In a right-angled triangle the ratio of any two sides depends only on the angle, not on the size of the triangle. Name those ratios โ sine, cosine, tangent and their reciprocals โ and learn to find all six from just one.
Stand a ladder against a wall and measure two things: how high it reaches and how long it is. The quotient $\dfrac{\text{height reached}}{\text{length of ladder}}$ tells you how steep the ladder is. A long ladder and a short ladder leaning at the same steepness give the same quotient. Builders, sailors and surveyors use such quotients all the time, so each of them has a name.
In this section you learn to name the three sides of a right-angled triangle from the point of view of an angle, define the six trigonometric ratios, see why they depend only on the angle, and practise the key skill of the chapter: when one ratio is known, find the other five.
Naming the sides: hypotenuse, opposite, adjacent
๐ Definition 8.1 โ Hypotenuse, opposite side, adjacent side
Let $\triangle ABC$ be right-angled at $B$, and look at the acute angle $\angle A$.
The hypotenuse is the side opposite the right angle: $AC$. It is the longest side.
The side opposite $\angle A$ is the side that does not touch $A$: $BC$.
The side adjacent to $\angle A$ is the side next to $A$ other than the hypotenuse: $AB$.
The words "opposite" and "adjacent" depend on which angle you are standing at. Stand at $C$ instead of $A$ and the two legs swap roles: $AB$ becomes the side opposite $\angle C$ and $BC$ becomes the side adjacent to $\angle C$. The hypotenuse never changes.
๐ Definition 8.2 โ Sine, cosine and tangent
For an acute angle $A$ of a right-angled triangle,
$$\sin A=\frac{\text{side opposite }\angle A}{\text{hypotenuse}},\qquad \cos A=\frac{\text{side adjacent to }\angle A}{\text{hypotenuse}},\qquad \tan A=\frac{\text{side opposite }\angle A}{\text{side adjacent to }\angle A}.$$
In $\triangle ABC$ right-angled at $B$: $\sin A=\dfrac{BC}{AC}$, $\cos A=\dfrac{AB}{AC}$, $\tan A=\dfrac{BC}{AB}$.
๐ Definition 8.3 โ Cosecant, secant and cotangent
The other three ratios are the reciprocals of the first three:
$$\operatorname{cosec}A=\frac{1}{\sin A}=\frac{\text{hypotenuse}}{\text{opposite}},\qquad \sec A=\frac{1}{\cos A}=\frac{\text{hypotenuse}}{\text{adjacent}},\qquad \cot A=\frac{1}{\tan A}=\frac{\text{adjacent}}{\text{opposite}}.$$
Reading the symbols. $\sin A$ is one symbol, read "sine of $A$". It is not "$\sin$ times $A$", and you cannot cancel the $\sin$ in $\dfrac{\sin A}{\sin B}$. A power is written $\sin^2A$, meaning $(\sin A)^2$; never write $\sin A^2$. Many students remember the first three ratios with the phrase "Some People Have Curly Brown Hair, Till Politics Brings Rapture" โ Sine = Perpendicular over Hypotenuse, Cosine = Base over Hypotenuse, Tangent = Perpendicular over Base. Make up your own if it helps.
Why the ratios depend only on the angle
๐ Theorem 8.1 โ The ratios depend only on the angle
If two right-angled triangles have the same acute angle $A$, then the six trigonometric ratios of $A$ are the same in both triangles, whatever their sizes.
Proof. Let $\triangle ABC$ and $\triangle PQR$ be right-angled at $B$ and $Q$, with $\angle A=\angle P$. Then $\angle B=\angle Q=90^\circ$, so the two triangles are similar by the AA criterion (Section 6.3): $\triangle ABC\sim\triangle PQR$. Corresponding sides of similar triangles are proportional, so
$$\frac{BC}{QR}=\frac{AC}{PR}\quad\Longrightarrow\quad\frac{BC}{AC}=\frac{QR}{PR}.$$
The left-hand side is $\sin A$ in the first triangle and the right-hand side is $\sin P=\sin A$ in the second. The same reasoning with $\dfrac{AB}{PQ}=\dfrac{AC}{PR}$ and $\dfrac{BC}{QR}=\dfrac{AB}{PQ}$ shows that $\cos A$ and $\tan A$ agree, and the reciprocals then agree too. $\blacksquare$
This is why the ladder at a fixed steepness always gives $0.8$: every such ladder, the wall and the ground form similar right-angled triangles. It is also why we can write "$\sin A$" without saying which triangle: the angle alone decides the value.
