When should you use the sine rule rather than the cosine rule?

Use it whenever the question gives you an opposite pair , a side together with the angle facing it. Two sides and an angle opposite one of them gives another angle, and two angles with a side gives another side.

What is the ambiguous case of the sine rule?

Two different triangles can sometimes fit the same two sides and non-included angle, one with an acute angle and one with an obtuse one. The inverse sine only ever gives the acute answer, so the diagram or the question has to tell you which is wanted.

When do you use the cosine rule to find a length?

Use it when you know two sides and the angle between them , and want the side opposite that angle. That is exactly the case the sine rule cannot handle, because there is no opposite pair to work with.

What must you have before the area formula can be used at all?

Two sides and the angle between them, which is the same setup the cosine rule needs for a length. Without that arrangement you may have to use the sine or cosine rule first to find a missing piece.

Where do you start when asked for the bearing of A from B?

Start at B , the point named after the word from, and draw a North line there. Then measure clockwise from that North line round to the line joining B to A.

What is a bearings question usually really testing?

A bearings question is normally a set-up for trigonometry, so the work is done with Pythagoras, right-angled trigonometry, or the sine or cosine rule. Missing distances and angles usually have to be found before the bearing itself can be worked out.

What two things do you match against each other to choose a triangle rule?

What the question gives you and what it asks for, taken together, decide the rule. An opposite pair points to the sine rule, while the angle between two sides points to the cosine rule or the area formula.

When does the area formula come into the choice of rule?

When you have two sides with the angle between them and the question asks for the area . That is the same information the cosine rule uses for a length, so the difference lies only in what is wanted.

True or False? A harder triangle question may need more than one rule in the same solution.

True. You may need the sine rule to find an angle and then the cosine rule or the area formula to finish. The area formula in particular often needs a missing side or angle found first, since it demands the angle between two known sides.

How can adding a line to a diagram open up a question?

Drawing a perpendicular creates a right angle that was not marked, which lets you use Pythagoras or the right-angled ratios. Dropping a vertical from the apex of a triangle down to its base is the usual move.

Which National 4 tools are still needed in these questions?

Pythagoras' theorem and SOHCAHTOA , the right-angled trigonometric ratios. Both apply as soon as a right angle appears, whether it was marked in the question or drawn in by you.

National 5MathsSQAFlashcardsTrigonometric SkillsTrigonometry with Triangles

Trigonometry with Triangles (SQA National 5 Maths): Flashcards

Exam code: X847 75

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How are the sides and angles labelled when using the sine or cosine rule?

