Grade 10 Trigonometry

Trig ratios in a triangle

Three fractions built from the sides of a right-angled triangle, and the angle fixes all three.

right-angled triangle

A triangle with one angle of exactly $90^{\circ }$.

hypotenuse

The side opposite the right angle. It is always the longest side.

opposite side

The side across from the marked angle. It does not touch that angle.

adjacent side

The side that touches the marked angle and is not the hypotenuse.

ratio

One length divided by another, written as a fraction.

inverse function

The rule that runs backwards. Give it a ratio and it gives you the angle, using ${\sin }^{-1}$, ${\cos }^{-1}$ or ${\tan }^{-1}$.

Start in a right-angled triangle and mark one of the two smaller angles. Call it $\theta$. Now name the three sides from that angle. The hypotenuse sits opposite the right angle. The opposite side lies across from $\theta$. The adjacent side is the third one, and it touches $\theta$.

Three ratios come out of those names. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. Tangent is the only one that leaves the hypotenuse out.

Now picture a bigger triangle with the same angles. Every side is longer, but each ratio gives the same number. Doubling the top and the bottom of a fraction changes nothing. So the ratio belongs to the angle, not to the size. Working in metres instead of centimetres changes nothing either.

A calculator finishes the job, but only in DEG mode. Check that before every question. An angle in, a ratio out: press sin, cos or tan. A ratio in, an angle out: use the inverse function.

Rules to remember

• $\sin \theta =\frac{\text{opp}}{\text{hyp}}$
• $\cos \theta =\frac{\text{adj}}{\text{hyp}}$
• $\tan \theta =\frac{\text{opp}}{\text{adj}}$

Examples

Traps

• Swapping the opposite and adjacent sides when the triangle is rotated or flipped on the page. Find the marked angle first, then the hypotenuse. Opposite and adjacent follow from those two.
• Writing $\tan \theta$ as adjacent over opposite. Tangent is opposite over adjacent. Say the ratio in words before you fill in numbers.
• Expecting a bigger triangle to give bigger ratio values. The ratio depends only on the angle. A $6$ cm and $8$ cm triangle gives the same $\frac{3}{4}$.
• Leaving the calculator in radian mode and trusting the answer. Set the mode to DEG before you start. Check that $\sin 30^{\circ }$ gives $0,5$.

Learn and practise “Trig ratios in a triangle” in the app →

Any angle and reciprocals

Angles bigger than a right angle still have ratios, and every ratio has a partner turned upside down.

terminal arm

The arm you land on after turning the angle anticlockwise from the positive $x$-axis.

quadrant

One of the four parts the axes cut the page into, numbered $1$ to $4$ anticlockwise from the top right.

reciprocal

What you get by turning a fraction upside down. The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

reciprocal ratios

The three partners $\text{cosec}\theta$, $\sec \theta$ and $\cot \theta$, one over sine, cosine and tangent.

Not every angle fits inside a right-angled triangle. So draw the angle on a set of axes instead. One arm lies along the positive $x$-axis, and you turn anticlockwise to reach the terminal arm. Take any point $(x;y)$ on that arm. Let $r$ be its distance from where the axes cross.

The three ratios now read $\sin \theta =\frac{y}{r}$, $\cos \theta =\frac{x}{r}$ and $\tan \theta =\frac{y}{x}$. Inside a right-angled triangle $r$ is just the hypotenuse, so nothing has really changed. What is new is that $x$ and $y$ may be negative.

A distance is never negative, so $r$ is always positive. Every sign comes from $x$ and $y$ alone. In quadrant $1$ all three ratios are positive. In quadrant $2$ only sine is. In quadrant $3$ only tangent is. In quadrant $4$ only cosine is.

Each ratio also has a partner. The reciprocal ratios are cosec, sec and cot. Watch the crossover: sec pairs with cos, and cosec pairs with sin. Cosec is not the inverse function. It does not turn a ratio back into an angle.

To simplify, rewrite every reciprocal as sine, cosine or tangent first. A whole ratio flips, top and bottom together. And $\cos$ is not a factor you may cancel. In $\cos 40^{\circ }$ the symbol and its angle travel as one thing.

Rules to remember

• $\sin \theta =\frac{y}{r}$, $\cos \theta =\frac{x}{r}$, $\tan \theta =\frac{y}{x}$
• $\text{cosec}\theta =\frac{1}{\sin \theta }$
• $\cot \theta =\frac{1}{\tan \theta }$

Examples

Traps

• Giving the distance $r$ a negative sign when the point sits in the second or third quadrant. $r$ is a distance, so it is always positive. Only $x$ and $y$ can be negative.
• Assuming every ratio is positive as long as the angle is under $180^{\circ }$. In quadrant $2$ only sine is positive. Read the signs of $x$ and $y$ first.
• Pairing sec with sin and cosec with cos. They cross over. Sec is one over cos, and cosec is one over sin.
• Cancelling the $\cos$ in $\cos 40^{\circ }$ against another $\cos$ in the expression. A ratio and its angle are one quantity. Only whole ratios cancel.