๐ Theorem 8.2 โ Tangent and cotangent as quotients
For every acute angle $A$,
$$\tan A=\frac{\sin A}{\cos A},\qquad \cot A=\frac{\cos A}{\sin A}.$$
Proof. In $\triangle ABC$ right-angled at $B$: $\dfrac{\sin A}{\cos A}=\dfrac{BC/AC}{AB/AC}=\dfrac{BC}{AC}\times\dfrac{AC}{AB}=\dfrac{BC}{AB}=\tan A$. The second equation follows by taking reciprocals. $\blacksquare$
๐ Theorem 8.3 โ Sizes of the ratios of an acute angle
For every acute angle $A$: $\;0\lt\sin A\lt1$, $\;0\lt\cos A\lt1$, $\;\sec A\gt1$ and $\operatorname{cosec}A\gt1$. The ratios $\tan A$ and $\cot A$ are positive and can be any positive number, large or small.
Why. The hypotenuse is the longest side of a right-angled triangle, so a leg divided by the hypotenuse is a positive number smaller than $1$; the reciprocals $\operatorname{cosec}A$ and $\sec A$ are therefore larger than $1$. A leg divided by the other leg can be anything: $\tan A$ is huge when the opposite side is much longer than the adjacent side (a very steep ladder) and tiny in the opposite case. In particular a statement like "$\sin A=\tfrac32$" can never be true for an acute angle.
โ ๏ธ Common mistakes
1. Mixing up opposite and adjacent. Always put your finger on the angle first. The side opposite is the side the angle "looks at"; the side adjacent is the leg that forms the angle together with the hypotenuse.
2. Using the right angle's sides. The six ratios are defined for the acute angles of the triangle. Do not use the right angle itself in this section.
3. Writing $\sin A^2$ or cancelling "$\sin$". $\sin^2A=(\sin A)^2$, and $\sin A+\sin B\ne\sin(A+B)$. Treat $\sin A$ as a single number.
4. Reading the reciprocals wrongly. $\operatorname{cosec}A$ is the reciprocal of $\sin A$ (not of $\cos A$), and $\sec A$ is the reciprocal of $\cos A$. Pairing the "co" names is a trap.
๐งญ Problem-solving strategy โ all six ratios from one
Draw a right-angled triangle and mark the angle $A$. Write the given ratio as a fraction of two sides, for example $\tan A=\tfrac{3}{4}$ means opposite $=3k$, adjacent $=4k$.
Find the missing side with the Pythagoras theorem: $\text{hypotenuse}^2=\text{opposite}^2+\text{adjacent}^2$. The common factor $k$ cancels in every ratio, so you may take $k=1$.
Read off all six ratios from the three sides, taking care over which side is opposite and which is adjacent.
Check: the sine and cosine must be less than $1$, and $\sin A\div\cos A$ must equal $\tan A$.
โ๏ธ Example 1 โ The six ratios of an angle
In $\triangle ABC$, right-angled at $B$, $AB=12$ cm and $BC=5$ cm. Find $AC$ and the six trigonometric ratios of $\angle A$.
๐ก Show solutionโถ
Hypotenuse $AC^2=AB^2+BC^2=144+25=169$, so $AC=13$ cm.
Name the sides From $A$: opposite $=BC=5$, adjacent $=AB=12$, hypotenuse $=AC=13$.
Note The numbers are not the same as for $\angle A$, because "opposite" and "adjacent" changed places. Never copy ratios from one angle to the other.
โ๏ธ Example 3 โ A ladder against a wall
A $6.5$ m ladder reaches $6$ m up a vertical wall. Find the distance of its foot from the wall, and $\sin\theta$, $\cos\theta$ and $\tan\theta$, where $\theta$ is the angle between the ladder and the ground.
๐ก Show solutionโถ
Triangle Wall, ground and ladder form a right-angled triangle with hypotenuse $6.5$ m (the ladder) and a vertical leg $6$ m. The vertical leg is opposite $\theta$ and the foot distance $x$ is adjacent to $\theta$.
Pythagoras $x^2=6.5^2-6^2=42.25-36=6.25$, so $x=2.5$ m.
Check $\dfrac{\sin\theta}{\cos\theta}=\dfrac{12}{5}=\tan\theta$ โ. The ladder is steep: $\tan\theta=2.4$ means it climbs $2.4$ m for every $1$ m of horizontal distance.
