Angles, Radians and Trigonometric Ratios

Degrees, Minutes and Seconds

An angle measures how far a line has turned about a point. One full turn is 360°, so a straight angle is 180° and a right angle is 90°. An angle smaller than 90° is acute; one between 90° and 180° is obtuse.

For finer measurement each degree is split into 60 minutes and each minute into 60 seconds:

1° = 60′ (minutes), 1′ = 60″ (seconds), so 1° = 3600″

These minutes and seconds measure turning, not time, but you add and subtract them exactly like the hours, minutes and seconds of a clock.

From degrees, minutes and seconds to a decimal. Divide the minutes by 60 and the seconds by 3600.

• 25°30′ = 25 + 30/60 = 25.5°
• 18°45′36″ = 18 + 45/60 + 36/3600 = 18 + 0.75 + 0.01 = 18.76°

From a decimal to degrees and minutes. Multiply the decimal part by 60.

• 32.4° = 32° + (0.4 × 60)′ = 32°24′
• 47.85° = 47°51′, because 0.85 × 60 = 51

Reading 32.4° as 32°40′ is the slip to watch for. The .4 is four-tenths of a degree, and four-tenths of 60 minutes is 24 minutes.

Adding and subtracting. Work unit by unit, carrying or borrowing 60 as you would with time. To find the complement of 36°48′25″, write 90° as 89°59′60″ and subtract:

89°59′60″ − 36°48′25″ = 53°11′35″

In a triangle with two angles 63°24′ and 48°51′, the two add up to 111°75′, which is 112°15′. The third angle is 180° − 112°15′ = 67°45′.

Before an angle in degrees and minutes can be changed into radians (the next unit), it has to be written as a single number of degrees, either as a decimal or as a fraction. 25°40′, for example, is $25\frac{2}{3}$ degrees, which is $\frac{77}{3}$ degrees.

The Radian

Degrees are not the only unit for angles. Draw a circle of radius r and measure along its edge an arc whose length is also r. The angle this arc makes at the centre is called one radian (1 rad).

The whole circumference is $2\pi r$, so it holds $2\pi$ such arcs, and a full turn is $2\pi$ radians. Half a turn gives the rule behind every conversion:

$$\pi \text{ rad} = 180^\circ \qquad 1 \text{ rad} = \frac{180^\circ}{\pi} \qquad 1^\circ = \frac{\pi}{180} \text{ rad}$$

Degrees to radians: multiply by $\frac{\pi}{180}$. So $150^\circ = 150 \times \frac{\pi}{180} = \frac{5\pi}{6}$.

Radians to degrees: multiply by $\frac{180}{\pi}$, which is the same as writing 180° in place of $\pi$. So $\frac{7\pi}{9} = \frac{7 \times 180^\circ}{9} = 140^\circ$.

These pairs are worth knowing by heart:

Minutes in the angle. Turn the minutes into a fraction of a degree first. 22°30′ is $22\frac{1}{2} = \frac{45}{2}$ degrees, and $\frac{45}{2} \times \frac{\pi}{180} = \frac{\pi}{8}$.

Radians without $\pi$. An angle of 2 radians has no $\pi$ in it, so use $\pi = \frac{22}{7}$: $2 \times \frac{180}{\pi} = 360 \times \frac{7}{22} \approx 114.5^\circ$. In the same way 1 rad $= \frac{630}{11}$ degrees, a little over 57°.

An angle written without the degree sign, such as $\theta = \frac{\pi}{4}$ or $\theta = 2$, is in radians.

[SSC Tip] In a degree-to-radian question, one of the options can be the right fraction with the $\pi$ dropped, such as $\frac{5}{6}$ for 150°. Your answer keeps the $\pi$ unless the question asks you to put $\pi = \frac{22}{7}$.

Arc Length and Sector Area with the Angle in Radians

An arc as long as the radius makes 1 radian at the centre, so an arc that makes $\theta$ radians is $\theta$ times the radius:

$$l = r\theta \qquad \theta = \frac{l}{r} \qquad (\theta \text{ in radians})$$

The sector cut off by that arc has area

$$A = \frac{1}{2}r^2\theta = \frac{1}{2}lr$$

Chapter 13 gave the same results with the angle in degrees, where the $\frac{\theta}{360}$ and the $2\pi$ appear separately. In radians they are already folded into $\theta$, which is why these formulas are so short.

