Chapter 8 · Introduction to Trigonometry

8.2 Trigonometric Ratios of Some Specific Angles

Five angles — 0°, 30°, 45°, 60° and 90° — have trigonometric ratios you can write down exactly. Build the table from two triangles in your geometry box, then learn to use it for values, for angles and for unknown sides.

Open your geometry box. Besides the ruler and the protractor there are two set squares. One has angles $45^\circ,45^\circ,90^\circ$; the other has $30^\circ,60^\circ,90^\circ$. Almost every right-angled triangle you meet in practice — a roof, a ramp, a ladder, a kite string — is built from those angles, because they are the easiest to draw and to cut. For these angles the ratios of Section 8.1 are not messy decimals; they are neat numbers such as $\tfrac12$, $\tfrac{\sqrt3}{2}$ and $1$.

In this section you will derive those values from the two set-square triangles (so that you can rebuild the table whenever you forget it), add the limiting cases $0^\circ$ and $90^\circ$, and then use the table to evaluate expressions, to find angles and to find unknown sides of right-angled triangles. Calculators give decimals for any angle; the exact values here are what examiners expect.

The $45^\circ$ triangle: half a square

📐 Theorem 8.4 — Ratios of $45^\circ$
$$\sin45^\circ=\frac{1}{\sqrt2},\qquad\cos45^\circ=\frac{1}{\sqrt2},\qquad\tan45^\circ=1,$$ and so $\operatorname{cosec}45^\circ=\sqrt2$, $\sec45^\circ=\sqrt2$, $\cot45^\circ=1$.

Proof. Let $\triangle ABC$ be right-angled at $B$ with $\angle A=45^\circ$. The angle sum is $180^\circ$, so $\angle C=180^\circ-90^\circ-45^\circ=45^\circ$. Equal angles mean equal opposite sides, so $BC=AB$; call this common length $a$. By Pythagoras, $AC^2=a^2+a^2=2a^2$, so $AC=a\sqrt2$. From $A$: opposite $=a$, adjacent $=a$, hypotenuse $=a\sqrt2$, so $$\sin45^\circ=\frac{a}{a\sqrt2}=\frac1{\sqrt2},\qquad\cos45^\circ=\frac{a}{a\sqrt2}=\frac1{\sqrt2},\qquad\tan45^\circ=\frac aa=1.\qquad\blacksquare$$ The size $a$ cancels, as Theorem 8.1 promised. This triangle is exactly half of a square cut along a diagonal, so its sides are always in the ratio $1:1:\sqrt2$.

The $30^\circ$ and $60^\circ$ triangle: half an equilateral triangle

📐 Theorem 8.5 — Ratios of $30^\circ$ and $60^\circ$
$$\sin30^\circ=\frac12,\quad\cos30^\circ=\frac{\sqrt3}{2},\quad\tan30^\circ=\frac{1}{\sqrt3};\qquad\sin60^\circ=\frac{\sqrt3}{2},\quad\cos60^\circ=\frac12,\quad\tan60^\circ=\sqrt3.$$ The reciprocals follow: $\operatorname{cosec}30^\circ=2$, $\sec30^\circ=\dfrac{2}{\sqrt3}$, $\cot30^\circ=\sqrt3$; $\operatorname{cosec}60^\circ=\dfrac{2}{\sqrt3}$, $\sec60^\circ=2$, $\cot60^\circ=\dfrac1{\sqrt3}$.

Proof. Start with an equilateral triangle $PQR$ of side $2a$, so every angle is $60^\circ$. Draw the altitude $PS$ from $P$ to $QR$. The right-angled triangles $PSQ$ and $PSR$ are congruent (RHS: common side $PS$, equal hypotenuses $PQ=PR$), so $S$ is the midpoint of $QR$, giving $QS=a$, and $PS$ bisects the angle at $P$, giving $\angle QPS=30^\circ$.