Interactive 1 โ Change the Angle and the Size
The triangle is right-angled at $B$. Move the angle $A$ slider (or press Play) and watch $\sin A$, $\cos A$ and $\tan A$ change. Then keep $A$ fixed and move the hypotenuse slider: the sides all grow, but the three ratios do not move. The opposite side is red, the adjacent side blue, and the hypotenuse green.
35ยฐ6.0
$\sin A=\frac{\text{opp}}{\text{hyp}}$โ
$\cos A=\frac{\text{adj}}{\text{hyp}}$โ
$\tan A=\frac{\text{opp}}{\text{adj}}$โ
opp, adj, hyp (cm)โ
Finding the other ratios when one is known
A single ratio fixes the angle, and so it fixes all the other ratios. We never need to know the angle itself: the triangle of the strategy box is enough.
โ๏ธ Example 4 โ Given a cotangent
If $20\cot A=21$ and $A$ is acute, find $\sin A$ and $\sec A$.
๐ก Show solutionโถ
Draw $\cot A=\dfrac{21}{20}=\dfrac{\text{adjacent}}{\text{opposite}}$, so take adjacent $=21$ and opposite $=20$ in a triangle right-angled at $B$.
Hypotenuse $AC^2=21^2+20^2=441+400=841$, so $AC=29$.
Answer $\sin A=\dfrac{20}{29}$ and $\sec A=\dfrac{29}{21}$.
Check $\operatorname{cosec}\theta$ is the reciprocal of $\sin\theta=\tfrac{9}{41}$ โ, and $\cot\theta\times\tan\theta=\tfrac{40}{9}\times\tfrac{9}{40}=1$ โ.
โ๏ธ Example 6 โ A condition on the sides
In $\triangle PQR$, right-angled at $Q$, $PQ=9$ cm and $PR-QR=1$ cm. Find $\sin P$, $\cos P$ and $\tan P$.
๐ก Show solutionโถ
Unknown Let $QR=x$ cm. Then $PR=(x+1)$ cm.
Pythagoras $PR^2=PQ^2+QR^2$ gives $(x+1)^2=81+x^2$, so $x^2+2x+1=81+x^2$, which gives $2x=80$ and $x=40$. Hence $QR=40$ cm and $PR=41$ cm.
Ratios at $P$ Opposite $\angle P$ is $QR=40$; adjacent is $PQ=9$; hypotenuse $PR=41$. So $\sin P=\dfrac{40}{41}$, $\cos P=\dfrac{9}{41}$, $\tan P=\dfrac{40}{9}$.
Check $9^2+40^2=81+1600=1681=41^2$ โ.
โ๏ธ Example 7 โ Can such an angle exist?
Decide whether an acute angle can have (a) $\sin A=\dfrac43$, (b) $\sec A=\dfrac54$. If it can, find $\tan A$.
๐ก Show solutionโถ
(a) $\sin A=\dfrac43\gt1$. By Theorem 8.3 the sine of an acute angle is less than $1$, so no such acute angle exists. (It would need a leg longer than the hypotenuse.)
(b) $\sec A=\dfrac54\gt1$ is allowed. Take hypotenuse $=5$, adjacent $=4$; opposite $=\sqrt{25-16}=3$.
Answer $\tan A=\dfrac{3}{4}$.
Interactive 2 โ Six Ratios from the Three Sides
Choose a right-angled triangle with whole-number sides and choose the angle to stand at, $A$ or $C$ (the right angle is at $B$). The sides are coloured by their role as seen from that angle: opposite, adjacent, hypotenuse. The six ratios below are simply those sides in fractions. Switch the angle and watch the roles โ and the numerators โ change places. Play steps through every triangle and both angles.
$\sin$ and $\operatorname{cosec}$โ
$\cos$ and $\sec$โ
$\tan$ and $\cot$โ
check $\sin\div\cos$โ
Exercises
I
Ratios from the sides. In each part $\triangle ABC$ is right-angled at $B$ (lengths in cm).
1 $AB=3$, $BC=4$. Find $AC$, then $\sin A$, $\cos A$ and $\tan A$.
2 $AB=5$, $BC=12$. Find all six ratios of $\angle C$.
3 $AC=17$, $BC=8$. Find $AB$, $\tan A$ and $\cos C$.
4 $AB=6$, $BC=8$. Find $\sin A$ and $\tan C$.
5 $AC=25$, $AB=7$. Find $\sin C$ and $\sec A$.
