Solution of Triangles

Unit 7 · Trigonometry · 4 teaching hours · 19 min read

Solution of Triangles

What the syllabus asks for 2
  • Solution of triangles (simple cases)
  • Heights and distances

Unit 6 built the tools; this unit uses them. To solve a triangle is to find all six parts — three sides and three angles — from the three that are given. Which law you reach for depends entirely on which three you were given, and there are only four patterns. Three of them have exactly one answer. The fourth, two sides and a non-included angle, may have two answers, one, or none at all, and knowing why is worth several marks every year.

What the exam asks

"Solve the triangle in which …" — identify the pattern (SSS, SAS, ASA/AAS, SSA), use the law that matches, and quote angles to the nearest minute. Always finish with the check A+B+C = 180^{\circ} and "largest angle faces the largest side". Applications are height-and-distance problems: draw the figure first, mark every given, then solve the triangle inside it.

7.1 Solution of triangles (simple cases)

Three parts fix a triangle provided at least one of them is a side. (Three angles do not: AAA gives the shape but not the size, since every similar triangle has the same angles.) That leaves four data patterns.

Given Name First step Answers
three sides SSS cosine law, or half angle law exactly one
two sides and the angle between them SAS cosine law, or tangent law exactly one
one side and two angles ASA / AAS third angle, then sine law exactly one
two sides and an angle not between them SSA sine law 0, 1 or 2

figure
Figure 1. The four patterns. Blue marks the given parts; the label names the law that starts the solution.

Case 1 — three sides (SSS)

Use the cosine law for one angle, then either the cosine law again or the sine law for a second, and subtract from 180^{\circ} for the third. If logarithm tables are being used, or if an angle is close to 90^{\circ} (where the cosine changes slowly and reading it back is inaccurate), the half angle law is safer.

7.1 SSS by the cosine law

Problem. Solve the triangle a = 5 cm, b = 7 cm, c = 8 cm.

Solution.

\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc} = \frac{49+64-25}{2 \times 7 \times 8} = \frac{88}{112} = 0.7857 \ \Rightarrow\ A = 38^{\circ}13'

\cos B = \frac{c^{2}+a^{2}-b^{2}}{2ca} = \frac{64+25-49}{2 \times 8 \times 5} = \frac{40}{80} = \frac12 \ \Rightarrow\ B = 60^{\circ}

C = 180^{\circ} - 38^{\circ}13' - 60^{\circ} = 81^{\circ}47'

Check. The largest angle C faces the largest side c = 8 ✓, and \cos C = \frac{25+49-64}{2 \times 5 \times 7} = \frac{10}{70} = 0.1429, which is indeed \cos 81^{\circ}47'. ✓

7.2 The same triangle by the half angle law

Problem. Solve a = 5, b = 7, c = 8 using half angles, and find the area.

Solution. s = \frac{5+7+8}{2} = 10, so s-a = 5, s-b = 3, s-c = 2.

\tan\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \sqrt{\frac{3 \times 2}{10 \times 5}} = \sqrt{0.12} = 0.3464

so \frac{A}{2} = 19^{\circ}6' and A = 38^{\circ}13' — as before.

\tan\frac{B}{2} = \sqrt{\frac{(s-c)(s-a)}{s(s-b)}} = \sqrt{\frac{2 \times 5}{10 \times 3}} = \frac{1}{\sqrt{3}} \ \Rightarrow\ \frac{B}{2} = 30^{\circ},\ B = 60^{\circ}

C = 81^{\circ}47'. Area, by Heron,

\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{10 \times 5 \times 3 \times 2} = \sqrt{300} = 10\sqrt{3} = 17.32\ \text{cm}^{2}

Check. \frac12 ca\sin B = \frac12(8)(5)\sin 60^{\circ} = 20(0.8660) = 17.32 ✓.

Case 2 — two sides and the included angle (SAS)

The cosine law gives the third side at once; the sine law (or another cosine) then gives a second angle. The tangent law gives both unknown angles without ever finding the third side, which is why it was the standard method before calculators.