✏️ Example 1 — Using the values
Evaluate $2\tan^245^\circ+\cos^230^\circ-\sin^260^\circ$.
Substitute $\tan45^\circ=1$, $\cos30^\circ=\dfrac{\sqrt3}2$, $\sin60^\circ=\dfrac{\sqrt3}{2}$. Remember $\tan^245^\circ=(\tan45^\circ)^2$.
Square $2(1)^2+\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{\sqrt3}2\right)^2=2+\dfrac34-\dfrac34$.
Answer $2$.
Interactive 1 — Build the Values from the Two Triangles

Choose $30^\circ$, $45^\circ$ or $60^\circ$ (or press Play). The dashed outline shows where the triangle comes from: half of a square for $45^\circ$, half of an equilateral triangle for $30^\circ$ and $60^\circ$. The sides are drawn with their exact lengths, coloured as opposite, adjacent and hypotenuse for the marked angle, and the readouts divide them to give the exact value and its decimal.

sine = opp ÷ hyp—
cosine = adj ÷ hyp—
tangent = opp ÷ adj—
sides (opp : adj : hyp)—

The angles $0^\circ$ and $90^\circ$

A right-angled triangle cannot have an angle of $0^\circ$ or $90^\circ$ besides its right angle, so these two values are fixed by a limiting argument instead. Take $\triangle ABC$ right-angled at $B$ with the hypotenuse $AC$ fixed, and let the angle $A$ get smaller and smaller. The side $BC$ opposite to it shrinks towards $0$ and $AB$ grows towards $AC$. When $A$ reaches $0^\circ$, the "triangle" has collapsed into a segment with $BC=0$ and $AB=AC$. Do the same with $A$ growing towards $90^\circ$: now $AB$ shrinks to $0$ and $BC$ grows to $AC$.

📘 Definition 8.4 — Ratios of $0^\circ$ and $90^\circ$
Following the collapsing triangle (opposite $=0$, adjacent $=$ hypotenuse, for $0^\circ$; adjacent $=0$, opposite $=$ hypotenuse, for $90^\circ$) we define $$\sin0^\circ=0,\ \ \cos0^\circ=1,\ \ \tan0^\circ=0,\ \ \sec0^\circ=1;\qquad\sin90^\circ=1,\ \ \cos90^\circ=0,\ \ \operatorname{cosec}90^\circ=1,\ \ \cot90^\circ=0.$$ The ratios $\operatorname{cosec}0^\circ$, $\cot0^\circ$, $\tan90^\circ$ and $\sec90^\circ$ would need a division by $0$, so they are not defined.
🧭 Problem-solving strategy — using the table of standard angles
  1. Replace every ratio by its exact value from the table. Write each value in brackets before squaring: $\sin^2 30^\circ=\left(\tfrac12\right)^2=\tfrac14$.
  2. Simplify with fractions and surds. Rationalise the denominator at the end, or keep $\tfrac1{\sqrt3}$ if that is easier to compare.
  3. To find an angle, solve for the ratio first (e.g. $\tan\theta=\sqrt3$), then look for that value in the table.
  4. To find an unknown side of a right-angled triangle, choose the ratio that contains the known side and the unknown side, write it as an equation, and solve. Use $\sqrt3\approx1.732$ and $\sqrt2\approx1.414$ for decimals.

The values for the five angles, in one table. The sines are $\dfrac{\sqrt0}2,\dfrac{\sqrt1}2,\dfrac{\sqrt2}2,\dfrac{\sqrt3}2,\dfrac{\sqrt4}2$ — a popular way to remember them — and the cosines are the same list read backwards. The tangent is $\sin\div\cos$ and the other three are reciprocals.