6 Can an acute angle have (a) $\sin A=1.2$, (b) $\operatorname{cosec}A=0.9$, (c) $\tan A=50$, (d) $\sec A=1$? Give a reason for each.
3 $AB=\sqrt{289-64}=\sqrt{225}=15$ cm. $\tan A=\dfrac{BC}{AB}=\dfrac{8}{15}$. For $\angle C$ the adjacent side is $BC$, so $\cos C=\dfrac{BC}{AC}=\dfrac{8}{17}$.
4 $AC=\sqrt{36+64}=10$. $\sin A=\dfrac{BC}{AC}=\dfrac{8}{10}=\dfrac45$. From $C$, opposite $=AB=6$ and adjacent $=BC=8$, so $\tan C=\dfrac{6}{8}=\dfrac34$.
5 $BC=\sqrt{625-49}=\sqrt{576}=24$. $\sin C=\dfrac{AB}{AC}=\dfrac{7}{25}$ (the side opposite $C$ is $AB$). $\sec A=\dfrac{AC}{AB}=\dfrac{25}{7}$ (the side adjacent to $A$ is $AB$).
6 (a) No: a sine of an acute angle is less than $1$ (Theorem 8.3). (b) No: $\operatorname{cosec}A\gt1$. (c) Yes: $\tan A$ may be any positive number, e.g. a very steep ladder. (d) No: $\sec A=\dfrac{\text{hyp}}{\text{adj}}$ and the hypotenuse is strictly longer than a leg, so $\sec A\gt1$.
1 A $13$ m ladder has its foot $5$ m from a wall. How high up the wall does it reach? Find $\sin\theta$, $\cos\theta$ and $\tan\theta$ for the angle $\theta$ between the ladder and the ground.
2 A kite flies at a height of $24$ m with its string taut, $25$ m long. The string is held at ground level. Find the horizontal distance from the flier to the point directly below the kite, and the sine and tangent of the angle the string makes with the ground.
3 A slide $5$ m long has its top $3$ m above the ground. Find the horizontal distance it covers and $\sin\theta$, $\cos\theta$, $\tan\theta$ of its inclination $\theta$.
4 In $\triangle ABC$, right-angled at $B$, $AB=x$ cm, $BC=(x+1)$ cm and $AC=(x+2)$ cm. Find $x$ and $\sin A$.
5 In $\triangle PQR$, right-angled at $Q$, $PQ=24$ cm and $PR-QR=8$ cm. Find $\sin P$, $\cos P$ and $\tan P$.
6 For the triangle with $AB=15$, $BC=8$, $AC=17$ (right angle at $B$), compute $\sin A\div\cos A$ and compare it with $\tan A$.
7 Explain, using the sides of a right-angled triangle, why $\sin A$ and $\cos A$ can never be greater than $1$ for an acute angle, whereas $\tan A$ can.
8 $\triangle ABC$ has $\angle B=90^\circ$, $AB=6$ cm, $BC=8$ cm; $\triangle PQR$ has $\angle Q=90^\circ$, $PQ=9$ cm, $QR=12$ cm. Show that $\sin A=\sin P$, and explain why this is no coincidence.
๐ก Show solutionsโถ
1 Height $=\sqrt{13^2-5^2}=\sqrt{144}=12$ m. $\sin\theta=\dfrac{12}{13}$, $\cos\theta=\dfrac{5}{13}$, $\tan\theta=\dfrac{12}{5}$.
2 Horizontal distance $=\sqrt{25^2-24^2}=\sqrt{49}=7$ m. $\sin\theta=\dfrac{24}{25}$, $\tan\theta=\dfrac{24}{7}$.
3 Horizontal distance $=\sqrt{25-9}=4$ m. $\sin\theta=\dfrac35$, $\cos\theta=\dfrac45$, $\tan\theta=\dfrac34$.
4 $x^2+(x+1)^2=(x+2)^2$: $2x^2+2x+1=x^2+4x+4$, so $x^2-2x-3=0$, i.e. $(x-3)(x+1)=0$. A length cannot be negative, so $x=3$: sides $3,4,5$ and $\sin A=\dfrac{BC}{AC}=\dfrac45$.
5 Let $QR=y$; $PR=y+8$. $(y+8)^2=24^2+y^2$: $16y+64=576$, so $y=32$ and $PR=40$ cm (check: $24^2+32^2=1600=40^2$ โ). $\sin P=\dfrac{32}{40}=\dfrac45$, $\cos P=\dfrac{24}{40}=\dfrac35$, $\tan P=\dfrac{32}{24}=\dfrac43$.
