Lesson 1 of 3Trigonometry of Acute Triangles

Unit 7 · Lesson 1 of 3

The Sine Law

Drawing an altitude splits any acute triangle into two right triangles, and writing the altitude two ways gives the sine law: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ . Use it to find a side or an angle whenever you know an angle together with the side opposite it.

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Key things to remember

1In $△ A BC$ , side $a$ is opposite $∠ A$ , side $b$ is opposite $∠ B$ , side $c$ is opposite $∠ C$ .
2Sine law: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ , or flipped, $\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ .
3It comes from writing the altitude two ways, $h = b sin A = a sin B$ , and dividing by $sin A sin B$ .
4Use it when you know an angle and its opposite side plus one more angle or side (AAS, ASA after the angle sum, or two sides and an angle opposite one of them).
5To find an angle, solve for its sine, then use $sin^{- 1}$ in degree mode. In an acute triangle the result is the angle you want; the two-triangle ambiguous case is Grade 11.
6The three angles add to $18 0^{\circ}$ : use this to find a third angle before or after the sine law.

Two cottages sit 800 m apart on the shore of a lake, and you want to know how far each one is from a small island. You can't stretch a measuring tape across the water, but standing at each cottage you can measure the angle between the shoreline and your line of sight to the island. The two cottages and the island form a triangle, and none of its angles is $9 0^{\circ}$ . SOH CAH TOA only works in a right triangle, so it can't solve this one directly. Surveyors, navigators and engineers meet triangles like this all the time. This lesson builds a tool for them, the sine law, out of the right-triangle trigonometry you already know.

An acute triangle is a triangle whose three angles are all less than $9 0^{\circ}$ . Every triangle you solve with the sine law or the cosine law in this unit is acute, apart from one right-angle example that shows how the cosine law connects to the Pythagorean theorem. (A triangle with a $9 0^{\circ}$ angle is a right triangle; one with an angle greater than $9 0^{\circ}$ is obtuse.)
Labelling convention: the vertices are capital letters $A$ , $B$ , $C$ , and each side gets the lower-case letter of the angle opposite it. Side $a$ is $BC$ , across from $∠ A$ ; side $b$ is $A C$ , across from $∠ B$ ; side $c$ is $A B$ , across from $∠ C$ .
An angle and the side opposite it form an opposite pair, such as $∠ A$ and side $a$ . The sine law is built from these pairs.
The angles of any triangle add to $18 0^{\circ}$ , so if you know two angles you always know the third.

Where the sine law comes from

The trick is to make the right angles yourself. In acute triangle $A BC$ , draw the altitude from $C$ : the line segment from $C$ that meets side $A B$ at a right angle. Call its length $h$ and the point where it lands $D$ . Because the triangle is acute, $D$ lands between $A$ and $B$ , and the altitude splits the triangle into two right triangles, $A C D$ on the left and $BC D$ on the right. Now SOH CAH TOA works in each piece. In the left triangle, the hypotenuse is $b$ and the side opposite $∠ A$ is $h$ . In the right triangle, the hypotenuse is $a$ and the side opposite $∠ B$ is $h$ .

Fig 1Acute triangle ABC with the altitude h from C to D on side AB. The left right triangle gives h = b sin A, the right one gives h = a sin B, and setting them equal leads to a / sin A = b / sin B.

sin A = \frac{h}{b} \Rightarrow h = b sin A sin B = \frac{h}{a} \Rightarrow h = a sin B

You now have two equations that both give $h$ . This is exactly the situation from solving a linear system by substitution: when two expressions are each equal to the same quantity, they are equal to each other, and the shared unknown $h$ disappears. So $b sin A = a sin B$ . Dividing both sides by $sin A sin B$ (neither sine is 0, because the angles are between $0^{\circ}$ and $9 0^{\circ}$ ) leaves $\frac{b}{sin B} = \frac{a}{sin A}$ . If you instead draw the altitude from $A$ to side $BC$ , the same steps give $\frac{b}{sin B} = \frac{c}{sin C}$ . Put together, all three ratios are equal. You don't need to reproduce this argument, but knowing it explains why the law is true and why it is all about opposite pairs.

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} or, flipped, \frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}

The sine law. Use the first form to find a side and the flipped form to find an angle.

Opposite pair	Side	Sine of the angle	Side ÷ sine
$a$ and $∠ A = 5 0^{\circ}$	10.0	$sin 5 0^{\circ} \approx 0.7660$	≈ 13.05
$b$ and $∠ B = 6 0^{\circ}$	11.3	$sin 6 0^{\circ} \approx 0.8660$	≈ 13.05
$c$ and $∠ C = 7 0^{\circ}$	12.3	$sin 7 0^{\circ} \approx 0.9397$	≈ 13.05

Measurements from one acute triangle with angles 50°, 60° and 70°. Each side divided by the sine of its opposite angle gives the same number (sides shown to one decimal; the ratios use the unrounded sides).

What the sine law needs

You need one complete opposite pair (an angle and the side across from it) plus one more measure: another angle, or another side. Write the equation using only two of the three ratios, the known pair and the pair with the unknown in it, then solve. If you know two sides and the angle between them, or all three sides, there is no complete pair, and you need the cosine law from the next lesson.

