Coaching Centre
Obtuse angles and the area of a triangle
Sine, cosine and tangent of angles between 90° and 180°, why sin θ = sin(180° − θ), and the area formula A = ½ab sin C for any triangle.
In a right-angled triangle every other angle is acute, so you never needed the sine of $130^\circ$. Once triangles can have an obtuse angle, you do. The good news: your calculator already knows the answers. This page explains what they mean, then uses them in the first of the three big formulas.
Trig ratios of obtuse angles
Type $\sin 130^\circ$ and $\sin 50^\circ$ into your calculator. Both give $0.7660\ldots$. Now try $\cos 130^\circ$ and $\cos 50^\circ$: one is $0.6428\ldots$ and the other is $-0.6428\ldots$.
Why? Picture an arm of length $1$ swinging up from the positive $x$-axis. Its height is the sine and its sideways distance is the cosine. At $50^\circ$ and at $130^\circ$ the arm is at exactly the same height, mirrored across the vertical axis. So the sines match, but at $130^\circ$ the arm points left, so the cosine is negative.
- For an obtuse angle, sine is positive, cosine is negative and tangent is negative.
- A negative cosine is a useful signal: in 4E it tells you the angle you found is obtuse.
Find $\sin 130^\circ$ and $\cos 130^\circ$ to four decimal places, and name the acute angle with the same sine.
The area of any triangle
You know area $= \tfrac{1}{2} \times \text{base} \times \text{height}$. The trouble is that the height is rarely given. But if you know two sides $a$ and $b$ and the angle $C$ between them, the height is hiding inside a right-angled triangle: $h = b\sin C$.
Read it as "half the product of two sides, times the sine of the angle between them". The angle must be the included angle, the one where the two sides meet. It works for obtuse angles too, because $\sin C$ is still positive.
Set $a = 9$, $b = 7$ and $C = 60^\circ$ to check Example 2. Now slide $C$ to $120^\circ$. The triangle changes shape, so why doesn't the area change? Look at the dashed ghost triangle.
Keep $a$ and $b$ fixed and find the angle that gives the biggest area. Explain why using the yellow height.
Find the area of a triangle with sides $9$ cm and $7$ cm and an included angle of $60^\circ$.
A corner garden bed in Hornsby is a triangle with edges $4.2$ m and $5.5$ m meeting at $115^\circ$. How many square metres of mulch does it need?
Finding an angle from the area
Rearrange the formula: $\sin C = \dfrac{2A}{ab}$. There are usually two angles with that sine, one acute and one obtuse, and both give a triangle with the right area.
A triangle has sides $8$ cm and $7$ cm and an area of $20$ cm². Find the possible sizes of the angle between those sides.
Your turn
Find the area of a triangle with sides $7.5$ m and $9$ m and an included angle of $118^\circ$, to one decimal place.
Show the working
$A = \tfrac{1}{2} \times 7.5 \times 9 \times \sin 118^\circ \approx 29.8$ m².
Which angle has the same sine as $30^\circ$?
$\sin(180^\circ - 30^\circ) = \sin 30^\circ$, so $\sin 150^\circ = \sin 30^\circ = 0.5$.
The value of $\cos 120^\circ$ is:
$\cos 120^\circ = -\cos 60^\circ = -0.5$. Cosine is negative for obtuse angles.
Find the area of a triangle with sides $10$ cm and $12$ cm and an included angle of $35^\circ$, to one decimal place.
$A = \tfrac{1}{2} \times 10 \times 12 \times \sin 35^\circ = 60\sin 35^\circ \approx 34.4$ cm².
Find the area of a triangle with sides $6$ m and $8$ m and an included angle of $140^\circ$, to one decimal place.
$A = \tfrac{1}{2} \times 6 \times 8 \times \sin 140^\circ = 24\sin 140^\circ \approx 15.4$ m².
Which angle has the same sine as $72^\circ$?
$180^\circ - 72^\circ = 108^\circ$, and supplementary angles share a sine.
A triangular block of land has two boundaries of $45$ m and $60$ m meeting at $102^\circ$. Find its area to the nearest square metre.
$A = \tfrac{1}{2} \times 45 \times 60 \times \sin 102^\circ = 1350\sin 102^\circ \approx 1320$ m².
$\sin C = 0.6$ and $C$ is obtuse. Find $C$ to one decimal place.
$\sin^{-1}(0.6) \approx 36.9^\circ$ is the acute angle. The obtuse angle is $180^\circ - 36.87^\circ \approx 143.1^\circ$.
Exam style: A triangular shade sail has an area of $12$ m². Two of its edges are $5$ m and $6$ m, and the angle between them is acute. Find that angle to one decimal place.
$12 = \tfrac{1}{2} \times 5 \times 6 \times \sin C = 15\sin C$, so $\sin C = 0.8$. The acute angle is $C = \sin^{-1}(0.8) \approx 53.1^\circ$.
Exam style: A triangular lawn has edges of $18$ m and $24$ m meeting at $128^\circ$. Turf costs $\$4.50$ per square metre. Find the cost of turfing the lawn, to the nearest cent.
Area $= \tfrac{1}{2} \times 18 \times 24 \times \sin 128^\circ = 216\sin 128^\circ \approx 170.21$ m². Cost $= 170.21 \times 4.50 \approx \$765.95$ (using the unrounded area).