$\angle A$$0^\circ$$30^\circ$$45^\circ$$60^\circ$$90^\circ$
$\sin A$$0$$\tfrac12$$\tfrac1{\sqrt2}$$\tfrac{\sqrt3}{2}$$1$
$\cos A$$1$$\tfrac{\sqrt3}{2}$$\tfrac1{\sqrt2}$$\tfrac12$$0$
$\tan A$$0$$\tfrac1{\sqrt3}$$1$$\sqrt3$not defined
$\operatorname{cosec}A$not defined$2$$\sqrt2$$\tfrac2{\sqrt3}$$1$
$\sec A$$1$$\tfrac2{\sqrt3}$$\sqrt2$$2$not defined
$\cot A$not defined$\sqrt3$$1$$\tfrac1{\sqrt3}$$0$
⚠️ Common mistakes

1. Treating $\sin$ as multiplication. $\sin60^\circ\ne2\sin30^\circ$: the left side is $\tfrac{\sqrt3}{2}\approx0.866$ and the right side is $1$. In the same way $\cos 2A\ne2\cos A$ and $\sin(A+B)\ne\sin A+\sin B$. A ratio of a sum or multiple of angles is not the sum or multiple of ratios.

2. Squares. $\sin^230^\circ$ means $\left(\sin30^\circ\right)^2=\tfrac14$, not $\sin\left(30^2\right)^\circ$ and not $\tfrac12$.

3. Writing a value for an undefined ratio. $\tan90^\circ$ is not "infinity" in this course; just say it is not defined.

4. Mixing the rows. $\sin30^\circ=\tfrac12$ but $\cos30^\circ=\tfrac{\sqrt3}2$. If you forget, redraw the half-equilateral triangle: the short side $1$ is opposite the small angle.

✏️ Example 2 — Tangents of $60^\circ$ and $30^\circ$
Evaluate $\dfrac{\tan60^\circ-\tan30^\circ}{1+\tan60^\circ\tan30^\circ}$.
Substitute $\tan60^\circ=\sqrt3$, $\tan30^\circ=\dfrac1{\sqrt3}$.
Numerator $\sqrt3-\dfrac1{\sqrt3}=\dfrac{3-1}{\sqrt3}=\dfrac2{\sqrt3}$.
Denominator $1+\sqrt3\times\dfrac1{\sqrt3}=1+1=2$.
Answer $\dfrac{2/\sqrt3}{2}=\dfrac1{\sqrt3}=\dfrac{\sqrt3}3\approx0.577$.
✏️ Example 3 — Finding the angle
Find the acute angle $\theta$ in each case: (a) $2\sin\theta=1$; (b) $\tan\theta=\sqrt3$; (c) $2\cos^2\theta-1=0$; (d) $\operatorname{cosec}\theta=\dfrac{2}{\sqrt3}$.
(a) $\sin\theta=\dfrac12$, which is $\sin30^\circ$. So $\theta=30^\circ$.
(b) $\tan\theta=\sqrt3=\tan60^\circ$, so $\theta=60^\circ$.
(c) $\cos^2\theta=\dfrac12$, so $\cos\theta=\dfrac1{\sqrt2}$ (an acute angle has a positive cosine). That is $\cos45^\circ$, so $\theta=45^\circ$.
(d) $\sin\theta=\dfrac{\sqrt3}{2}$, which is $\sin60^\circ$. So $\theta=60^\circ$.
✏️ Example 4 — Unknown sides
In $\triangle ABC$, right-angled at $B$, $AB=6$ cm and $\angle C=30^\circ$. Find $BC$ and $AC$.
Sides from $C$ Opposite $\angle C$ is $AB=6$ cm; adjacent is $BC$; hypotenuse is $AC$.
Find $BC$ The ratio with the known $AB$ and the wanted $BC$ is the tangent: $\tan30^\circ=\dfrac{AB}{BC}$, so $\dfrac{1}{\sqrt3}=\dfrac6{BC}$ and $BC=6\sqrt3\approx10.39$ cm.
Find $AC$ $\sin30^\circ=\dfrac{AB}{AC}$, so $\dfrac12=\dfrac6{AC}$ and $AC=12$ cm.
Check $AB^2+BC^2=36+108=144=12^2$ ✓.
✏️ Example 5 — A ladder
A $10$ m ladder leans against a vertical wall and makes an angle of $60^\circ$ with the ground. How high up the wall does it reach, and how far is its foot from the wall? (Use $\sqrt3=1.732$.)
Set up The ladder is the hypotenuse ($10$ m). The height $h$ is opposite the $60^\circ$ angle, and the foot distance $d$ is adjacent to it.
Height $\sin60^\circ=\dfrac h{10}$, so $h=10\times\dfrac{\sqrt3}2=5\sqrt3=5\times1.732=8.66$ m.
Distance $\cos60^\circ=\dfrac d{10}$, so $d=10\times\dfrac12=5$ m.
Check $8.66^2+5^2=75+25=100=10^2$ ✓.
✏️ Example 6 — Testing a formula with values
For $A=30^\circ$, (a) check that $\cos2A=2\cos^2A-1$; (b) decide whether $\sin2A=2\sin A$.
(a) Left side $\cos2A=\cos60^\circ=\dfrac12$.
(a) Right side $2\cos^230^\circ-1=2\times\dfrac34-1=\dfrac12$. They agree ✓ (for this one angle; it does not prove the formula for all angles).
(b) $\sin2A=\sin60^\circ=\dfrac{\sqrt3}2\approx0.866$ but $2\sin A=2\times\dfrac12=1$. They differ, so $\sin2A\ne2\sin A$ — one counter-example is enough to show a formula is false.
Interactive 2 — Sweep the Angle from 0° to 90°