Learn and practise “Any angle and reciprocals” in the app →

Special angles

Five angles whose ratios you can write down exactly, with no calculator in your hand.

special angle

One of $0^{\circ }$, $30^{\circ }$, $45^{\circ }$, $60^{\circ }$ and $90^{\circ }$, whose ratios you are expected to know.

exact value

A value written as a whole number, a fraction or a root, with nothing rounded off.

surd

A root that does not work out neatly, such as $\sqrt{3}$. It is left standing as it is.

undefined

No value exists here, because working it out would mean dividing by zero.

Take a square with sides of $1$ and cut it along a diagonal. Each half is a right-angled triangle with two sides of $1$. Its hypotenuse is $\sqrt{2}$, and the other two angles are both $45^{\circ }$. That gives $\sin 45^{\circ }=\frac{1}{\sqrt{2}}$ and $\tan 45^{\circ }=1$.

Now take a triangle with three sides of $2$ and cut it in half. You get a right-angled triangle with sides $1$, $\sqrt{3}$ and $2$, and angles of $30^{\circ }$ and $60^{\circ }$. From it, $\sin 30^{\circ }=\frac{1}{2}$ and $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$.

The special angles at the two ends are the easiest to forget. $\sin 0^{\circ }=0$ and $\cos 0^{\circ }=1$. $\sin 90^{\circ }=1$ and $\cos 90^{\circ }=0$. And $\tan 90^{\circ }$ is undefined, because it would need a division by zero.

Read the notation carefully. ${\sin }^2\theta$ is short for $(\sin \theta {)}^2$. Work the ratio out first, then square that answer. It does not mean the sine of ${\theta }^2$. So ${\sin }^{2}30^{\circ }={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$.

Doubling a ratio is not the same as doubling an angle. At $30^{\circ }$, $2\sin 30^{\circ }=1$, but $\sin 60^{\circ }$ is about $0,87$. They are not always different, though. At $0^{\circ }$ both of them come to $0$.

Rules to remember

• $\sin 30^{\circ }=\cos 60^{\circ }=\frac{1}{2}$
• $\tan 45^{\circ }=1$
• ${\sin }^2\theta$ means $(\sin \theta {)}^2$

Examples

Traps

• Swapping the values of $\sin 30^{\circ }$ and $\sin 60^{\circ }$. $\sin 30^{\circ }$ is the smaller one, $\frac{1}{2}$. A bigger angle gives a bigger sine here.
• Writing a number for $\tan 90^{\circ }$. $\tan 90^{\circ }$ is undefined. Write that word down, not $0$ and not a big number.
• Reading ${\sin }^2\theta$ as the sine of ${\theta }^2$. Work out $\sin \theta$ first, then square the answer you get.
• Treating $2\sin \theta$ and $\sin 2\theta$ as the same thing. Test them at $30^{\circ }$. One comes to $1$ and the other to about $0,87$.

Learn and practise “Special angles” in the app →

Solving right triangles

Two facts about a right-angled triangle are enough to unlock every other side and angle.

unknown

The side or the angle a question asks you to find. A letter holds its place until you know it.

subject of the formula

The letter left standing alone on one side of the equals sign.

solving a triangle

Working out every side and every angle that a question has not given you.

Mark the angle you are working from. Then name the hypotenuse, the opposite side and the adjacent side from where that angle stands. Now choose the ratio that joins the side you know to the side you want. Solving a triangle is that one choice, made again for each unknown.

When the unknown sits on top of the fraction, multiply. If $\tan 35^{\circ }=\frac{x}{8}$, then $x=8\tan 35^{\circ }$. The $8$ crosses over and multiplies, and $x$ becomes the subject of the formula.

When the unknown sits underneath, swap it with the ratio instead. If $\cos 70^{\circ }=\frac{9}{x}$, then $x=\frac{9}{\cos 70^{\circ }}$. Multiplying here would give an answer far too small.

To find an angle, run the same machine backwards. Two sides hand you a ratio, and the inverse function turns that ratio into the angle. Skip that step and you write $0,4$ where an angle belongs.

The hypotenuse lies opposite the right angle, however the triangle is turned. Given an angle and one side, sine or cosine reaches it. Given two sides and no angle, use $a^2+b^2=c^2$ instead.