Arc from the angle. On a circle of radius 21 cm, an angle of $\frac{2\pi}{3}$ at the centre cuts off an arc $l = 21 \times \frac{2\pi}{3} = 14\pi = 44$ cm (with $\pi = \frac{22}{7}$). The sector has area $\frac{1}{2} \times 44 \times 21 = 462$ cm².

Angle from the arc. An arc of 33 cm on a circle of radius 14 cm makes $\theta = \frac{33}{14}$ radian at the centre. In degrees that is

$$\frac{33}{14} \times \frac{180}{\pi} = \frac{33}{14} \times 180 \times \frac{7}{22} = 135^\circ$$

A swinging pendulum. The bob of a pendulum moves along an arc whose radius is the length of the pendulum. A pendulum 42 cm long that swings through 30° moves its bob along $42 \times \frac{\pi}{6} = 7\pi = 22$ cm.

[SSC Tip] $l = r\theta$ is true only with $\theta$ in radians. Put 120° straight into it for the circle above and the arc comes out as 21 × 120 = 2520 cm, far longer than the whole circumference of 132 cm. Change degrees to radians first.

Wheels, Turning and a Train on a Curve

One revolution of a wheel is one full turn: 360°, or $2\pi$ radians. So

$$\text{angle turned (in radians)} = 2\pi \times \text{number of revolutions}$$

A wheel that makes 480 revolutions a minute makes 8 a second, and turns through $8 \times 2\pi = 16\pi$ radians every second. Over a full minute that is $960\pi$ radians, so read carefully whether the question wants the angle per second or per minute.

A wheel rolling on the ground. As a wheel rolls without slipping, the length of road it covers equals the length of rim that has touched the road. With the angle in radians,

$$\text{distance rolled} = r\theta \qquad \theta = \frac{\text{distance}}{r}$$

A wheel of radius 35 cm that rolls 11 m (1100 cm) turns through $\frac{1100}{35} = \frac{220}{7}$ radians. With $\pi = \frac{22}{7}$ that is $10\pi$ radians, or 5 full revolutions. Counting revolutions gives the same: one revolution covers $2\pi \times 35 = 220$ cm, and 1100 ÷ 220 = 5.

A train on a circular track. The train runs along an arc of the track, so the angle it turns through, measured at the centre of the curve, is the distance travelled divided by the radius. Take a train at 72 km/h on a curve of radius 1.4 km. In metres per second its speed is $72 \times \frac{5}{18} = 20$, so in 11 seconds it covers 220 m and turns through

$$\theta = \frac{220}{1400} = \frac{11}{70} \text{ rad} = \frac{11}{70} \times 180 \times \frac{7}{22} = 9^\circ$$

Put the distance and the radius in the same unit before dividing: 220 m on a radius of 1.4 km is $\frac{220}{1400}$, not $\frac{220}{1.4}$.

Radian or degree in the answer. Questions on wheels and tracks may ask for the angle either way. The radian value is the bare ratio of two lengths; multiply it by $\frac{180}{\pi}$ only if degrees are wanted.

Clock Hands, Fractions of a Turn and Angles in a Ratio

Clock hands. The minute hand goes round once in 60 minutes and the hour hand once in 12 hours. So

• the minute hand turns 6° a minute, that is $\frac{\pi}{30}$ radian a minute;
• the hour hand turns 30° an hour, that is 0.5° a minute or $\frac{\pi}{360}$ radian a minute.

At h hours m minutes the hour hand is (30h + 0.5m)° past 12 and the minute hand is 6m° past 12. The angle between them is the difference, |30h − 5.5m|. If it comes out above 180°, take 360° minus it for the smaller angle.

At 2:30 the hour hand has turned 30 × 2 + 0.5 × 30 = 75° from 12 and the minute hand 180°. They are 105° apart, which is $105 \times \frac{\pi}{180} = \frac{7\pi}{12}$ radians. The hour hand is halfway between 2 and 3, not on the 2.