The hypotenuse is a fixed radius of length $1$. Slide the angle $A$ (or press Play). The opposite side is $\sin A$, the adjacent side is $\cos A$, and the bars on the right show $\sin A$, $\cos A$ and $\tan A$. Watch the sine climb from $0$ to $1$ while the cosine falls from $1$ to $0$, and the tangent grow without bound as $A$ nears $90^\circ$. The dots mark $30^\circ$, $45^\circ$ and $60^\circ$.

30°
$\sin A$—
$\cos A$—
$\tan A$—
exact value—

Exercises

I
Evaluating expressions. Give exact values.
  • 1 $2\sin30^\circ+3\tan45^\circ$.
  • 2 $4\cos^245^\circ-2\sin^230^\circ$.
  • 3 $\sec^260^\circ-\tan^260^\circ$.
  • 4 $\dfrac{3\sin90^\circ+\cos0^\circ}{\tan45^\circ+\cot45^\circ}$.
  • 5 $3\operatorname{cosec}^230^\circ-4\cot^260^\circ$.
  • 6 $\sin30^\circ\cos45^\circ\tan60^\circ$.
  • 7 Check that $\sin60^\circ=2\sin30^\circ\cos30^\circ$.
  • 8 Evaluate $\dfrac{1-\tan^230^\circ}{1+\tan^230^\circ}$ and compare with $\cos60^\circ$.
1 $2\times\dfrac12+3\times1=1+3=4$.
2 $4\times\dfrac12-2\times\dfrac14=2-\dfrac12=\dfrac32$.
3 $2^2-(\sqrt3)^2=4-3=1$.
4 Numerator $3(1)+1=4$; denominator $1+1=2$. Value $2$.
5 $3\times2^2-4\times\left(\dfrac1{\sqrt3}\right)^2=12-\dfrac43=\dfrac{32}{3}$.
6 $\dfrac12\times\dfrac1{\sqrt2}\times\sqrt3=\dfrac{\sqrt3}{2\sqrt2}=\dfrac{\sqrt6}{4}\approx0.612$.
7 Right side: $2\times\dfrac12\times\dfrac{\sqrt3}2=\dfrac{\sqrt3}2$; left side: $\dfrac{\sqrt3}{2}$ ✓.
8 $\tan^230^\circ=\dfrac13$, so the value is $\dfrac{1-1/3}{1+1/3}=\dfrac{2/3}{4/3}=\dfrac12=\cos60^\circ$.
II
Finding angles. In every part $\theta$ is acute.
  • 1 $2\sin\theta=\sqrt3$.
  • 2 $\cot\theta=1$.
  • 3 $4\cos^2\theta=3$.
  • 4 $\sec\theta=2$.
  • 5 $\tan^2\theta-3=0$.
  • 6 $3\tan\theta=\sqrt3$.
  • 7 For $\theta=30^\circ$ verify that $\dfrac{2\tan\theta}{1-\tan^2\theta}=\tan2\theta$.
  • 8 Find the acute angle $\theta$ for which $\sin\theta=\cos\theta$.
1 $\sin\theta=\dfrac{\sqrt3}2$, so $\theta=60^\circ$.
2 $\cot45^\circ=1$, so $\theta=45^\circ$.
3 $\cos^2\theta=\dfrac34$, $\cos\theta=\dfrac{\sqrt3}{2}$ (positive), so $\theta=30^\circ$.
4 $\cos\theta=\dfrac12$, so $\theta=60^\circ$.
5 $\tan^2\theta=3$, $\tan\theta=\sqrt3$ (positive), so $\theta=60^\circ$.
6 $\tan\theta=\dfrac{\sqrt3}{3}=\dfrac1{\sqrt3}$, so $\theta=30^\circ$.
7 Left side: $\dfrac{2/\sqrt3}{1-1/3}=\dfrac{2/\sqrt3}{2/3}=\dfrac{3}{\sqrt3}=\sqrt3$. Right side: $\tan60^\circ=\sqrt3$ ✓.
8 Dividing by $\cos\theta\ne0$ gives $\tan\theta=1$, so $\theta=45^\circ$. (Check: $\sin45^\circ=\cos45^\circ=\tfrac1{\sqrt2}$ ✓.)
III
Triangles and applications (use $\sqrt2=1.414$, $\sqrt3=1.732$ for decimals; answer to 2 decimal places where asked).
  • 1 In $\triangle ABC$, right-angled at $B$, $\angle A=30^\circ$ and $BC=5$ cm. Find $AC$ and $AB$.