7 $\sin A=\dfrac{\text{opp}}{\text{hyp}}$ and $\cos A=\dfrac{\text{adj}}{\text{hyp}}$ have the hypotenuse, the longest side, in the denominator, so each is less than $1$. $\tan A=\dfrac{\text{opp}}{\text{adj}}$ compares two legs, and one leg can be longer than the other.
8 $AC=\sqrt{36+64}=10$, so $\sin A=\dfrac{8}{10}=\dfrac45$. $PR=\sqrt{81+144}=15$, so $\sin P=\dfrac{12}{15}=\dfrac45$. Equal, because the legs are in the same ratio ($6:9=8:12=2:3$), so the triangles are similar (SAS, with the right angle between the legs) and $\angle A=\angle P$; by Theorem 8.1 the ratios agree.
IV
Challenge.
1 If $\tan A=\dfrac pq$ with $p,q\gt0$, express $\sin A$ and $\cos A$ in terms of $p$ and $q$.
2 In $\triangle ABC$, right-angled at $B$, $\tan A=\dfrac34$ and $AC=20$ cm. Find the perimeter and the area of the triangle.
3 If $\sec A=\dfrac{17}{8}$, find the value of $\dfrac{\sin A-\cos A}{\sin A+\cos A}$.
๐ก Show solutionsโถ
1 Take opposite $=p$, adjacent $=q$, hypotenuse $=\sqrt{p^2+q^2}$. So $\sin A=\dfrac{p}{\sqrt{p^2+q^2}}$ and $\cos A=\dfrac{q}{\sqrt{p^2+q^2}}$.
2 Opposite $=3k$, adjacent $=4k$, hypotenuse $=5k=20$, so $k=4$: $BC=12$ cm, $AB=16$ cm. Perimeter $=12+16+20=48$ cm; area $=\tfrac12\times16\times12=96$ cmยฒ.
3 $\cos A=\dfrac{8}{17}$, opposite $=\sqrt{289-64}=15$, so $\sin A=\dfrac{15}{17}$. The expression is $\dfrac{15/17-8/17}{15/17+8/17}=\dfrac{7}{23}$.
How this is examined: Board papers ask for one or two ratios from given sides (1โ2 marks), "given one ratio, find the others" (2โ3 marks), and a side-condition problem such as $PR+QR=\dots$ (3 marks). Always draw and label the triangle, write the formula before you substitute, and give the ratio as a fraction in lowest terms. JEE Main uses the same idea inside a ratio expression, where dividing by $\cos\theta$ saves work.
E1.Board ยท 1 mark In $\triangle ABC$, right-angled at $B$, $AB=6$ cm and $BC=8$ cm. Then $\sin C=$ (A) $\dfrac35$ (B) $\dfrac45$ (C) $\dfrac34$ (D) $\dfrac43$
๐ก Show solutionโถ
Hypotenuse $AC=\sqrt{36+64}=10$ cm.
Sine of $C$ The side opposite $\angle C$ is $AB=6$, so $\sin C=\dfrac{AB}{AC}=\dfrac{6}{10}=\dfrac35$.
Trap (B) $\dfrac45$ is $\sin A$ โ the angle at the other end. (D) $\dfrac43$ is $\tan A$ and (C) $\dfrac34$ is $\tan C$: both are ratios of legs, not of a leg to the hypotenuse. Answer: (A).
E2.Board ยท 1 markAssertion (A): No acute angle $\theta$ has $\operatorname{cosec}\theta=0.8$. Reason (R): In a right-angled triangle the hypotenuse is the longest side, so $\operatorname{cosec}\theta=\dfrac{\text{hypotenuse}}{\text{opposite}}$ is greater than $1$. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
๐ก Show solutionโถ
Check R True: the hypotenuse is longer than either leg, so $\operatorname{cosec}\theta\gt1$ (Theorem 8.3).
Check A $0.8\lt1$, so it is impossible for an acute angle. A is true, and it follows from R.
Answer (A).
E3.Board ยท 2 marks If $\operatorname{cosec}A=\dfrac{25}{7}$, find $\cos A$ and $\tan A$.
Answer The value is $\dfrac{5}{30}=\dfrac16$. (1 mark)
Shortcut Divide numerator and denominator by $\cos\theta$ (it is not zero): $\dfrac{5\tan\theta-3}{5\tan\theta+2}=\dfrac{4-3}{4+2}=\dfrac16$ โ. No hypotenuse is needed.
