For triangles with no right angle — pair every side with the angle facing it.
The problem
In triangle ABC, ∠A = 40°, ∠B = 75° and side a = 12. Find side b.
From bsin B = asin A, b = 12 × sin 75° ÷ sin 40° ≈ 12 × 0.966 ÷ 0.643 ≈ 18.0. D divides the wrong way round. ∠B (75°) is the bigger angle, so side b must be longer than 12 — check that before computing.
Lower-case letters are sides; capitals are the angles opposite them.
asin A = bsin B = csin C
You need a matching pair — a side and the angle opposite it — plus one more piece. Two angles and any side, or two sides and a non-included angle.
12sin 40° = bsin 75°
the a and b pairs
b = 12 · sin 75°sin 40°
cross-multiply
= 12 (0.9659)0.6428
calculator in degrees
b ≈ 18.0
11.59 ÷ 0.6428
sin 75 ÷ sin 40 is 1.50. It is not sin(75 − 40). The sine of a difference is not the difference of the sines.
∠C = 180 − 40 − 75
the angle sum
= 65°
no trigonometry needed
c = 12 sin 65 ÷ sin 40 ≈ 16.9
the last side
With two angles given, always take the third by subtraction first. It opens up every remaining ratio.
Two sides and a non-included angle can describe two different triangles.
sin⁻¹ from the calculator
180° minus that
say sin B = 0.8
from the ratio
B = 53.1° or 126.9°
both have the same sine
check each against the angle sum
discard any total over 180
Finding a side is never ambiguous. The second answer only appears when you use sin⁻¹, because two angles share every sine value.
Area = ½ab · sin C
two sides and the angle between them
a = 12, b = 18, C = 65°
from our triangle
½(12)(18)(0.9063)
sin 65°
≈ 97.9
square units
The angle has to sit between the two sides. With C = 90°, sin C = 1 and the formula collapses to the familiar ½ base × height.
In △ABC, ∠A = 30°, a = 4 and b = 10. Find ∠B.
sin B = b sin A ÷ a = 10 × ½ ÷ 4
the b pair and the a pair
sin B = 54 = 1.25
bigger than 1
no such triangle
no angle has a sine above 1
Different: side a is too short to reach the third side. A sine above 1 is the method telling us the triangle cannot be built.
In △ABC, ∠A = 60°, a = 6√3 and b = 6. Find ∠B and ∠C.
sin B = 6 × (√32) ÷ (6√3) = 12
b sin A ÷ a
B = 30° or 150°
the acute angle and its obtuse twin
60 + 150 = 210 > 180
the twin does not fit
∠B = 30°, ∠C = 90°
exactly one triangle
Different: the given angle faces the longer side (6√3 > 6), so the other angle must be smaller than 60° — one triangle only.
Answering “two triangles” every time sin⁻¹ is used is as wrong as ignoring the twin. Add the obtuse twin to the given angle: under 180° keeps it, 180° or more discards it.
UPCAT-style items pick angles whose sines you know. An obtuse angle has the same sine as its supplement.
| Angle | Same sine as | sin |
|---|---|---|
| 120° | 60° | √32 |
| 135° | 45° | √22 |
| 150° | 30° | ½ |
In △ABC, ∠A = 15°, ∠B = 135° and c = 7. Find b exactly.
∠C = 180 − 15 − 135 = 30°
c needs its own angle first
bsin 135° = 7sin 30°
the b pair and the c pair
b = 7 × (√22) ÷ (½)
sin 135° = sin 45°
b = 7√2
÷ ½ is × 2
Sense check: b faces the biggest angle, so it must be the longest side — 7√2 ≈ 9.9 is longer than c = 7 ✓. The same table handles areas: ½ab sin 150° is ½ab × ½. It also turns angles into a side ratio: the sides are in the ratio of the sines of the angles they face. Angles in the ratio 1 : 1 : 2 are 45°, 45° and 90°, so the sides are sin 45° : sin 45° : sin 90° = √22 : √22 : 1 = 1 : 1 : √2 — not 1 : 1 : 2.
Here c is given but ∠C is not. Using sin 15° or sin 135° under the 7 is the usual slip — find the angle opposite the known side before writing a ratio.
Law of Sines items use 30°, 45°, 60° and their obtuse partners, often with the angle opposite the given side left for you to find, or a story with two sighting angles from a known baseline. Get the third angle, pair each side with the angle it faces, and divide by sines exactly. Ten-second check: the longest side faces the largest angle.
The problem
In triangle PQR, ∠P = 55°, ∠R = 80° and side r = 20. Find side p.
psin P = rsin R gives p = 20 × sin 55° ÷ sin 80° ≈ 20 × 0.819 ÷ 0.985 ≈ 16.6. ∠P is smaller than ∠R, so p has to be shorter than 20 — A fails that check immediately.
