True or False? SOHCAHTOA works in any triangle.

False. Both SOHCAHTOA and Pythagoras need a right angle . For any other triangle you need the sine rule, the cosine rule, or the area formula.

Which sides and angles must you know to use the sine rule?

You need a matching pair , meaning a side and the angle opposite it, plus one more side or angle. Without a complete pair the rule gives you two unknowns in one equation and cannot be solved.

Which two situations need the cosine rule?

When you know two sides and the angle between them , and want the third side. And when you know all three sides , and want any angle. Neither situation gives you a side paired with its opposite angle.

How do you approach a problem that needs more than one triangle?

Work out which triangle contains the length or angle you have been asked for, then find whatever that triangle is missing by working in a neighbouring triangle first. Choose the rule triangle by triangle, so a right-angled one can use SOHCAHTOA while the next one uses the sine or cosine rule.

What is the difference between an angle of elevation and an angle of depression?

Both are measured from the horizontal . An angle of elevation is measured upwards to an object above you, and an angle of depression downwards to an object below you. Drawing the horizontal line in before you start stops you measuring from the vertical by mistake.

What is the first step in any 3D trigonometry problem?

Find a right-angled triangle inside the solid that contains what you know and what you want, then redraw it flat as a 2D triangle. Once it is drawn separately, the problem is ordinary Pythagoras or SOHCAHTOA.

In a pyramid, which triangle gives the angle between a sloping edge and the base?

The triangle formed by the sloping edge, the vertical height from the apex, and the line joining the centre of the base to that corner. The vertical from the apex of a symmetrical pyramid meets the base at its centre , so that third side is half a base diagonal.

True or False? SOHCAHTOA can be used in a 3D problem.

True. It applies to any right-angled triangle you can find inside the solid. What you cannot do is apply it to a triangle drawn on a 3D sketch without checking that the angle really is a right angle, since a diagram in perspective can make angles look wrong.

Why must you be careful which base length you use in a 3D problem?

Because the triangle almost always needs the base diagonal , not a base edge, and the two are easy to confuse on a perspective sketch. Labelling the length you have just calculated on your own 2D redrawing is what prevents it.

GCSEFurther MathsAQAFlashcards6. GeometryTrigonometry

Trigonometry (AQA GCSE Further Maths): Flashcards

Exam code: 8365

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FrontRight-Angled Trigonometry
Complete Pythagoras' theorem, where $c$ is the hypotenuse:
$a^{2} + b^{2} =______$

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Cards in this collection (27)