E5.Board ยท 3 marks In $\triangle PQR$, right-angled at $Q$, $PQ=7$ cm and $PR+QR=49$ cm. Find $\sin P$ and $\tan R$.
๐ก Show solutionโถ
Set up Let $QR=x$, so $PR=49-x$. By Pythagoras, $(49-x)^2=7^2+x^2$. (1 mark)
Solve $2401-98x+x^2=49+x^2$, so $98x=2352$ and $x=24$. Hence $QR=24$ cm and $PR=25$ cm. (1 mark)
Ratios $\sin P=\dfrac{QR}{PR}=\dfrac{24}{25}$. For $\angle R$, the opposite side is $PQ=7$ and the adjacent side is $QR=24$, so $\tan R=\dfrac{7}{24}$. (1 mark)
E6.JEE Main If $\tan\theta=\dfrac{7}{24}$ for an acute angle $\theta$, the value of $625\sin\theta\cos\theta$ is
Compute $\sin\theta\cos\theta=\dfrac{7}{25}\times\dfrac{24}{25}=\dfrac{168}{625}$, so $625\sin\theta\cos\theta=168$.
Answer $168$. (A common slip is to use $\tfrac{7}{24}$ directly as $\sin\theta$; the denominator $24$ is the adjacent side, not the hypotenuse.)
๐ฌ ResearchโTeaching NexusMini project ยท optional, about 40 minutes
Research question. If you draw right-angled triangles of very different sizes but with the same angle, do the measured ratios $\dfrac{\text{opposite}}{\text{hypotenuse}}$ and $\dfrac{\text{opposite}}{\text{adjacent}}$ really come out the same โ and how close do they get to the values a calculator gives for that angle?
Why it matters. Every ramp, roof, road gradient and staircase is specified by an angle or by a ratio of heights. A builder who knows that the steepness ratio does not depend on size can draw a small plan and build a large structure. The same fact lets surveyors and navigators work out distances they cannot measure directly.
What to do.
On plain paper draw a horizontal line and, at its left end $A$, use a protractor to draw a ray at $35^\circ$ to it.
Along the horizontal line mark points at $4$, $8$, $12$ and $16$ cm from $A$. At each point draw a line perpendicular to the horizontal line (use a set square) up to the ray. You now have four right-angled triangles with the same angle $A$.
For each triangle measure the opposite side and the hypotenuse to the nearest millimetre. Record them in a table with the adjacent side ($4$, $8$, $12$, $16$ cm).
In a spreadsheet or on a calculator find $\dfrac{\text{opp}}{\text{hyp}}$, $\dfrac{\text{adj}}{\text{hyp}}$ and $\dfrac{\text{opp}}{\text{adj}}$ for each triangle, and the average of each column.
Compare your averages with the calculator values of $\sin35^\circ$, $\cos35^\circ$ and $\tan35^\circ$ (use the "sin", "cos", "tan" keys in degree mode). Compute the percentage difference.
Repeat with a $60^\circ$ angle. Which ratios have grown and which have shrunk?
Tools: Protractor, set square, ruler, pencil, a calculator or spreadsheet
Hand in: Your drawing, a table of measurements and ratios, the comparison with the calculator values, and a short paragraph explaining any differences (measurement error, thick pencil linesโฆ).
Going further. Photograph a staircase or ramp at your school, measure its rise and run, and find the angle of the ramp from the ratio $\tfrac{\text{rise}}{\text{run}}$. Is it steeper than a $1:12$ wheelchair ramp (a ratio of $0.083$)?
Teacher/parent note: the activity is an empirical check of Theorem 8.1 (similar right triangles). Typical measurement error with a school protractor is about 1ยฐ, so the averaged ratios usually agree with the calculator values to within 2โ3%. Do not expect exact agreement; discuss why the largest triangle often gives the most accurate ratios.
๐งช Self-check quiz
0/5 correct
Q1. In $\triangle ABC$, right-angled at $B$, $AB=8$ cm and $BC=15$ cm. Then $\cos A=$
Q2. If $\tan A=\dfrac34$ for an acute angle $A$, then $\sec A=$
Q3. Which of these can be the sine of an acute angle?
Q4. $\operatorname{cosec}A$ equals
Q5. If $\sin\theta=\dfrac{5}{13}$ and $\theta$ is acute, then $\cot\theta=$