7.3 SAS by the cosine law

Problem. Solve the triangle a = 3 cm, c = 2 cm, B = 60^{\circ}.

Solution. B lies between a and c, so

b^{2} = a^{2}+c^{2}-2ac\cos B = 9 + 4 - 2(3)(2)\left(\tfrac12\right) = 13 - 6 = 7,\qquad b = \sqrt{7} = 2.646\ \text{cm}

\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc} = \frac{7+4-9}{2(2.646)(2)} = \frac{2}{10.583} = 0.1890 \ \Rightarrow\ A = 79^{\circ}6'

C = 180^{\circ} - 60^{\circ} - 79^{\circ}6' = 40^{\circ}54'

Check. a = 3 is the largest side and A = 79^{\circ}6' is the largest angle ✓.

7.4 The same triangle by the tangent law

Problem. Solve a = 3, c = 2, B = 60^{\circ} by Napier's analogy.

Solution. The unknown angles are A and C, and

\frac{A+C}{2} = 90^{\circ} - \frac{B}{2} = 90^{\circ}-30^{\circ} = 60^{\circ}

\tan\frac{A-C}{2} = \frac{a-c}{a+c}\cot\frac{B}{2} = \frac{3-2}{3+2}\cot 30^{\circ} = \frac{1.7321}{5} = 0.3464

\frac{A-C}{2} = 19^{\circ}6'

Adding and subtracting:

A = 60^{\circ}+19^{\circ}6' = 79^{\circ}6',\qquad C = 60^{\circ}-19^{\circ}6' = 40^{\circ}54'

Then b = \frac{a\sin B}{\sin A} = \frac{3(0.8660)}{0.9822} = 2.646 cm — the same triangle as Example 7.3.

Case 3 — one side and two angles (ASA or AAS)

Subtract to get the third angle, then use the sine law twice. This case is never ambiguous: the two angles fix the shape and the one side fixes the size.

7.5 ASA

Problem. Solve the triangle A = 60^{\circ}, C = 75^{\circ}, b = 10 cm, and find its area.

Solution. B = 180^{\circ}-60^{\circ}-75^{\circ} = 45^{\circ}.

a = \frac{b\sin A}{\sin B} = \frac{10\sin 60^{\circ}}{\sin 45^{\circ}} = \frac{10(0.8660)}{0.7071} = 12.25\ \text{cm}\ \ (= 5\sqrt{6})

c = \frac{b\sin C}{\sin B} = \frac{10\sin 75^{\circ}}{\sin 45^{\circ}} = \frac{10(0.9659)}{0.7071} = 13.66\ \text{cm}

\Delta = \tfrac12 ab\sin C = \tfrac12(12.25)(10)(0.9659) = 59.15\ \text{cm}^{2}

Check. Sides increase in the same order as their opposite angles: b = 10 < a = 12.25 < c = 13.66 for 45^{\circ} < 60^{\circ} < 75^{\circ} ✓.

Case 4 — two sides and a non-included angle (SSA): the ambiguous case

Suppose a, b and A are given. The sine law gives \sin B = \frac{b\sin A}{a} — but \sin B does not determine B: both B and 180^{\circ}-B have the same sine. Geometrically, the side a is a compass arc swung from C, and an arc can cut the base line twice, once, or not at all.

figure
Figure 2. Ambiguous case: a = 20, b = 30, A = 30^{\circ}. The arc of radius 20 centred at C meets the base at two points, giving B = 48^{\circ}35' or 131^{\circ}25'.

Compare a with the altitude b\sin A from C to the base, which is the shortest possible distance from C to the line.

figure
Figure 3. The four possibilities when a, b and A are given (with A acute in the first three).

The SSA test (with A acute)

Compute h = b\sin A. Then

  • a < h: no triangle (\sin B > 1);
  • a = h: one triangle, right-angled at B;
  • h < a < b: two triangles, B and 180^{\circ}-B;
  • a \ge b: one triangle (the obtuse option is rejected).

If A is obtuse there is one triangle when a > b and none otherwise.

7.6 Two triangles

Problem. In a triangle a = 20 cm, b = 30 cm and A = 30^{\circ}. Solve it.