Complete Pythagoras' theorem, where $c$ is the hypotenuse:
$a^{2} + b^{2} =______$
The completed theorem is:
$a^{2} + b^{2} = c^{2}$
The hypotenuse is the longest side and is always opposite the right angle, and the theorem holds only for right-angled triangles.
When rearranging Pythagoras, when do you add and when do you subtract?
Add inside the root to find the hypotenuse, so $c = \sqrt{a^{2} + b^{2}}$ .
Subtract to find one of the shorter sides, so $a = \sqrt{c^{2} - b^{2}}$ .
A right-angled triangle has shorter sides $x$ and 13. Find an expression for the hypotenuse $y$ .
Pythagoras gives $x^{2} + 13^{2} = y^{2}$ , so $y = \sqrt{x^{2} + 169}$ .
No plus-or-minus is needed because $y$ is a length, and you cannot square-root the two terms separately.
Complete the three trigonometric ratios:
$\sin θ = \frac{opposite}{______}$
$\cos θ = \frac{______}{hypotenuse}$
$\tan θ = \frac{opposite}{______}$
The completed ratios are:
$\sin θ = \frac{opposite}{hypotenuse}$
$\cos θ = \frac{adjacent}{hypotenuse}$
$\tan θ = \frac{opposite}{adjacent}$
SOHCAHTOA is the usual way of remembering which ratio is which.
Which sides change label when you pick a different angle?
The opposite and adjacent swap over, because both are defined relative to the angle $θ$ .
The hypotenuse never changes, since it is always the side opposite the right angle.
How do you find an angle when you know two sides?
Choose the ratio that uses those two sides, then apply the inverse function.
For example $\cos y = \frac{8}{23}$ gives $y = \cos^{- 1} (\frac{8}{23}) = 69.6 °$ to 3 significant figures.
True or False?
SOHCAHTOA works in any triangle.
False.
Both SOHCAHTOA and Pythagoras need a right angle.
For any other triangle you need the sine rule, the cosine rule, or the area formula.
Complete the exact values:
$\sin 30 ° =______$
$\cos 30 ° =______$
$\tan 45 ° =______$
The completed values are:
$\sin 30 ° = \frac{1}{2}$
$\cos 30 ° = \frac{\sqrt{3}}{2}$
$\tan 45 ° = 1$
Also worth knowing: $\sin 60 ° = \frac{\sqrt{3}}{2}$ , $\tan 60 ° = \sqrt{3}$ , and $\tan 90 °$ is not defined.
Which two triangles give you all the exact trig values you need?
A right-angled triangle with hypotenuse 2 and shortest side 1, whose third side is $\sqrt{3}$ and whose angles are $30 °$ and $60 °$ .
And a right-angled isosceles triangle with two sides of 1, whose hypotenuse is $\sqrt{2}$ and whose other angles are both $45 °$ .
How are $\sin θ$ and $\cos θ$ related between $0 °$ and $90 °$ ?
The values of $\sin θ$ from $0 °$ to $90 °$ match those of $\cos θ$ from $90 °$ back down to $0 °$ .
So $\sin 30 °$ and $\cos 60 °$ are both $\frac{1}{2}$ , which halves what there is to remember.
Complete the sine rule:
$\frac{a}{\sin A} = \frac{b}{______} = \frac{______}{\sin C}$
The completed rule is:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Each side is paired with the angle opposite it, so side $a$ is opposite angle $A$ .
You are expected to know this formula; it is not given in the exam.
Which sides and angles must you know to use the sine rule?
You need a matching pair, meaning a side and the angle opposite it, plus one more side or angle.
Without a complete pair the rule gives you two unknowns in one equation and cannot be solved.
Complete the cosine rule:
$a^{2} = b^{2} + c^{2} -______$
The completed rule is:
$a^{2} = b^{2} + c^{2} - 2 b c \cos A$
The angle $A$ is the one between sides $b$ and $c$ , and it sits opposite the side $a$ you are finding.
This formula also has to be memorised.
Which two situations need the cosine rule?
When you know two sides and the angle between them, and want the third side.
And when you know all three sides, and want any angle.
Neither situation gives you a side paired with its opposite angle.
How do you rearrange the cosine rule to find an angle?
Make $\cos A$ the subject:
$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
Then apply $\cos^{- 1}$ . Note that the side opposite the angle you want is the one that is subtracted.
Complete the formula for the area of a triangle:
$Area = \frac{1}{2} a b \sin______$
The completed formula is:
$Area = \frac{1}{2} a b \sin C$
This one is also on the list you must memorise, and it works in any triangle, not just a right-angled one.
True or False?
In $Area = \frac{1}{2} a b \sin C$ , the angle $C$ must lie between the sides $a$ and $b$ .
True.
The angle has to be the included angle, the one enclosed by the two sides you are using.
Using any other angle gives a wrong area, and it is one of the most common errors in this topic.
A triangle has two sides of length $2 x$ with an angle of $30 °$ between them. Find the exact value of $d^{2}$ , where $d$ is the third side.
The cosine rule gives $d^{2} = {(2 x)}^{2} + {(2 x)}^{2} - 2 (2 x) (2 x) \cos 30 °$ .
Using $\cos 30 ° = \frac{\sqrt{3}}{2}$ this becomes $8 x^{2} - 4 \sqrt{3} x^{2}$ , so:
$d^{2} = 4 x^{2} (2 - \sqrt{3})$
Keeping the surd rather than rounding is what makes the answer exact.
How do you approach a problem that needs more than one triangle?
Work out which triangle contains the length or angle you have been asked for, then find whatever that triangle is missing by working in a neighbouring triangle first.
Choose the rule triangle by triangle, so a right-angled one can use SOHCAHTOA while the next one uses the sine or cosine rule.
What is the difference between an angle of elevation and an angle of depression?
Both are measured from the horizontal. An angle of elevation is measured upwards to an object above you, and an angle of depression downwards to an object below you.
Drawing the horizontal line in before you start stops you measuring from the vertical by mistake.
What is the first step in any 3D trigonometry problem?
Find a right-angled triangle inside the solid that contains what you know and what you want, then redraw it flat as a 2D triangle.
Once it is drawn separately, the problem is ordinary Pythagoras or SOHCAHTOA.
Complete the formula for the longest diagonal $d$ of a cuboid with edges $a$ , $b$ and $c$ :
$d^{2} = a^{2} + b^{2} +______$
The completed formula is:
$d^{2} = a^{2} + b^{2} + c^{2}$
This diagonal runs from one corner of the cuboid to the opposite corner, passing through the inside of the solid.
Why does the cuboid diagonal formula have three squared terms?
Because Pythagoras is used twice. The first use gives the diagonal across the base, $a^{2} + b^{2}$ , and the second uses that diagonal with the height.
Substituting the first result into the second is what merges the two steps into one formula, and the two-step method still works in solids that are not cuboids.
In a pyramid, which triangle gives the angle between a sloping edge and the base?
The triangle formed by the sloping edge, the vertical height from the apex, and the line joining the centre of the base to that corner.
The vertical from the apex of a symmetrical pyramid meets the base at its centre, so that third side is half a base diagonal.
True or False?
SOHCAHTOA can be used in a 3D problem.
True.
It applies to any right-angled triangle you can find inside the solid.
What you cannot do is apply it to a triangle drawn on a 3D sketch without checking that the angle really is a right angle, since a diagram in perspective can make angles look wrong.
A cuboid measures 3 cm by 4 cm by 12 cm. Find the angle between the longest diagonal and the base.
The base diagonal is $\sqrt{3^{2} + 4^{2}} = 5$ cm, and this is the adjacent side of the triangle you need, with the height 12 cm opposite.
So $\tan θ = \frac{12}{5}$ , giving:
$θ = 67.4 °$
to 3 significant figures. As a check, the diagonal itself is $\sqrt{5^{2} + 12^{2}} = 13$ cm.
Why must you be careful which base length you use in a 3D problem?
Because the triangle almost always needs the base diagonal, not a base edge, and the two are easy to confuse on a perspective sketch.
Labelling the length you have just calculated on your own 2D redrawing is what prevents it.

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