Area of triangle = ½ab·sin C: Worked examples

This is the drill page for the area formula. We will hit every kind of situation this one formula can face — friendly angles, obtuse angles, the sneaky ambiguous case, degenerate flat triangles, a real-world word problem, and an exam twist. First we map the territory, then we walk every cell of the map with a full solution.

Before we start, one reminder of the tool, stated in plain words so nothing is assumed:

The scenario matrix

Every problem this formula can pose falls into one of these cells. Our worked examples below are each tagged with the cell they cover, so together they leave no gap.

Cell	What varies	Why it needs its own example
A. Acute included angle	$0^{\circ} < C < 9 0^{\circ}$	The plain, direct plug-in.
B. Right angle	$C = 9 0^{\circ}$	$sin 9 0^{\circ} = 1$ ; formula must reduce to $\frac{1}{2} ab$ .
C. Obtuse included angle	$9 0^{\circ} < C < 18 0^{\circ}$	Foot of the height lands outside the base — does the sign survive?
D. Solve for the angle (ambiguous)	given Area, find $C$	$sin C = sin (18 0^{\circ} - C)$ → two answers.
E. Solve for a side	given Area, find $a$	Treat formula as an equation, rearrange.
F. Degenerate / limiting	$C \to 0^{\circ}$ or $C \to 18 0^{\circ}$	Triangle flattens; area $\to 0$ . Sanity boundary.
G. Real-world word problem	units, "included angle" hidden in prose	Reader must extract $a, b, C$ before computing.
H. Exam twist (equilateral / find side then area, or cross-product link)	combine with another idea	Tests whether you see the formula inside a bigger problem.

Cell A — Acute included angle

Worked example Example 1 (Cell A)

A triangle has sides $a = 9$ and $b = 12$ with the included angle $C = 4 0^{\circ}$ . Find the area.

Forecast: The angle is fairly gentle (less than $4 5^{\circ}$ ), so $sin C$ is smallish (around $0.64$ ). Guess: the area is a fair bit less than the full " $\frac{1}{2} \cdot 9 \cdot 12 = 54$ " you'd get if it were a right angle. Somewhere in the 30s?

Step 1. Confirm $C$ is the included angle. The two sides are $a = 9, b = 12$ ; the given angle is $C$ , whose letter is neither $a$ nor $b$ . ✔ Why this step? The formula only works when the angle sits between the two sides. Checking first stops the most common mistake.

Step 2. Compute $sin 4 0^{\circ} \approx 0.6428$ . Why this step? $sin C$ is the "across-ness" factor that turns the slanted side into a height.

Step 3. Substitute: $Area = \frac{1}{2} (9) (12) (0.6428) = 54 \times 0.6428 \approx 34.71.$ Why this step? Straight application — half of base times the height $b sin C$ .

Verify: If $C$ were $9 0^{\circ}$ we'd get exactly $54$ . Since $4 0^{\circ} < 9 0^{\circ}$ , the area should be less than $54$ , and $34.71 < 54$ . ✔ Units: sides in the same length unit → area in that unit squared.

Cell B — Right angle (the sanity anchor)

Cell C — Obtuse included angle

Worked example Example 3 (Cell C)

$a = 5$ , $b = 7$ , $C = 13 0^{\circ}$ . Find the area.

Forecast: $13 0^{\circ}$ is $5 0^{\circ}$ past the right angle. Since $sin$ is symmetric about $9 0^{\circ}$ , $sin 13 0^{\circ} = sin 5 0^{\circ}$ , so the area should equal the area you'd get with $C = 5 0^{\circ}$ . Guess: a bit less than the $17.5$ maximum.

Step 1. Note that the perpendicular height dropped from vertex $A$ lands outside the base line (see the figure — the dashed base is extended, and the violet height meets it beyond vertex $C$ ). Why this step? This is the scary part of obtuse triangles. We must confirm the formula still holds even though the picture looks different.

Step 2. In that external right triangle the height is $h = b sin (18 0^{\circ} - C)$ , but because $sin (18 0^{\circ} - C) = sin C$ (sine is positive and mirror-symmetric across $9 0^{\circ}$ — see Unit Circle and Sine), we simply get $h = b sin C$ again. Why this step? It proves the sign never flips: for any $0 < C < 18 0^{\circ}$ , $sin C > 0$ , so the formula is unchanged.

Step 3. $sin 13 0^{\circ} \approx 0.7660$ , so $Area = \frac{1}{2} (5) (7) (0.7660) = 17.5 \times 0.7660 \approx 13.41.$

Verify: $sin 13 0^{\circ}$ should equal $sin 5 0^{\circ} \approx 0.7660$ . ✔ And $13.41 < 17.5$ (the right-angle max), as required. ✔

Cell D — Solve for the angle (the ambiguous case)

Worked example Example 4 (Cell D)

A triangle has $a = 6$ , $b = 10$ , and area $= 15$ . Find the included angle $C$ .

Forecast: We're running the formula backwards. Careful: $sin$ gives the same value for an acute angle and its obtuse partner, so expect two possible triangles.