From noon to 3:40 the hour hand turns 220 × 0.5 = 110° = $\frac{11\pi}{18}$ rad, while the minute hand turns 220 × 6 = 1320° = $\frac{22\pi}{3}$ rad, three full turns and 240° more. The tip of a minute hand 7 cm long moves along an arc: in 30 minutes the hand turns $\pi$ radians, so the tip moves $7\pi = 22$ cm.

An angle as a part of a full turn. Divide by 360° (or by $2\pi$ for radians) and multiply by 100 for a percentage.

• 72° is $\frac{72}{360}$ = 20% of a full turn.
• $\frac{3\pi}{4}$ is $\frac{3}{8}$ of a turn, that is 37.5%.
• One radian is $\frac{1}{2\pi} = \frac{7}{44}$ of a turn, a little under 16%.
• Going the other way, 40% of a full turn is 144°, which is $\frac{4\pi}{5}$.

Angles in a ratio. The angles of a triangle add up to 180° ($\pi$ radians) and those of a quadrilateral to 360° ($2\pi$). Angles of a triangle in the ratio 2 : 3 : 7 are $\frac{2}{12}$, $\frac{3}{12}$ and $\frac{7}{12}$ of 180°: that is 30°, 45° and 105°, or $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{7\pi}{12}$.

The Six Trigonometric Ratios

Trigonometry starts with a right triangle. In triangle ABC below, the right angle is at B and the angle we study, $\theta$, is at A. Name the sides from A's point of view:

• the hypotenuse AC faces the right angle and is the longest side;
• the opposite side BC is across from $\theta$ (also called the perpendicular);
• the adjacent side AB is the other side that touches $\theta$ (also called the base).

The six ratios are

$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB}$$

$$\operatorname{cosec}\theta = \frac{AC}{BC} \qquad \sec\theta = \frac{AC}{AB} \qquad \cot\theta = \frac{AB}{BC}$$

With P for perpendicular, B for base and H for hypotenuse, the first three are P/H, B/H and P/B. Many students keep this order with the line “Pandit Badri Prasad Har Har Bole”: the first and fourth words give P/H for sin, the second and fifth B/H for cos, the third and sixth P/B for tan.

An example. If AB = 12, BC = 5 and AC = 13, then

$$\sin A = \frac{5}{13}, \ \cos A = \frac{12}{13}, \ \tan A = \frac{5}{12}, \ \operatorname{cosec} A = \frac{13}{5}, \ \sec A = \frac{13}{12}, \ \cot A = \frac{12}{5}$$

Change the angle, and opposite and adjacent swap. Standing at C, the opposite side is AB and the adjacent one is BC, so $\sin C = \frac{12}{13}$ and $\cos C = \frac{5}{13}$. The hypotenuse is the same for both acute angles.

Size does not matter. A right triangle with sides 10, 24 and 26 has the same angles as the 5, 12, 13 triangle, and so the same ratios: $\frac{10}{26} = \frac{5}{13}$. A ratio fixes the angle, not the size of the triangle.

Reading the symbols. $\sin\theta$ is one quantity, not sin multiplied by $\theta$. $\sin^2\theta$ means $(\sin\theta)^2$, and $\sin 2\theta$ is not the same as $2\sin\theta$.

Reciprocal Pairs, and Tan from Sin and Cos

Turn each of the first three ratios upside down and you get the other three:

$$\operatorname{cosec}\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta}$$

So $\sin\theta \operatorname{cosec}\theta = 1$, $\cos\theta \sec\theta = 1$ and $\tan\theta \cot\theta = 1$.

The pairs are sin with cosec and cos with sec: in each pair one name has “co” and the other does not. Pairing sec with sin is the mistake to watch for. If $\sin\theta = \frac{3}{5}$, then $\operatorname{cosec}\theta = \frac{5}{3}$, while $\sec\theta$ is a different number, $\frac{5}{4}$ (the adjacent side is 4).

Tan from sin and cos. Divide sin by cos and the hypotenuse cancels:

$$\frac{\sin\theta}{\cos\theta} = \frac{BC/AC}{AB/AC} = \frac{BC}{AB} = \tan\theta \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}$$

If $\sin\theta = \frac{7}{25}$ and $\cos\theta = \frac{24}{25}$, then $\tan\theta = \frac{7}{24}$ at once, with no triangle drawn.