Solution. h = b\sin A = 30(0.5) = 15, and 15 < 20 < 30, so two triangles exist. By the sine law,

\sin B = \frac{b\sin A}{a} = \frac{30 \times 0.5}{20} = 0.75

B_1 = 48^{\circ}35' \qquad\text{or}\qquad B_2 = 180^{\circ}-48^{\circ}35' = 131^{\circ}25'

Both are admissible because A + B < 180^{\circ} in each case.

First triangle: C_1 = 180^{\circ}-30^{\circ}-48^{\circ}35' = 101^{\circ}25', so

c_1 = \frac{a\sin C_1}{\sin A} = \frac{20(0.9803)}{0.5} = 39.21\ \text{cm}

Second triangle: C_2 = 180^{\circ}-30^{\circ}-131^{\circ}25' = 18^{\circ}35', so

c_2 = \frac{20(0.3189)}{0.5} = 12.75\ \text{cm}

Check. In the figure above, AB_1 = 12.75 and AB_2 = 39.21, and 25.98 \pm 13.23 gives exactly those two feet. ✓

7.7 No triangle

Problem. Can a triangle have a = 10 cm, b = 30 cm and A = 30^{\circ}?

Solution. \sin B = \frac{b\sin A}{a} = \frac{30 \times 0.5}{10} = 1.5. No angle has a sine greater than 1, so no such triangle exists. Equivalently h = b\sin A = 15 > 10 = a: the side a is too short to reach the base line. ∎

7.8 One triangle

Problem. Solve a = 40 cm, b = 30 cm, A = 30^{\circ}.

Solution. Here a > b, so only one triangle exists.

\sin B = \frac{30 \times 0.5}{40} = 0.375 \ \Rightarrow\ B = 22^{\circ}1'

The second value 157^{\circ}59' is rejected: it would make A+B = 187^{\circ}59' > 180^{\circ}. (Equivalently, b < a forces B < A, so B cannot be obtuse.)

C = 180^{\circ}-30^{\circ}-22^{\circ}1' = 127^{\circ}59'

c = \frac{a\sin C}{\sin A} = \frac{40(0.7883)}{0.5} = 63.06\ \text{cm}

Do not throw away the obtuse root too soon

When the sine law gives \sin B = k, write both B and 180^{\circ}-B, then test each: keep it only if A+B < 180^{\circ}. Students lose marks for silently taking the calculator's answer. Conversely, if the given angle is opposite the larger side, the obtuse root is always impossible — say so, do not just drop it.

7.2 Heights and distances

Every height-and-distance problem is a triangle with one known side (a measured baseline) and two known angles, so Case 3 solves it. Draw the figure, mark the angles of elevation from the horizontal, and name the triangle you intend to use.

figure
Figure 4. Angle of elevation 30^{\circ} at P; after walking 40 m to Q it is 60^{\circ}. Drawn to scale: h = 20\sqrt{3} = 34.64 m.

7.9 Height of a tower

Problem. From a point P on level ground the angle of elevation of the top of a tower is 30^{\circ}. Walking 40 m towards the tower to a point Q, the elevation becomes 60^{\circ}. Find the height of the tower.

Solution. Let S be the top and T the foot. In triangle PQS:

  • \angle SPQ = 30^{\circ};
  • \angle SQT = 60^{\circ} is an exterior angle of triangle PQS at Q, so \angle PQS = 180^{\circ}-60^{\circ} = 120^{\circ};
  • hence \angle PSQ = 180^{\circ}-30^{\circ}-120^{\circ} = 30^{\circ}.

Triangle PQS therefore has two equal angles, so QS = PQ = 40 m. Now in the right triangle QTS,

h = ST = QS\sin 60^{\circ} = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} = 34.64\ \text{m}

Check. QT = h\cot 60^{\circ} = 20 m and PT = h\cot 30^{\circ} = 60 m, and PT - QT = 40 m ✓.

figure
Figure 5. Width of a river: the baseline AB = 100 m is measured along one bank and the angles to the tree at C are read at A and B.