Step 1. Write the formula as an equation: $15 = \frac{1}{2} (6) (10) sin C = 30 sin C .$ Why this step? The known area becomes a constraint on the unknown $sin C$ .

Step 2. Solve for $sin C$ : $sin C = \frac{15}{30} = 0.5.$ Why this step? Isolate the sine so we can un-do it.

Step 3. Ask "which angle has this sine?" The calculator gives $arcsin (0.5) = 3 0^{\circ}$ . But look at the figure: the two violet sides can open to $3 0^{\circ}$ or swing past vertical to $15 0^{\circ}$ and still cast the same height. So $C = 3 0^{\circ} or C = 18 0^{\circ} - 3 0^{\circ} = 15 0^{\circ} .$ Why this step? $sin θ = sin (18 0^{\circ} - θ)$ — this is the ambiguous case. Reporting only $3 0^{\circ}$ throws away a real triangle.

Verify: Both must give area $15$ . $30 sin 3 0^{\circ} = 30 (0.5) = 15$ ✔ and $30 sin 15 0^{\circ} = 30 (0.5) = 15$ ✔. Both angles are valid unless a picture or extra fact rules one out.

Cell E — Solve for a side

Cell F — Degenerate / limiting case

Worked example Example 6 (Cell F)

$a = 8$ , $b = 5$ . What is the area as the included angle $C$ shrinks toward $0^{\circ}$ , and what if $C = 18 0^{\circ}$ ?

Forecast: If the two sides fold flat onto each other, there's no triangle left — zero area both ways.

Step 1. As $C \to 0^{\circ}$ : the sides point the same way, the triangle squashes into a line segment (top-left of figure). $sin 0^{\circ} = 0$ , so $Area = \frac{1}{2} (8) (5) (0) = 0.$ Why this step? $sin$ of $0$ kills the "across-ness" — nothing rises off the base.

Step 2. As $C \to 18 0^{\circ}$ : the sides point in opposite directions along one line (bottom of figure). $sin 18 0^{\circ} = 0$ , so again $Area = \frac{1}{2} (8) (5) (0) = 0.$ Why this step? A straight line has no interior area, from either extreme.

Step 3. The maximum in between is at $C = 9 0^{\circ}$ : $Area = \frac{1}{2} (8) (5) (1) = 20$ . Why this step? Confirms area rises from $0$ , peaks at the right angle, and falls back to $0$ — a smooth $sin$ hump.

Verify: $sin 0^{\circ} = 0$ , $sin 18 0^{\circ} = 0$ , $sin 9 0^{\circ} = 1$ . Areas $0, 0, 20$ respectively. ✔ The formula's whole "life story" over $0^{\circ}$ to $18 0^{\circ}$ is one arch of the sine curve scaled by $\frac{1}{2} ab = 20$ .

Cell G — Real-world word problem

Worked example Example 7 (Cell G)

A triangular garden plot has two fences of lengths $18 m$ and $24 m$ meeting at a corner where they make an angle of $6 5^{\circ}$ . How much ground does the plot cover?

Forecast: The two fences are the two sides; the corner angle is the included angle. This is a pure Cell-A computation dressed in words. Expect a few hundred square metres.

Step 1. Extract the maths: $a = 18 m$ , $b = 24 m$ , $C = 6 5^{\circ}$ . Why this step? The hard part of word problems is spotting that "the corner where they meet" is the included angle — exactly the sandwiched angle the formula demands.

Step 2. $sin 6 5^{\circ} \approx 0.9063$ . Why this step? Convert the lean of the fences into a height factor.

Step 3. Substitute with units: $Area = \frac{1}{2} (18) (24) (0.9063) = 216 \times 0.9063 \approx 195.76 m^{2} .$

Verify: Max possible for these fences (at $9 0^{\circ}$ ) is $\frac{1}{2} (18) (24) = 216 m^{2}$ . Our $6 5^{\circ}$ answer $195.76$ is just under that. ✔ Units: metre $\times$ metre = square metres — correct for area.

Cell H — Exam twist (combine with another rule)

Recall Quick self-test on the matrix

Which cell has two valid answers, and why? ::: Cell D (solve for angle) — because $sin C = sin (18 0^{\circ} - C)$ gives an acute and an obtuse solution. What is the largest area two fixed sides can enclose, and at what angle? ::: $\frac{1}{2} ab$ , achieved at $C = 9 0^{\circ}$ . As $C \to 0^{\circ}$ or $C \to 18 0^{\circ}$ , what is the area? ::: $0$ — the triangle collapses to a line. In a word problem, what must you identify before using the formula? ::: The two sides and the angle between them (the included angle).

Connections

Area of triangle = ½ab·sin C — the parent formula these examples drill.
Sine Rule — pairs with Cell D when you must find remaining angles/sides.
Cosine Rule — finds a missing third side before area if only two sides given without the angle.
Heron's Formula — the go-to when you know all three sides instead.
Right-Angled Trigonometry (SOHCAHTOA) — source of $h = b sin C$ used in every cell.
Cross Product — Example 9's vector twin.
Unit Circle and Sine — why $sin C > 0$ for obtuse $C$ (Cell C) and why Cell D is ambiguous.