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How are the sides and angles labelled when using the sine or cosine rule?
Angles take upper case letters and the side opposite each angle takes the matching lower case letter.
So side $a$ lies opposite angle $A$ , and labelling the triangle this way is the first step in every question.
When should you use the sine rule rather than the cosine rule?
Use it whenever the question gives you an opposite pair, a side together with the angle facing it.
Two sides and an angle opposite one of them gives another angle, and two angles with a side gives another side.
How do you use the sine rule to find a missing length?
Put the sides on top, equate just two of the three parts, and solve for the side you want.
With $8.1$ opposite $27 °$ and $109.4 °$ opposite $y$ , that gives $y = \frac{8.1}{\sin 27} \times \sin 109.4 = 16.8$ .
Why is the sine rule flipped when you are looking for an angle?
Writing it as $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ puts the angles on top, so the unknown is easier to reach.
The two forms say exactly the same thing, and only two of the three parts are ever needed at once.
A triangle has $12.3$ opposite angle $x$ and $8.1$ opposite $27 °$ , so complete the substitution.
$\frac{\sin x}{______} = \frac{\sin 27}{______}$
The completed substitution is:
$\frac{\sin x}{12.3} = \frac{\sin 27}{8.1}$
Each side sits under the sine of the angle it faces, which gives $x = 43.6 °$ to one decimal place.
What is the ambiguous case of the sine rule?
Two different triangles can sometimes fit the same two sides and non-included angle, one with an acute angle and one with an obtuse one.
The inverse sine only ever gives the acute answer, so the diagram or the question has to tell you which is wanted.
True or False?
If an angle found by the sine rule turns out to be obtuse, it is $180 °$ minus the acute answer.
True.
The sine of an acute angle equals the sine of $180 °$ minus that angle, which is the same symmetry used to solve trigonometric equations.
So an inverse sine of $40 °$ leaves $140 °$ as the other possibility.
When do you use the cosine rule to find a length?
Use it when you know two sides and the angle between them, and want the side opposite that angle.
That is exactly the case the sine rule cannot handle, because there is no opposite pair to work with.
How do you adapt the cosine rule to a differently labelled triangle?
Swap the letters so that the lower case letter on the left matches the upper case one inside the cosine.
So $b^{2} = a^{2} + c^{2} - 2 a c \cos B$ and $c^{2} = a^{2} + b^{2} - 2 a b \cos C$ are equally valid.
A triangle has sides $8.1$ and $12.3$ with $109 °$ between them, so complete the third side.
$b^{2} = 8 . 1^{2} + 12 . 3^{2} - 2 \times 8.1 \times 12.3 \times \cos 109$
$b =______cm to 3 significant figures$
The completed working is:
$b = \sqrt{8 . 1^{2} + 12 . 3^{2} - 2 \times 8.1 \times 12.3 \times \cos 109} = 16.8 cm$
Remember to take the square root at the end, since the formula gives $b^{2}$ rather than $b$ .
When do you use the cosine rule to find an angle?
Use it when you know all three sides and want any one of the angles.
Rearranged as $\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ , it gives the angle sitting between $b$ and $c$ .
In the angle form of the cosine rule which side appears after the minus sign?
The side opposite the angle you are finding, so $\cos A$ has $a^{2}$ after the minus sign.
That gives $\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ when the angle wanted is $C$ .
A triangle has sides $3.8$ , $4.2$ and $7.1$ so how is the angle between the two shorter sides found?
Label the side opposite that angle as $a = 7.1$ , then use $\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ .
Substituting gives $\frac{3 . 8^{2} + 4 . 2^{2} - 7 . 1^{2}}{2 \times 3.8 \times 4.2}$ , so $A = 125.0 °$ to one decimal place.
In $Area = \frac{1}{2} a b \sin C$ where must the angle $C$ sit?
The angle must lie between the two sides you use, so $C$ sits between $a$ and $b$ .
Swapping letters gives $\frac{1}{2} a c \sin B$ and $\frac{1}{2} b c \sin A$ , and each still has its angle in between.
What does the area formula become when the angle between the sides is a right angle?
Since $\sin 90 ° = 1$ , the formula collapses to $\frac{1}{2} a b$ .
That is the familiar half base times height, so the general formula contains the right-angled one as a special case.
True or False?
In the Formulae List version $A = \frac{1}{2} a b \sin C$ the $A$ stands for the angle at $A$ .
False.
The $A$ there stands for Area, and has nothing to do with an angle labelled $A$ .
The only angle in the formula is $C$ , the one sitting between the two sides used.
A triangle has sides $8.1$ and $12.3$ with $109 °$ between them, so complete its area.
$Area = \frac{1}{2} \times 8.1 \times 12.3 \times \sin______=______{cm}^{2}$
The completed area is:
$Area = \frac{1}{2} \times 8.1 \times 12.3 \times \sin 109 = 47.1 {cm}^{2}$
The angle used is the one between the two given sides, and the answer is to three significant figures.
What must you have before the area formula can be used at all?
Two sides and the angle between them, which is the same setup the cosine rule needs for a length.
Without that arrangement you may have to use the sine or cosine rule first to find a missing piece.
How do you find a side when the area and one other side are known?
Substitute everything you know into the area formula and rearrange to make that side the subject.
An area of $160$ with a side of $30$ and an angle of $26 °$ gives $c = \frac{2 \times 160}{30 \times \sin 26} = 24.3$ cm.
How do you find the angle when the area and both sides are known?
Rearrange the formula to $\sin C = \frac{2 \times Area}{a b}$ , then take the inverse sine.
It is the inverse sine that is needed here, not the inverse cosine, because the formula contains a sine.
Complete the three rules that every bearing must follow.
$measured from______$
$measured______$
$written with______digits$
The completed rules are:
$measured from North, measured clockwise, written with three digits$
So an angle of $59 °$ is written as $059 °$ when it is a bearing.
Where do you start when asked for the bearing of A from B?
Start at B, the point named after the word from, and draw a North line there.
Then measure clockwise from that North line round to the line joining B to A.
How do you get the bearing of B from A once you know the bearing of A from B?
Add $180 °$ if the bearing you have is less than $180 °$ , and subtract $180 °$ if it is more.
Either way the answer stays between $000 °$ and $360 °$ , which is why the rule has two cases.
True or False?
A bearing is always measured clockwise, even when the shorter turn would be anticlockwise.
True.
Bearings are always measured clockwise from North, however far round that takes you.
A direction just west of North is therefore about $350 °$ , not $010 °$ measured the other way.
What bearings do the four main compass directions have?
North is $000 °$ , East is $090 °$ , South is $180 °$ and West is $270 °$ .
Each quarter turn clockwise adds another $90 °$ .
What is a bearings question usually really testing?
A bearings question is normally a set-up for trigonometry, so the work is done with Pythagoras, right-angled trigonometry, or the sine or cosine rule.
Missing distances and angles usually have to be found before the bearing itself can be worked out.
What two things do you match against each other to choose a triangle rule?
What the question gives you and what it asks for, taken together, decide the rule.
An opposite pair points to the sine rule, while the angle between two sides points to the cosine rule or the area formula.
When does the area formula come into the choice of rule?
When you have two sides with the angle between them and the question asks for the area.
That is the same information the cosine rule uses for a length, so the difference lies only in what is wanted.
True or False?
A harder triangle question may need more than one rule in the same solution.
True.
You may need the sine rule to find an angle and then the cosine rule or the area formula to finish.
The area formula in particular often needs a missing side or angle found first, since it demands the angle between two known sides.
What can you try when none of the three rules seems to apply?
Use the fact that the angles of a triangle add to $180 °$ to find a missing angle first.
That often creates the opposite pair the sine rule needs, or the angle between two sides the other rules need.
How can adding a line to a diagram open up a question?
Drawing a perpendicular creates a right angle that was not marked, which lets you use Pythagoras or the right-angled ratios.
Dropping a vertical from the apex of a triangle down to its base is the usual move.
A rocket is seen at $72 °$ from C and $62 °$ from A, which are $400$ m apart, so how is its height found?
Find the third angle as $180 - 72 - 62 = 46 °$ , then use the sine rule to get the slant distance $A R = 528.8$ m.
Dropping a perpendicular from the rocket gives a right angle, so the height is $528.8 \times \sin 62 = 467$ m.
Which National 4 tools are still needed in these questions?
Pythagoras' theorem and SOHCAHTOA, the right-angled trigonometric ratios.
Both apply as soon as a right angle appears, whether it was marked in the question or drawn in by you.

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