7.10 Width of a river

Problem. A and B are two points 100 m apart on one straight bank of a river. A tree C stands on the opposite bank. It is found that \angle CAB = 60^{\circ} and \angle CBA = 45^{\circ}. Find AC, BC and the width of the river.

Solution. \angle ACB = 180^{\circ}-60^{\circ}-45^{\circ} = 75^{\circ} (Case 3, ASA). By the sine law, with c = AB = 100,

AC = \frac{c\sin B}{\sin C} = \frac{100\sin 45^{\circ}}{\sin 75^{\circ}} = \frac{100(0.7071)}{0.9659} = 73.21\ \text{m}

BC = \frac{c\sin A}{\sin C} = \frac{100(0.8660)}{0.9659} = 89.66\ \text{m}

The width is the perpendicular distance from C to the bank AB:

\text{width} = AC\sin 60^{\circ} = 73.21 \times 0.8660 = 63.40\ \text{m}

Check. The same width from the other triangle: BC\sin 45^{\circ} = 89.66(0.7071) = 63.40 m ✓.

7.11 Area of a field

Problem. A triangular field in Chitwan has sides 120 m, 170 m and 250 m. Find its area in square metres and in ropani (1 ropani = 508.72 m²).

Solution. s = \frac{120+170+250}{2} = 270 m, so s-a = 150, s-b = 100, s-c = 20.

\Delta = \sqrt{270 \times 150 \times 100 \times 20} = \sqrt{81\,000\,000} = 9000\ \text{m}^{2}

\frac{9000}{508.72} = 17.69\ \text{ropani}

Check. The field is nearly degenerate (120+170 = 290, only 40 m more than 250), so a small area for such long sides is exactly what we expect. ✓

7.12 A two-bearing problem

Problem. Two roads leave a junction J at an angle of 60^{\circ}. A cyclist rides 3 km along one and a walker 2 km along the other. How far apart are they, and what angle does the line joining them make with the cyclist's road?

Solution. This is SAS with the 60^{\circ} angle included, exactly the triangle of Example 7.3 measured in kilometres: a = 3, c = 2, B = 60^{\circ}.

b^{2} = 9+4-2(3)(2)\left(\tfrac12\right) = 7,\qquad b = 2.65\ \text{km}

The required angle is the one at the cyclist's end, opposite the walker's distance c = 2:

\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab} = \frac{9+7-4}{2(3)(2.646)} = 0.7559 \ \Rightarrow\ C = 40^{\circ}54'

So they are 2.65 km apart, and the joining line makes 40^{\circ}54' with the cyclist's road.

Marking points in a solution

State the case (SSS, SAS, …) in words before computing — it shows the examiner you chose the law deliberately. Keep five figures during the work and round only at the end. Give angles in degrees and minutes (1^{\circ} = 60'): 0.5^{\circ} is 30', not 50'.

Chapter summary

  • Three parts solve a triangle if at least one is a side; AAA gives shape only.
  • SSS: cosine law \cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}, or half angles \tan\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}.
  • SAS: a^{2} = b^{2}+c^{2}-2bc\cos A; or Napier, \tan\frac{A-B}{2} = \frac{a-b}{a+b}\cot\frac{C}{2} with \frac{A+B}{2} = 90^{\circ}-\frac{C}{2}.
  • ASA/AAS: third angle first, then \frac{a}{\sin A} = \frac{b}{\sin B}.
  • SSA is ambiguous: with h = b\sin A and A acute — none if a<h, one (right-angled) if a = h, two if h<a<b, one if a \ge b.
  • Area: \Delta = \frac12 ab\sin C = \sqrt{s(s-a)(s-b)(s-c)}.
  • Heights and distances: draw the figure, use the exterior-angle trick to find the angle at the near point, then solve the triangle on the baseline.