Rules to remember

• $\tan \theta =\frac{x}{a}$ gives $x=a\tan \theta$
• $\cos \theta =\frac{a}{x}$ gives $x=\frac{a}{\cos \theta }$
• $\sin \theta =k$ gives $\theta ={\sin }^{-1}k$

Examples

Traps

• Multiplying when the unknown is underneath the fraction. Swap the two. From $\cos 70^{\circ }=\frac{9}{x}$ you get $x=\frac{9}{\cos 70^{\circ }}$.
• Picking a ratio that does not contain the side you are looking for. Write down what you know and what you want. Pick the ratio holding both.
• Writing the ratio value down as though it were the angle. $0,4$ is a ratio. The inverse key turns it into about $21,8^{\circ }$.
• Using the ratio belonging to the other angle in the triangle. Opposite and adjacent are decided by the marked angle. Check which angle that is.

Learn and practise “Solving right triangles” in the app →

Heights and distances

An angle measured from a horizontal line turns a tape measure on the ground into the height of a mast.

horizontal

Level with the ground, the way still water lies.

line of sight

The straight line from your eye to the thing you are looking at.

angle of elevation

The angle you look up through, measured from a horizontal line.

angle of depression

The angle you look down through, also measured from a horizontal line.

Both angles start at a horizontal line, never a vertical one. Sketch the picture even when the question gives none. Draw the horizontal in first, then the line of sight, and mark the angle between them. Looking up gives an angle of elevation. Looking down gives an angle of depression.

An angle of depression is drawn outside the triangle, above the line of sight. It matches the angle of elevation looking back the other way, because the two horizontals are parallel. Carry it to the far end first.

A trig ratio only works in a triangle with a right angle. A pole on level ground gives you one. Mark that right angle and the given side, and the unknown has a name. When two triangles share a side, work that side out first.

Rules to remember

• Angle of elevation up equals angle of depression back down
• $\tan (\text{angle})=\frac{\text{height}}{\text{distance}}$

Examples

Traps

• Measuring an angle of elevation or depression from the vertical line, not the horizontal. Both angles are measured from a horizontal line. Sketch that line in before you start.
• Dropping the angle of depression straight into the triangle as one of its angles. It sits outside. Carry it to the far end, where it becomes the angle of elevation.
• Using a trig ratio in a triangle with no right angle in it. Find the right angle first. Without one, none of the three ratios applies.
• Taking a side of the other triangle as the hypotenuse. Take one triangle at a time. Name its own three sides before you choose a ratio.

Learn and practise “Heights and distances” in the app →

Trigonometric equations

The unknown is an angle hiding inside a ratio, so free the ratio first.

trigonometric equation

An equation whose unknown sits inside a ratio, such as $\sin x=0,5$.

coefficient

The number a term is multiplied by. In $2\sin x$ the coefficient is $2$.

parameter

A letter standing for a fixed number, not for the unknown you are solving for.

In a trigonometric equation the unknown is an angle, hiding inside a ratio. The plan never changes. Get the ratio alone on one side first. Then read the angle off, using a special angle you know or the inverse function on a calculator.

A coefficient in front divides into the ratio, not into the angle. From $2\sin x=1$, divide both sides by $2$ and get $\sin x=\frac{1}{2}$. So $x=30^{\circ }$. Halving the angle instead would be a different equation altogether.

When the angle is a multiple of $x$, the inverse gives back that whole multiple. From $\sin 2x=1$ you get $2x=90^{\circ }$, so $x=45^{\circ }$. Stopping at $2x$ leaves the job half done. And $\cos 2x$ is not the same as $2\cos x$.

Some equations carry a parameter, a letter that is not the unknown. Treat it as an ordinary number. In $a\cos \theta =\frac{a}{2}$, divide both sides by $a$ and it vanishes. That leaves $\cos \theta =\frac{1}{2}$, so $\theta =60^{\circ }$.

Grade 10 keeps these answers between $0^{\circ }$ and $90^{\circ }$. Check yours lands in that range before you write it down. The calculator hands you one angle, and the range is what tells you it is the one wanted.

Rules to remember

• $a\sin \theta =b$ gives $\sin \theta =\frac{b}{a}$
• $\cos 2x=c$ gives $2x={\cos }^{-1}c$
• $2\sin \theta$ is not $\sin 2\theta$

Examples

Traps

• Reading $2\sin x=1$ as the sine of $\frac{x}{2}$. The $2$ multiplies the ratio, so the ratio is what gets halved.
• Answering $0,5$ when the question asked for an angle. $0,5$ is only the ratio. The angle it belongs to is $30^{\circ }$.
• Stopping at $2x=90^{\circ }$ and writing that down as the answer. The inverse gives the whole angle inside the ratio. Divide by the $2$ as well.
• Solving for the parameter instead of for the angle. Read the instruction again. Solve for $\theta$ means $\theta$ must end up alone.

Learn and practise “Trigonometric equations” in the app →