Coaching Centre
The sine rule
Labelling a triangle, the sine rule for finding a side or an angle, when to use it (a matching side and angle pair), and the obtuse-angle trap.
The sine rule says that in any triangle, each side divided by the sine of the angle opposite it gives the same number. That one fact lets you find missing sides and angles without a right angle anywhere.
Labelling the triangle
Name the corners with capital letters $A$, $B$, $C$. Then name each side with the lower-case letter of the corner it faces: side $a$ is opposite angle $A$, and so on. A side and the angle opposite it are a matching pair. The longest side is always opposite the largest angle, and the shortest side opposite the smallest angle. That is your quickest sense check.
The rule
Why is it true? Drop a perpendicular of height $h$ from $C$ onto side $c$. From one right-angled triangle, $h = b\sin A$. From the other, $h = a\sin B$. So $b\sin A = a\sin B$, and dividing both sides by $\sin A\sin B$ gives $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$.
You only ever use two of the three fractions at once. Put the unknown on top: use the side form to find a side, and the flipped form to find an angle. That saves a rearranging step.
Drag the corners around. However you stretch the triangle, the three readouts stay equal. Make the triangle as close to right-angled at $C$ as you can, then compare the shared value with side $c$. What do you notice?
Make angle $A$ the biggest angle. Which side is now the longest?
Finding a side
In triangle $ABC$, $A = 48^\circ$, $B = 62^\circ$ and $a = 10$ cm. Find $b$ to two decimal places.
Two surveyors stand $200$ m apart at $A$ and $B$ on one bank of the Parramatta River. A gum tree $C$ stands on the far bank. Angle $CAB = 72^\circ$ and angle $CBA = 58^\circ$. Find the distance $AC$ and the width of the river, assuming the bank is straight.
Finding an angle
In triangle $ABC$, $a = 12$ m, $b = 9$ m and $A = 64^\circ$. Find $B$.
Your turn
In triangle $ABC$, $A = 55^\circ$, $C = 80^\circ$ and $c = 14$ cm. Find $a$ to one decimal place.
Show the working
$(c, C)$ is a matching pair. $\dfrac{a}{\sin 55^\circ} = \dfrac{14}{\sin 80^\circ}$, so $a = \dfrac{14\sin 55^\circ}{\sin 80^\circ} \approx 11.6$ cm. Sense check: $A$ is smaller than $C$, so $a$ is shorter than $c$.
Which of these is a correct form of the sine rule?
Each side must be divided by the sine of the angle opposite it: $a$ with $A$, $c$ with $C$.
In triangle $ABC$, $A = 40^\circ$, $B = 75^\circ$ and $a = 8$ cm. Find $b$ to two decimal places.
$b = \dfrac{8\sin 75^\circ}{\sin 40^\circ} \approx 12.02$ cm.
In triangle $ABC$, $a = 15$, $b = 10$ and $A = 110^\circ$. Find $B$ to one decimal place.
$\sin B = \dfrac{10\sin 110^\circ}{15} \approx 0.6265$, so $B \approx 38.8^\circ$. A triangle can only have one obtuse angle, and $A$ already is, so $B$ is acute.
Which set of information lets you use the sine rule straight away?
With two angles you can find the third, so every side has a known opposite angle and you have a matching pair. Three sides, or two sides with the included angle, need the cosine rule. Three angles fix the shape but not the size.
In triangle $ABC$, $A = 35^\circ$, $B = 85^\circ$ and $c = 20$ m. Find $a$ to two decimal places.
$C = 180^\circ - 35^\circ - 85^\circ = 60^\circ$. Then $a = \dfrac{20\sin 35^\circ}{\sin 60^\circ} \approx 13.25$ m.
In triangle $ABC$, $b = 7.2$, $c = 9$ and $B = 50^\circ$. Angle $C$ is acute. Find $C$ to one decimal place.
$\sin C = \dfrac{9\sin 50^\circ}{7.2} \approx 0.9576$, so $C \approx 73.2^\circ$ (the acute value, as given).
In a triangle, the longest side is opposite:
From the sine rule, a bigger angle (up to $90^\circ$, and an obtuse angle is always the largest) gives a bigger sine and so a longer opposite side.
Exam style: Two lifeguards stand $500$ m apart at $A$ and $B$ on a straight beach. A surfer $S$ is out to sea. Angle $SAB = 63^\circ$ and angle $SBA = 48^\circ$. Find the distance $AS$ to one decimal place.
Angle $S = 180^\circ - 63^\circ - 48^\circ = 69^\circ$. $AS$ is opposite angle $B$. $AS = \dfrac{500\sin 48^\circ}{\sin 69^\circ} \approx 398.0$ m.
Exam style: From point $P$, the angle of elevation of the top $T$ of a tower is $30^\circ$. After walking $50$ m straight towards the tower to $Q$, the angle of elevation is $45^\circ$. Find the height of the tower to one decimal place.
In triangle $PQT$: angle $P = 30^\circ$, angle $PQT = 180^\circ - 45^\circ = 135^\circ$, so angle $T = 15^\circ$. Sine rule: $QT = \dfrac{50\sin 30^\circ}{\sin 15^\circ} \approx 96.59$ m. Then in the right-angled triangle at $Q$: height $= 96.59\sin 45^\circ \approx 68.3$ m.