  • 2 In $\triangle ABC$, right-angled at $B$, $\angle C=60^\circ$ and $AC=14$ cm. Find $BC$ and $AB$.
  • 3 The hypotenuse of an isosceles right-angled triangle is $12$ cm. Find the length of each equal side.
  • 4 A $12$ m ladder makes $60^\circ$ with the ground. How high does it reach, and how far is its foot from the wall?
  • 5 A kite string $50$ m long is taut and makes $30^\circ$ with the horizontal ground. Find the height of the kite.
  • 6 A slide is inclined at $30^\circ$ to the ground and its top is $2.4$ m above the ground. Find its length.
  • 7 A rectangular field has a diagonal of $16$ m that makes $30^\circ$ with the longer side. Find the sides and the area.
  • 8 An equilateral triangle has side $8$ cm. Use $\sin60^\circ$ to find its altitude and then its area.
1 $BC$ is opposite $A$. $\sin30^\circ=\dfrac{BC}{AC}$: $AC=\dfrac{5}{1/2}=10$ cm. $\tan30^\circ=\dfrac{BC}{AB}$: $AB=5\sqrt3\approx8.66$ cm.
2 $BC$ is adjacent to $C$: $\cos60^\circ=\dfrac{BC}{14}$, so $BC=7$ cm. $\sin60^\circ=\dfrac{AB}{14}$, so $AB=7\sqrt3\approx12.12$ cm.
3 Each equal side $x$ satisfies $\sin45^\circ=\dfrac{x}{12}$, so $x=\dfrac{12}{\sqrt2}=6\sqrt2\approx8.49$ cm.
4 Height $=12\sin60^\circ=6\sqrt3\approx10.39$ m; foot distance $=12\cos60^\circ=6$ m.
5 Height $=50\sin30^\circ=25$ m.
6 $\sin30^\circ=\dfrac{2.4}{L}$, so $L=2.4\div\dfrac12=4.8$ m.
7 Longer side $=16\cos30^\circ=8\sqrt3\approx13.86$ m; shorter side $=16\sin30^\circ=8$ m. Area $=8\sqrt3\times8=64\sqrt3\approx110.85$ m². (Check: $13.86^2+8^2\approx192+64=256=16^2$ ✓.)
8 Altitude $=8\sin60^\circ=4\sqrt3\approx6.93$ cm. Area $=\tfrac12\times8\times4\sqrt3=16\sqrt3\approx27.71$ cm².
IV
Challenge.
  • 1 Without a calculator, arrange $\sin30^\circ$, $\sin45^\circ$, $\sin60^\circ$ in increasing order, and likewise $\tan30^\circ$, $\tan45^\circ$, $\tan60^\circ$.
  • 2 In $\triangle ABC$, right-angled at $B$, $AB=\sqrt3\,BC$. Find $\angle A$ and $\angle C$.
  • 3 The shortest side of a right-angled triangle with angles $30^\circ$, $60^\circ$, $90^\circ$ is $6$ cm. Find the other two sides and the perimeter.
1 $\sin30^\circ=0.5$, $\sin45^\circ=\tfrac1{\sqrt2}\approx0.707$, $\sin60^\circ=\tfrac{\sqrt3}2\approx0.866$: increasing order as listed. $\tan30^\circ\approx0.577\lt\tan45^\circ=1\lt\tan60^\circ\approx1.732$. (The sine and tangent grow as the angle grows.)
2 $\tan A=\dfrac{BC}{AB}=\dfrac{1}{\sqrt3}$, so $\angle A=30^\circ$ and $\angle C=90^\circ-30^\circ=60^\circ$.
3 The sides are in the ratio $1:\sqrt3:2$, so the sides are $6$, $6\sqrt3\approx10.39$ and $12$ cm. Perimeter $=18+6\sqrt3\approx28.39$ cm.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main

