Area of a Segment: Is Your Calculator in Radian Mode?

The question

The diagram shows a sector $OAB$ of a circle with centre $O$ and radius $8\ \text{cm}$ . The angle $AOB$ is $1.2$ radians. The shaded region is the segment between the chord $AB$ and the arc $AB$ .

Sector OAB with centre O, radii OA and OB of 8 cm and angle AOB of 1.2 radians. The chord AB is drawn, and the segment between chord AB and arc AB is shaded. — Sector OAB with radius 8 cm and angle 1.2 rad; the segment above chord AB is shaded.

(a) Find the perimeter of the shaded segment. [3 marks]

(b) Find the area of the shaded segment. [3 marks]

Give your answers correct to 3 significant figures.

The common wrong answer

This is the working I see most often for part (b). The method is right:

\text{area of segment} = \text{area of sector} - \text{area of triangle } OAB

\frac{1}{2}(8)^2(1.2) - \frac{1}{2}(8)^2\sin 1.2 = 38.4 - 32\sin 1.2

Then the calculator, still in degree mode from the last trigonometry question, gives $\sin 1.2 = 0.02094\ldots$ , so

38.4 - 32 \times 0.02094 = 38.4 - 0.670 = 37.7\ \text{cm}^2

The formula is correct, but $\sin 1.2$ has been worked out as the sine of $1.2^\circ$ , a tiny angle, instead of $1.2$ radians (about $69^\circ$ ). The triangle comes out as only $0.670\ \text{cm}^2$ , so the “segment” is almost the whole sector. Look at the diagram: triangle $OAB$ clearly takes up most of the sector, so that can’t be right.

The same slip spoils part (a): $16\sin 0.6$ in degree mode gives a chord of $0.168\ \text{cm}$ , when the diagram shows $AB$ is a little longer than the $8\ \text{cm}$ radii.

The correct working

Set your calculator to radian mode first. The angle is $\theta = 1.2$ rad and $r = 8\ \text{cm}$ .

(a) Arc length, using $s = r\theta$ :

s = 8 \times 1.2 = 9.6\ \text{cm}

For the chord, split triangle $OAB$ down the middle into two right-angled triangles, each with angle $0.6$ rad at $O$ :

AB = 2 \times 8\sin 0.6 = 16\sin 0.6 = 9.0343\ \text{cm}

(The cosine rule gives the same: $AB^2 = 8^2 + 8^2 - 2(8)(8)\cos 1.2 = 81.618$ , so $AB = 9.0343\ \text{cm}$ .)

Perimeter of segment:

9.6 + 9.0343 = 18.634

So the perimeter is $18.6\ \text{cm}$ .

(b) Area of sector, using $A = \frac{1}{2}r^2\theta$ :

\frac{1}{2}(8)^2(1.2) = 38.4\ \text{cm}^2

Area of triangle $OAB$ , using $\frac{1}{2}ab\sin C$ :

\frac{1}{2}(8)(8)\sin 1.2 = 32\sin 1.2 = 29.8253\ \text{cm}^2

Area of segment:

38.4 - 29.8253 = 8.5747

So the area is $8.57\ \text{cm}^2$ .

Why it works

The formulas $s = r\theta$ and $A = \frac{1}{2}r^2\theta$ only work when $\theta$ is in radians, and in this question the angle is given in radians. That means every trigonometric function of that angle must be worked out in radians as well. Your calculator doesn’t know which unit the question uses: $\sin 1.2$ in degree mode is the sine of $1.2^\circ$ , a completely different angle from $1.2$ rad.

You could convert instead ( $1.2$ rad $\approx 68.75^\circ$ ) and use degree mode for the sine, but that’s an extra step and an extra chance to round too early. Staying in radians throughout is simpler.

Examiners mention this one regularly: candidates who set out a correct method lose the accuracy marks because their calculator was in degree mode, or because they mixed degrees and radians in the same working.

How to spot it next time

When you see “radians” or an angle with no degree sign (like $1.2$ or $\frac{\pi}{3}$ ), check that your calculator shows R or RAD before you press anything.
Quick test: $\sin 1.2$ should be about $0.932$ . If you get $0.0209$ , you’re in degree mode.
Sense-check against the diagram. A segment cut off by an angle of about $69^\circ$ is only a small part of the sector (here about $22\%$ ), and a chord across that angle is about as long as the radius, not a fraction of a centimetre.