Practice questions

Group A — Multiple choice (1 mark each)

  1. To solve a triangle given its three sides you should start with [1]athe sine lawbthe cosine lawcthe projection lawdthe angle sum
  2. A triangle can not be solved from [1]aSSSbSAScAAAdASA
  3. If a = 20, b = 30, A = 30^{\circ}, the number of possible triangles is [1]a0b1c2dinfinitely many
  4. In a triangle b = 4, c = 5 and A = 60^{\circ}; then a equals [1]a\sqrt{21}b\sqrt{41}c\sqrt{61}d9
  5. If \sin B = 0.8 and b < a, then B is [1]a53^{\circ}8' onlyb126^{\circ}52' onlyceitherdneither
  6. From the top of a tower h m high the angle of depression of a point on the ground is 45^{\circ}; the point is at a distance [1]a\frac{h}{2}bhch\sqrt{3}d2h
Answers to Group A

1. (b) — the sine law would need an angle, which SSS does not give.

2. (c) — AAA fixes only the shape; all similar triangles share it.

3. (c) — h = b\sin A = 15 and 15 < 20 < 30, the two-triangle case.

4. (a) — a^{2} = 16+25-2(4)(5)\left(\frac12\right) = 21.

5. (a) — b < a gives B < A, so B cannot be obtuse.

6. (b) — \tan 45^{\circ} = h/d = 1, so d = h.

Group B — Short answer (5 marks each)

  1. Solve the triangle a = 5 cm, b = 7 cm, c = 8 cm and find its area. [5]
  2. Solve the triangle in which a = 3 cm, c = 2 cm and B = 60^{\circ}, using the tangent law for the angles. [5]
  3. Solve the triangle A = 60^{\circ}, C = 75^{\circ}, b = 10 cm. [5]
  4. Show that no triangle exists with a = 10 cm, b = 30 cm, A = 30^{\circ}, but that exactly one exists with a = 40 cm, b = 30 cm, A = 30^{\circ}; solve the second. [5]
  5. The angle of elevation of the top of a tower from a point on level ground is 30^{\circ}; on walking 40 m towards the tower it becomes 60^{\circ}. Find the height of the tower. [5]
  6. A triangular plot has sides 120 m, 170 m and 250 m. Find its area and the radius of its circumcircle. [5]
Answers to Group B

1. \cos A = \frac{49+64-25}{112} = 0.7857, so A = 38^{\circ}13'; \cos B = \frac{64+25-49}{80} = 0.5, so B = 60^{\circ}; and C = 81^{\circ}47'. With s = 10, \Delta = \sqrt{10 \times 5 \times 3 \times 2} = 10\sqrt{3} = 17.32 cm².

2. \frac{A+C}{2} = 90^{\circ}-30^{\circ} = 60^{\circ} and \tan\frac{A-C}{2} = \frac{3-2}{3+2}\cot 30^{\circ} = \frac{1.7321}{5} = 0.3464, so \frac{A-C}{2} = 19^{\circ}6'. Hence A = 79^{\circ}6', C = 40^{\circ}54', and b = \frac{a\sin B}{\sin A} = \frac{3(0.8660)}{0.9822} = 2.65 cm (= \sqrt{7}).

3. B = 45^{\circ}; a = \frac{10\sin 60^{\circ}}{\sin 45^{\circ}} = 12.25 cm; c = \frac{10\sin 75^{\circ}}{\sin 45^{\circ}} = 13.66 cm. (Area, if wanted: \frac12 ab\sin C = 59.15 cm².)

4. First: \sin B = \frac{30 \times 0.5}{10} = 1.5 > 1, impossible, so no triangle. Second: a = 40 > b = 30, so exactly one. \sin B = \frac{30 \times 0.5}{40} = 0.375 gives B = 22^{\circ}1' (the obtuse root 157^{\circ}59' fails A+B<180^{\circ}). Then C = 127^{\circ}59' and c = \frac{40\sin C}{\sin 30^{\circ}} = \frac{40(0.7883)}{0.5} = 63.06 cm.

5. With P, Q the two points and S the top: \angle SPQ = 30^{\circ}, \angle PQS = 180^{\circ}-60^{\circ} = 120^{\circ}, so \angle PSQ = 30^{\circ} and the triangle PQS is isosceles with QS = PQ = 40 m. Then h = QS\sin 60^{\circ} = 40\left(\frac{\sqrt{3}}{2}\right) = 20\sqrt{3} = 34.64 m.