Quick simplifications. Write every ratio in terms of sin and cos, then cancel.

• $\sin\theta \cot\theta = \sin\theta \times \frac{\cos\theta}{\sin\theta} = \cos\theta$
• $\cos\theta \tan\theta = \cos\theta \times \frac{\sin\theta}{\cos\theta} = \sin\theta$
• $\sec\theta \cot\theta = \frac{1}{\cos\theta} \times \frac{\cos\theta}{\sin\theta} = \operatorname{cosec}\theta$
• $\tan\theta \cos\theta \operatorname{cosec}\theta = \frac{\sin\theta}{\cos\theta} \times \cos\theta \times \frac{1}{\sin\theta} = 1$

Pythagoras in ratio form. In the triangle, $BC^2 + AB^2 = AC^2$. Dividing by $AC^2$ gives

$$\sin^2\theta + \cos^2\theta = 1$$

It is a handy check on any pair of values you work out: $\left(\frac{7}{25}\right)^2 + \left(\frac{24}{25}\right)^2 = \frac{49 + 576}{625} = 1$. The full set of identities, and how to simplify with them, is in Identities and Simplification.

One Ratio Given: Find All the Others

A question may give one ratio and ask for another, or for an expression in several of them. The method does not change:

1. Write the given ratio as side over side and put those two numbers on a right triangle.
2. Find the third side by Pythagoras.
3. Read off whatever is asked.

Take $\tan\theta = \frac{8}{15}$. The opposite side is 8 and the adjacent side 15, so the hypotenuse is $\sqrt{64 + 225} = \sqrt{289} = 17$.

$$\sin\theta = \frac{8}{17} \quad \cos\theta = \frac{15}{17} \quad \operatorname{cosec}\theta = \frac{17}{8} \quad \sec\theta = \frac{17}{15} \quad \cot\theta = \frac{15}{8}$$

The 8 and 15 need not be the real lengths. Sides 8k, 15k and 17k give the same angle, and k cancels in every ratio.

Pythagorean triplets save the square root. These whole-number right triangles are worth recognising:

with their multiples, such as 6, 8, 10 or 15, 20, 25. The largest number is the hypotenuse. So $\cos\theta = \frac{21}{29}$ means adjacent 21, hypotenuse 29 and opposite 20, giving $\tan\theta = \frac{20}{21}$. And $\sec\theta = \frac{97}{65}$ means $\cos\theta = \frac{65}{97}$, the opposite side is 72 and $\sin\theta = \frac{72}{97}$.

When the third side is a surd. $\sin\theta = \frac{2}{3}$ gives an adjacent side of $\sqrt{9 - 4} = \sqrt{5}$, so $\cos\theta = \frac{\sqrt{5}}{3}$ and $\tan\theta = \frac{2}{\sqrt{5}}$.

With letters. If $\tan\theta = \frac{a}{b}$, the hypotenuse is $\sqrt{a^2 + b^2}$ and

$$\sin\theta = \frac{a}{\sqrt{a^2 + b^2}} \qquad \cos\theta = \frac{b}{\sqrt{a^2 + b^2}}$$

[SSC Tip] Before drawing, check which two sides the given ratio uses. $\sec\theta = \frac{17}{8}$ puts 17 on the hypotenuse and 8 on the adjacent side. Putting the 8 opposite instead swaps every sine with its cosine, and the swapped value makes a ready-made wrong option.

Expressions in the Ratios

Once sin and cos are known, any expression in the six ratios is arithmetic. Find the values from the triangle, put them in, and simplify.

A fraction in sin and cos. Take $\sec\theta = \frac{5}{3}$. Then $\cos\theta = \frac{3}{5}$, the opposite side is 4 and $\sin\theta = \frac{4}{5}$. So

$$\frac{3\sin\theta + 2\cos\theta}{3\sin\theta - 2\cos\theta} = \frac{\frac{12}{5} + \frac{6}{5}}{\frac{12}{5} - \frac{6}{5}} = \frac{18}{6} = 3$$

Every term has a 5 underneath, and it cancels. Whenever each term on top and below has exactly one sin or one cos, you can use the sides themselves: 4 for sin and 3 for cos give $\frac{12 + 6}{12 - 6} = 3$ at once. Values from a Condition, Maximum and Minimum does the same job by dividing through by $\cos\theta$.