How this is examined: the table of standard angles is the single most-used fact of Class 10 trigonometry. Expect an evaluation (1–3 marks), a "find the angle" equation (2 marks), a verification of a formula for $30^\circ$ or $60^\circ$ (3 marks), and a case study that needs one or two of these values. Write each value from the table before simplifying — marks are for the substitution step.

E1. Board · 1 mark The value of $\sin60^\circ\cos30^\circ-\sin30^\circ\cos60^\circ$ is  (A) $0$  (B) $\dfrac12$  (C) $1$  (D) $\dfrac{\sqrt3}{2}$

Substitute $\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}-\dfrac12\times\dfrac12=\dfrac34-\dfrac14=\dfrac12$.
Trap (C) $1$ comes from adding the two products instead of subtracting; (D) is just $\sin60^\circ$. Answer: (B).

E2. Board · 1 mark Assertion (A): $\tan90^\circ$ is not defined. Reason (R): $\tan90^\circ=\dfrac{\sin90^\circ}{\cos90^\circ}=\dfrac10$, and division by zero has no meaning.
(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.

Check $\sin90^\circ=1$ and $\cos90^\circ=0$, so $\tan90^\circ=\dfrac{1}{0}$, which is not defined. Both statements are true, and R is exactly the reason for A.
Answer (A).

E3. Board · 2 marks Evaluate $2\operatorname{cosec}^230^\circ+3\sec^245^\circ-\cot^260^\circ$.

Values $\operatorname{cosec}30^\circ=2$, $\sec45^\circ=\sqrt2$, $\cot60^\circ=\dfrac1{\sqrt3}$. (1 mark)
Evaluate $2(4)+3(2)-\dfrac13=8+6-\dfrac13=\dfrac{41}{3}$. (1 mark)

E4. Board · 3 marks Verify that (i) $\cos60^\circ=1-2\sin^230^\circ$ and (ii) $\cos60^\circ=\dfrac{1-\tan^230^\circ}{1+\tan^230^\circ}$.

LHS $\cos60^\circ=\dfrac12$. (1 mark)
(i) $1-2\left(\dfrac12\right)^2=1-\dfrac12=\dfrac12$ ✓. (1 mark)
(ii) $\tan^230^\circ=\dfrac13$, so $\dfrac{1-\frac13}{1+\frac13}=\dfrac{2/3}{4/3}=\dfrac12$ ✓. (1 mark)

E5. JEE Main The number of acute angles $\theta$ satisfying $\tan\theta=\cot\theta$ is  (A) $0$  (B) $1$  (C) $2$  (D) $3$

Solve $\tan\theta=\dfrac1{\tan\theta}$ gives $\tan^2\theta=1$. For an acute angle $\tan\theta\gt0$, so $\tan\theta=1$ and $\theta=45^\circ$ — only one.
Trap (C) comes from writing $\tan\theta=\pm1$; the value $-1$ is impossible for an acute angle. Answer: (B).