6. s = 270 m; \Delta = \sqrt{270 \times 150 \times 100 \times 20} = 9000 m². Then R = \frac{abc}{4\Delta} = \frac{120 \times 170 \times 250}{4 \times 9000} = \frac{5\,100\,000}{36\,000} = 141.67 m.

Group C — Long answer (8 marks each)

  1. Explain what is meant by solving a triangle and list the four cases of given data. Solve the triangle a = 20 cm, b = 30 cm, A = 30^{\circ} completely, explaining why it has two solutions, and state the condition for each of the other possibilities. [8]
  2. Derive the tangent law \tan\frac{A-B}{2} = \frac{a-b}{a+b}\cot\frac{C}{2} and use it to solve the triangle a = 9 cm, b = 7 cm, C = 60^{\circ}. Find the area of that triangle. [8]
  3. A and B are points 100 m apart on a straight bank of a river and C is a tree on the opposite bank, with \angle CAB = 60^{\circ} and \angle CBA = 45^{\circ}. Find AC, BC, the width of the river and the area of triangle ABC. [8]
Answers to Group C

1. Solving a triangle means finding all six parts from three given ones, at least one of which must be a side. The four cases are SSS and SAS (cosine or tangent law), ASA/AAS (angle sum, then sine law) and SSA (sine law, possibly ambiguous). Here h = b\sin A = 30 \times 0.5 = 15 and 15 < a = 20 < b = 30, which is the two-triangle case: the arc of radius a centred at C cuts the base twice. \sin B = \frac{b\sin A}{a} = 0.75, so B = 48^{\circ}35' or 131^{\circ}25', and both satisfy A+B < 180^{\circ}. First: C = 101^{\circ}25', c = \frac{20\sin C}{\sin 30^{\circ}} = \frac{20(0.9803)}{0.5} = 39.21 cm. Second: C = 18^{\circ}35', c = \frac{20(0.3189)}{0.5} = 12.75 cm. Other possibilities (for acute A): a < h gives no triangle, a = h gives one right-angled triangle, and a \ge b gives exactly one triangle.

2. Derivation. By the sine law a = 2R\sin A, b = 2R\sin B, so

\frac{a-b}{a+b} = \frac{\sin A-\sin B}{\sin A+\sin B} = \frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}} = \cot\frac{A+B}{2}\tan\frac{A-B}{2}

and since \frac{A+B}{2} = 90^{\circ}-\frac{C}{2}, \cot\frac{A+B}{2} = \tan\frac{C}{2}, giving \tan\frac{A-B}{2} = \frac{a-b}{a+b}\cot\frac{C}{2}. Application. \frac{A+B}{2} = 60^{\circ} and \tan\frac{A-B}{2} = \frac{2}{16}\cot 30^{\circ} = 0.2165, so \frac{A-B}{2} = 12^{\circ}13', A = 72^{\circ}13', B = 47^{\circ}47'. Third side: c^{2} = 81+49-2(9)(7)\left(\frac12\right) = 67, c = 8.19 cm. Area: \Delta = \frac12 ab\sin C = \frac12(9)(7)(0.8660) = 27.28 cm².

3. \angle ACB = 180^{\circ}-60^{\circ}-45^{\circ} = 75^{\circ}, so with AB = c = 100 m the sine law gives

AC = \frac{100\sin 45^{\circ}}{\sin 75^{\circ}} = 73.21\ \text{m},\qquad BC = \frac{100\sin 60^{\circ}}{\sin 75^{\circ}} = 89.66\ \text{m}

Width = AC\sin 60^{\circ} = 73.21(0.8660) = 63.40 m (check: BC\sin 45^{\circ} = 63.40 m ✓). Area = \frac12 \times AB \times \text{width} = \frac12(100)(63.40) = 3170 m², and equally \frac12(AC)(BC)\sin 75^{\circ} = \frac12(73.21)(89.66)(0.9659) = 3170 m².

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