“Trigonometry: Solutions of Triangles”

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your ECAT exam.

Solving a triangle means finding all unknown sides and angles when some of them are given. The three primary tools are the sine rule, the cosine rule, and the area formula. The choice depends on what data is given.

The Sine Rule: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$ where $R$ is the circumradius. Use this when you know:

• Two angles and one side (AAS or ASA)
• Two sides and an angle opposite one of them (SSA — the ambiguous case)

The Cosine Rule: $$a^2 = b^2 + c^2 - 2bc \cos A$$ $$b^2 = a^2 + c^2 - 2ac \cos B$$ $$c^2 = a^2 + b^2 - 2ab \cos C$$

Use this when you know:

• Two sides and the included angle (SAS)
• All three sides (SSS) — to find an angle

Area of a Triangle: $$\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B$$ Also: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ (Heron’s formula) where $s = \frac{a+b+c}{2}$ (semi-perimeter)

⚡ ECAT exam tips:

• The SSA case (two sides and a non-included angle) can have 0, 1, or 2 solutions — this is the “ambiguous case.” If $b \sin A < a < b$: two possible triangles. If $a = b \sin A$: one right triangle. If $a < b \sin A$: no triangle.
• Always check that the sum of angles is $180°$ — it’s a useful consistency check
• When using the sine rule for an angle, use the inverse sine carefully — $\sin \theta = \sin(180° - \theta)$. Both could be valid.

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🟡 Standard — Regular Study (2d–2mo)

For ECAT students who want genuine understanding of triangle solutions.

Case 1 — AAS/ASA (Two Angles + One Side):

Given $A = 50°$, $B = 60°$, $a = 10$.

Step 1: Find $C = 180° - 50° - 60° = 70°$. Step 2: Use sine rule: $b/\sin 60° = 10/\sin 50°$, so $b = 10 \sin 60° / \sin 50° ≈ 10 \times 0.866 / 0.766 = 11.31$. Step 3: $c/\sin 70° = 10/\sin 50°$, so $c = 10 \sin 70° / \sin 50° ≈ 10 \times 0.940 / 0.766 = 12.27$.

Case 2 — SAS (Two Sides + Included Angle):

Given $b = 7$, $c = 5$, $A = 40°$. Find $a$, $B$, $C$.

Step 1: Cosine rule: $a^2 = 7^2 + 5^2 - 2(7)(5)\cos 40° = 49 + 25 - 70 \times 0.766 = 74 - 53.62 = 20.38$. So $a = 4.52$.

Step 2: Cosine rule for angle B: $\cos B = (a^2 + c^2 - b^2)/(2ac) = (20.38 + 25 - 49)/(2 \times 4.52 \times 5) = (-3.62)/(45.2) = -0.0801$. So $B ≈ 94.59°$.

Step 3: $C = 180° - 40° - 94.59° = 45.41°$.

Case 3 — SSS (All Three Sides):

Given $a = 8$, $b = 6$, $c = 5$. Find all angles.

Step 1: $s = (8+6+5)/2 = 9.5$. Step 2: Use Heron’s formula: $\Delta = \sqrt{9.5 \times 1.5 \times 3.5 \times 4.5} = \sqrt{224.44} = 14.98$. Step 3: $\sin A = 2\Delta/(bc) = 29.96/30 = 0.9987$. $A = 87.1°$. Step 4: $\cos A = (b^2 + c^2 - a^2)/(2bc) = (36 + 25 - 64)/(60) = -3/60 = -0.05$. Wait — negative cos? That suggests obtuse angle. Let’s use $\cos B$: $(a^2 + c^2 - b^2)/(2ac) = (64 + 25 - 36)/(80) = 53/80 = 0.6625$. $B = 48.4°$. $C = 180 - 87.1 - 48.4 = 44.5°$. Actually $\cos A = (36+25-64)/(60) = -3/60 = -0.05$. So $A = 92.9°$.

The Ambiguous Case (SSA) — Detailed:

Given $a = 8$, $b = 10$, $A = 30°$.

Step 1: Find $h = b \sin A = 10 \times \sin 30° = 10 \times 0.5 = 5$.

Case (i): If $a < h$: no triangle — the side $a$ is too short to reach the base. Case (ii): If $a = h$: exactly one right triangle. Case (iii): If $h < a < b$: two triangles possible — the apex can be placed on either side of the altitude. Case (iv): If $a \geq b$: one triangle.