E6. Board · Case study A park slide. A playground slide is $4$ m long and makes an angle of $30^\circ$ with the ground. Its top platform is reached by a straight ladder fixed on the other side of the platform, making $60^\circ$ with the ground. (Use $\sqrt3=1.732$.)
(i) Find the height of the platform and the horizontal distance covered by the slide. (2 marks)
(ii) Find the length of the ladder. (1 mark)
(iii) Find the distance between the foot of the slide and the foot of the ladder. (1 mark)

(i) The slide is the hypotenuse. Height $=4\sin30^\circ=2$ m. Horizontal distance $=4\cos30^\circ=2\sqrt3=3.46$ m.
(ii) $\sin60^\circ=\dfrac{2}{L}$, so $L=\dfrac{2}{\sqrt3/2}=\dfrac{4}{\sqrt3}=2.31$ m.
(iii) The ladder's horizontal distance is $\dfrac{2}{\tan60^\circ}=\dfrac{2}{\sqrt3}=1.15$ m (or $L\cos60^\circ=\dfrac{2}{\sqrt3}$). The two feet are on opposite sides of the platform, so the distance is $2\sqrt3+\dfrac{2}{\sqrt3}=\dfrac{8}{\sqrt3}\approx4.62$ m.
🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. Can a plain set square from your geometry box be used as a height-measuring instrument — and how close do the heights it gives for a tree, a flagpole or a building come to the true height?

Why it matters. Foresters, surveyors and builders have always needed heights they cannot reach with a tape. A $45^\circ$ sight line gives a height equal to the distance walked, and a $30^\circ$ sight line gives $\tfrac1{\sqrt3}\approx0.577$ times the distance. The exact values of this section turn a pocket tool into a measuring device.

What to do.

  1. Choose a tall object whose height you can check (a school flagpole, a two-storey building whose floor height is known, or a lamp post).
  2. Hold the $45^\circ$ set square so that one short side is horizontal at eye level and the hypotenuse points up. Walk backwards until the top of the object lies exactly on the hypotenuse. Measure the distance $d_1$ from your feet to the base of the object.
  3. The height of the object above your eye is $d_1\tan45^\circ=d_1$. Add your eye height $e$ (measure it) to get the total height.
  4. Repeat with the $30^\circ$ set square (the $30^\circ$ angle at your eye). Measure the distance $d_2$. The height above your eye is $d_2\tan30^\circ=\dfrac{d_2}{\sqrt3}$.
  5. Compare the two answers with each other and with the true height. Which one is more accurate, and why might that be?

Tools: $45^\circ$ and $30^\circ$ set squares, a measuring tape, a helper

Hand in: A labelled sketch, a table of distances and calculated heights, the percentage difference from the true height, and a note on the sources of error (eye height, leaning, ground slope).

Going further. Use the $60^\circ$ angle of the other set square. What distance do you now need to walk for the same height, and why is the measurement more sensitive to small errors in distance?

Teacher/parent note: this activity previews Chapter 9. The only mathematics needed is $\tan45^\circ=1$ and $\tan30^\circ=\tfrac1{\sqrt3}$ from Theorem 8.4 and Theorem 8.5. Typical errors come from not holding the set square level and from forgetting to add the eye height.

🧪 Self-check quiz

0/5 correct

Q1. The value of $\sin30^\circ+\cos60^\circ$ is

Q2. $\tan60^\circ=$

Q3. Which of the following is not defined?

Q4. In $\triangle ABC$, right-angled at $B$, $\angle A=30^\circ$ and $BC=7$ cm. Then $AC=$

Q5. $\sec^245^\circ=$

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