Finding a side and finding an angle

Take triangle $A BC$ with $∠ A = 3 5^{\circ}$ , $∠ B = 6 5^{\circ}$ and $A C = 18$ cm, and find $BC$ . First name the sides: $A C$ is across from $∠ B$ , so $b = 18$ ; $BC$ is across from $∠ A$ , so you want $a$ . The pair ( $∠ B$ , $b$ ) is complete, and $a$ pairs with the known $∠ A$ . So $\frac{a}{sin 3 5 ^{\circ}} = \frac{18}{sin 6 5 ^{\circ}}$ , and multiplying both sides by $sin 3 5^{\circ}$ gives $a = \frac{18 sin 3 5 ^{\circ}}{sin 6 5 ^{\circ}} \approx 11.4$ cm. When the unknown is an angle, use the flipped form so the unknown $sin$ is on top, solve for the sine, then press $sin^{- 1}$ (the inverse sine) to get the angle. Set your calculator to degree mode first: in radian mode, $sin 35$ comes out as about $- 0.43$ instead of $0.57$ .

Why acute triangles only, for now

An angle and its supplement have the same sine (for example

sin 4 0^{\circ} = sin 14 0^{\circ}

), but for a positive sine value

sin^{- 1}

only ever returns an angle from

0^{\circ}

9 0^{\circ}

. In an acute triangle every angle is under

9 0^{\circ}

, so the angle

sin^{- 1}

gives is the right one. When you know two sides and an angle that is not between them, some sets of measurements fit two different triangles, one with an obtuse angle. That situation is called the ambiguous case of the sine law, and you will study it in Grade 11.

Loading interactive…

Choose Solve, then Acute, and pick AAS or ASA: type in two angles and a side, and the triangle is drawn to scale and solved step by step with the sine law. Practice mode gives triangles with one unknown to find.

Rounding rule for this unit

Keep every intermediate value unrounded in your calculator (use its memory or the previous-answer key), and round only the final answer: sides to one decimal place and angles to the nearest degree. Rounding a sine to two decimals partway through can shift a side by a few tenths.

VideoAn Ontario teacher uses the sine law to find an unknown side in an acute triangle.Watch on YouTube

VideoThe second part of the lesson uses the sine law and inverse sine to find an unknown angle.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

△ A BC

∠ A = 3 5^{\circ}

∠ B = 6 5^{\circ}

, and

A C = 18

cm. Determine

BC

to one decimal place.

Example 2

△ PQR

∠ P = 4 8^{\circ}

∠ Q = 7 1^{\circ}

, and

PQ = 25

m. Find

QR

to one decimal place.

Example 3

In acute

△ D EF

∠ D = 6 8^{\circ}

d = 15

cm, and

e = 12

cm. Find

∠ E

to the nearest degree, then

∠ F

Example 4

Cottages

A

and

B

are 800 m apart on a straight shoreline. The angle between the shoreline and the line of sight to island

I

5 4^{\circ}

A

and

6 7^{\circ}

B

. How far is the island from cottage

B

, to the nearest tenth of a metre?

Where marks get lost

Pairing a side with the wrong angle. Side $a$ must go with $∠ A$ , the angle across from it, not an angle at one of its ends. Always name each side by its opposite angle before writing the equation.
Trying the sine law with ASA before finding the third angle. With $∠ A$ , $∠ B$ and side $c$ , no pair is complete until you compute $∠ C = 18 0^{\circ} - ∠ A - ∠ B$ .
Leaving the calculator in radian mode. $sin 35$ in radians is about $- 0.43$ ; in degrees it is about $0.57$ . A negative sine or a side longer than makes sense is the warning sign.
Forgetting the last step when finding an angle: $sin B = 0.74$ is not the answer. Press $sin^{- 1}$ to turn the sine into an angle.
Rounding the sine to two decimals partway through, then rounding again at the end, which can change the final tenth. Keep the full calculator value until the last line.

Flashcards 8

Acute triangle

Side a in triangle ABC

Sine law (finding a side)

Sine law (finding an angle)

+ 4 more

Practice

△ A BC

, which side is opposite

∠ B

In acute

△ A BC

∠ A = 4 0^{\circ}

∠ B = 7 5^{\circ}

, and

b = 20

cm. Find side

a

, to one decimal place.

In acute

△ A BC

∠ A = 7 0^{\circ}

a = 16

m, and

c = 13

m. Find

∠ C

, to the nearest degree.

△ X Y Z

∠ X = 5 2^{\circ}

∠ Y = 6 3^{\circ}

, and

X Y = 14

km. Find

Y Z

, to one decimal place.

In which situation can you use the sine law straight away to find side

b

, with no other step first?

Put the steps of the sine law's development in order.

1Write h two ways using the sine ratio: h = b sin A in one right triangle and h = a sin B in the other.
2Divide both sides by sin A sin B to get a / sin A = b / sin B.
3Since both expressions equal h, set them equal: b sin A = a sin B.
4Draw the altitude h from C, at right angles to side AB, splitting the triangle into two right triangles.

Triangle ABC has angle A = 35°, angle B = 65° and AC = 18 cm. Match each item with its value or name.

Angle C

The side opposite angle A

The side opposite angle C

The given side, AC

Which statement about using

sin^{- 1}

with the sine law is correct?

You compute

a = \frac{18 sin 35}{sin 65}

and your calculator shows about

- 9.3

. What most likely went wrong?

△ A BC

∠ A = 3 5^{\circ}

and

∠ B = 6 5^{\circ}

. What is

∠ C

, in degrees?

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