triangle has sides 6 and 10 with a 30° angle between them. What is its area?
Area = (1/2)ab sin C = (1/2)(6)(10)(0.5) = 15.
Question 9 of 30
In a triangle, A = 40°, B = 60° and a = 10. What is c, to one decimal place?
The third angle is 80°. Then c = 10 sin 80° ÷ sin 40° = 10(0.9848) ÷ 0.6428 ≈ 15.3.
Question 10 of 30
When must the Law of Cosines be used rather than the Law of Sines?
Both SSS and SAS leave no side paired with a known opposite angle, so the Law of Sines cannot start.
Question 11 of 30
Two angles of a triangle and the side between them are known. Which law finds another side?
With two angles we get the third from the angle sum, and then every side is paired with a known angle: law of sines. Answering not enough information forgets the angle sum gives the third angle.
Question 12 of 30
Which formula gives the area of a triangle from sides a and b and the angle C between them?
The height onto a is b sin C, so the area is (1/2)ab sin C. Answering ab sin C drops the half; (1/2)ab cos C uses the wrong ratio.
Question 13 of 30
A triangle has angles 40°, 65° and 75°. Which side is the longest?
The longest side faces the largest angle, 75°. Answering the side opposite 40° picks the shortest side.
Question 14 of 30
In a triangle, cos C works out to 0.3. What kind of angle is C?
A positive cosine means C is less than 90°: acute. A negative cosine would mean obtuse, and 0 would mean a right angle.
Question 15 of 30
In a triangle, A = 50° and B = 70°. What is C?
C = 180° − 50° − 70° = 60°. Answering 120° is A + B, not the remaining angle.
Question 16 of 30
A boy stands 30 m from a tree and sees its top at an angle of elevation of 45°. How tall is the tree above his eye level?
A right triangle: tan 45° = h/30, so h = 30 × 1 = 30 m. Answering 21.2 m uses sin 45° instead of tan 45°; 42.4 m divides by cos 45°.
Question 17 of 30
In a triangle, a = 4, b = 6 and C = 60°. What is c?
c² = 16 + 36 − 2(4)(6)(0.5) = 52 − 24 = 28, so c = √28 = 2√7 ≈ 5.29. Answering 2√13 forgets the cosine term (√52); 10 just adds the sides.
Question 18 of 30
A triangle has sides 5, 6 and 7. What is the cosine of the angle opposite the side 7?
cos C = (25 + 36 − 49)/(2 × 5 × 6) = 12/60 = 0.2. Answering −0.2 subtracts in the wrong order; 0.4 divides by ab instead of 2ab.
Question 19 of 30
In a triangle, B = 30°, C = 120° and b = 4. What is c?
c = b sin C / sin B = 4(√3/2) ÷ (1/2) = 4√3 ≈ 6.93. Sensible: C is the larger angle, so c > b ✓. Answering 4√3/3 turns the ratio upside down.
Question 20 of 30
A triangle has sides 6 and 7 with a 60° angle between them. What is its area, to one decimal place?
Area = (1/2)(6)(7) sin 60° = 21 × 0.866 ≈ 18.2. Answering 21.0 drops the sine; 36.4 drops the half; 10.5 uses sin 30°.
Question 21 of 30
In a triangle, A = 52°, b = 9 and c = 13, and A is the angle between sides b and c. Which law finds a?
Two sides and the included angle (SAS) with no matched pair: law of cosines, a² = b² + c² − 2bc cos A. Answering law of sines needs side a to go with angle A, which is exactly what is missing.
Question 22 of 30
From the top of a 40 m lighthouse, the angle of depression to a boat is 30°. How far is the boat from the foot of the lighthouse, to one decimal place?
The angle of elevation from the boat is also 30°, so tan 30° = 40/d and d = 40 ÷ 0.5774 ≈ 69.3 m. Answering 23.1 m multiplies by tan 30° instead of dividing; 80.0 m divides by sin 30°.
Question 23 of 30
In a triangle, a = 6, b = 8 and A = 30°. How many different triangles fit these measurements?
sin B = 8 sin 30° / 6 = 4/6 ≈ 0.667, so B ≈ 41.8° or 180° − 41.8° = 138.2°. Both leave room: 30° + 138.2° < 180°. So two triangles. Answering one keeps only the acute value of B.
Question 24 of 30
In a triangle, A = 45°, C = 105° and b = 10. What is a, to one decimal place?
B = 180° − 45° − 105° = 30°, so b and B are a pair. a = 10 sin 45° / sin 30° = 7.071 ÷ 0.5 ≈ 14.1. Answering 19.3 finds c instead; 7.1 forgets to divide by sin 30°.
Question 25 of 30
A triangle has sides 7, 9 and 12. What is its largest angle, to one decimal place?
Opposite 12: cos C = (49 + 81 − 144)/(2 × 7 × 9) = −14/126 ≈ −0.111, so C ≈ 96.4°, obtuse. Answering 83.6° drops the minus sign; 90.0° assumes 49 + 81 = 144, which is false.
Question 26 of 30
Two boats leave a port at the same time. One travels 10 km and the other 16 km, on courses 60° apart. How far apart are they?
d² = 100 + 256 − 2(10)(16)(0.5) = 356 − 160 = 196, so d = 14 km. Answering 18.9 km forgets the cosine term (√356); 22.7 km adds 160 instead of subtracting it.
Question 27 of 30
A triangular garden has sides of 12 m and 15 m meeting at 120°. What is its area, to one decimal place?
Area = (1/2)(12)(15) sin 120° = 90 × 0.866 ≈ 77.9 m². Answering 45.0 m² uses 0.5, the size of cos 120°, instead of the sine; 155.9 m² drops the half.
Question 28 of 30
A triangle has sides 5, 7 and 8. What is the angle opposite the side 7?
cos = (25 + 64 − 49)/(2 × 5 × 8) = 40/80 = 0.5, so the angle is 60°. Answering 120° reads the cosine as −0.5; 30° confuses cos 60° = 0.5 with sin 30° = 0.5.
Question 29 of 30
From a point on level ground, the top of a building is at an elevation of 60°. From 20 m farther back it is at 30°. How tall is the building, to one decimal place?
In the triangle formed by the two viewing points and the top, the angles are 30°, 120° and 30°, so the near point is 20 m from the top. Then h = 20 sin 60° ≈ 17.3 m. Answering 20.0 m stops at that slant distance; 10.0 m uses sin 30°.
Question 30 of 30
In triangle ABC, a = 9, B = 40° and C = 80°. What is b, to one decimal place?
A = 180° − 40° − 80° = 60°. b = 9 sin 40° / sin 60° = 5.785 ÷ 0.866 ≈ 6.7. Answering 12.1 turns the ratio upside down; 10.2 finds c instead.
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