For $a = 8, b = 10, h = 5$: $5 < 8 < 10$ → two solutions.

$\sin B = b \sin A / a = 10 \times 0.5 / 8 = 0.625$. So $B_1 = 38.68°$ or $B_2 = 180° - 38.68° = 141.32°$.

If $B_2 = 141.32°$, then $C = 180° - 30° - 141.32° = 8.68°$ (very small). If $B_1 = 38.68°$, then $C = 180° - 30° - 38.68° = 111.32°$.

Both give valid triangles.

⚡ Common student mistakes:

1. Not checking the ambiguous case when given SSA — missing the second triangle
2. Using sine rule to find an acute angle when the obtuse angle is also possible
3. Forgetting to use the correct units — angles in degrees throughout
4. Rounding too early in multi-step calculations — keep full precision until the final answer

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for ECAT mastery of triangle solution methods.

Derivation of Sine Rule:

In any triangle $ABC$ with circumradius $R$: Area $\Delta = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B = \frac{1}{2}ab \sin C$. Also $\Delta = \frac{abc}{4R}$ (a well-known formula — can be derived by noting $\sin A = a/(2R)$ from the circumcircle). Equating: $\frac{1}{2}bc \sin A = \frac{abc}{4R}$, so $\sin A / a = 1/(2R)$. Similarly for $\sin B$ and $\sin C$. Hence $a/\sin A = b/\sin B = c/\sin C = 2R$.

Derivation of Cosine Rule:

Place triangle with side $a$ along the x-axis from $(0,0)$ to $(a,0)$. Place vertex $C$ at coordinates $(b \cos A, b \sin A)$ where side $b = AC$ and $c = BC$. Then $c^2 = (b \cos A - a)^2 + (b \sin A)^2 = b^2 \cos^2 A - 2ab \cos A + a^2 + b^2 \sin^2 A = a^2 + b^2 - 2ab \cos A$. $\blacksquare$

Napier’s Analogies:

For any triangle: $$\frac{\sin A - \sin B}{\sin A + \sin B} = \frac{\tan\left(\frac{A-B}{2}\right)}{\cot\left(\frac{C}{2}\right)}$$

These are used in advanced surveying and navigation problems. ECAT rarely tests Napier’s analogies directly.

Area from Two Sides and Included Angle:

As noted: $\Delta = \frac{1}{2}bc \sin A$. This can be extended: $\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}ca \sin B$.

From $\Delta = \frac{abc}{4R}$ and $\Delta = rs$ (where $r$ = inradius, $s$ = semiperimeter), we get: $r = \frac{4R \sin(A/2) \sin(B/2) \sin(C/2)}{1}$ — an important relationship between circumradius and inradius.

The $m-n$ Theorem (useful for some ECAT problems):

For any triangle, if a point on side $BC$ divides it into segments $m$ and $n$ from $B$ and $C$ respectively: $$AB^2 \cdot n + AC^2 \cdot m = (AN^2 + BN \cdot NC)(m+n)$$ where $AN$ is the cevian to point $N$ on $BC$. This is less commonly tested.

Complete Solution Strategy:

1. Draw the triangle and label all given quantities
2. Identify the case: AAS/ASA, SAS, SSS, or SSA
3. For AAS/ASA: find the third angle, then use sine rule
4. For SAS: find the third side by cosine rule, then find the smaller of the remaining angles by cosine rule (more precise than sine rule, which is ambiguous)
5. For SSS: find the largest angle first (it’s opposite the longest side — and if $\cos A < 0$, the angle is obtuse), then use sine rule for remaining angles
6. For SSA: calculate $h = b \sin A$, compare with $a$, determine number of solutions

Special Triangle Relationships:

• Equilateral triangle: all sides equal, all angles = 60°. Area = $\frac{\sqrt{3}}{4}a^2$. Inradius = $a/(2\sqrt{3})$, circumradius = $a/\sqrt{3}$.
• Right triangle: if $A = 90°$, then $a^2 = b^2 + c^2$ (Pythagoras), $R = a/2$.

ECAT Previous Year Patterns:

• Sine rule and cosine rule applications: very common
• Area calculations (Heron’s formula): very common
• Ambiguous case: frequently tested
• Mixed problems: given some angles and some sides

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