Brackets of the form (1 + x)(1 − x). Multiply out first: $(1 + \sin\theta)(1 - \sin\theta) = 1 - \sin^2\theta = \cos^2\theta$, and in the same way $(1 + \cos\theta)(1 - \cos\theta) = \sin^2\theta$. So if $\cot\theta = \frac{5}{12}$,

$$\frac{(1 + \sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(1 - \cos\theta)} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = \frac{25}{144}$$

The long way agrees: the hypotenuse is 13, $\sin\theta = \frac{12}{13}$, $\cos\theta = \frac{5}{13}$, and $\frac{1 - \frac{144}{169}}{1 - \frac{25}{169}} = \frac{25}{144}$.

Sums of two ratios. With $\cos\theta = \frac{20}{29}$ the opposite side is 21, so

$$\sec\theta + \tan\theta = \frac{29}{20} + \frac{21}{20} = \frac{50}{20} = \frac{5}{2} \qquad \operatorname{cosec}\theta - \cot\theta = \frac{29}{21} - \frac{20}{21} = \frac{3}{7}$$

Squares and products. With $\sin\theta = \frac{3}{5}$ and $\cos\theta = \frac{4}{5}$: $2\sin\theta\cos\theta = \frac{24}{25}$ and $\cos^2\theta - \sin^2\theta = \frac{16 - 9}{25} = \frac{7}{25}$.

Before putting values into a long expression, check that $\sin^2\theta + \cos^2\theta = 1$ for them. If it fails, a side of your triangle is wrong.

Ratios in a Right Triangle with Sides Given

In a triangle question, find the right angle first: the side facing it is the hypotenuse. “Right-angled at B” means AC is the hypotenuse, and the two acute angles are A and C.

Two sides known. Find the third by Pythagoras and read off. In triangle PQR, right-angled at Q, with PR = 89 and PQ = 39, the third side is $QR = \sqrt{7921 - 1521} = \sqrt{6400} = 80$. Then $\sin P = \frac{QR}{PR} = \frac{80}{89}$, $\cos R = \frac{80}{89}$ and $\tan R = \frac{PQ}{QR} = \frac{39}{80}$.

A sum of two sides given. Triangle ABC is right-angled at B, with AB = 12 cm and AC + BC = 18 cm. Write Pythagoras as a difference of squares and factorise:

$$AC^2 - BC^2 = AB^2 = 144 \quad\Rightarrow\quad (AC - BC)(AC + BC) = 144 \quad\Rightarrow\quad AC - BC = \frac{144}{18} = 8$$

Adding and subtracting AC + BC = 18 and AC − BC = 8 gives AC = 13 cm and BC = 5 cm. Now $\sin A = \frac{5}{13}$, $\tan A = \frac{5}{12}$ and $\cos C = \frac{5}{13}$. A difference of two sides is handled the same way.

Sides written in x. If the sides of a right triangle are x, x + 7 and x + 8, the longest is the hypotenuse:

$$x^2 + (x + 7)^2 = (x + 8)^2 \quad\Rightarrow\quad x^2 - 2x - 15 = 0 \quad\Rightarrow\quad (x - 5)(x + 3) = 0$$

so x = 5 (a length cannot be −3). The sides are 5, 12 and 13, and the smallest angle, which faces the side 5, has sine $\frac{5}{13}$.

One angle and the other. BC is opposite A but next to C, so the same fraction serves both angles:

$$\sin A = \cos C = \frac{BC}{AC} \qquad \tan A = \cot C = \frac{BC}{AB}$$

The two acute angles add up to 90°, and Standard Angles and Complementary Angles builds a whole method on this.

Mark the angle the question asks about before reading off a side. In a triangle right-angled at B, $\sin A$ and $\sin C$ use different sides, and picking the wrong one gives an answer that looks just as tidy as the right one.

Isosceles Triangles and Triangles with an Altitude

A triangle without a right angle has no hypotenuse, so its ratios cannot be read off directly. Draw an altitude to make right triangles, then work inside them.

Isosceles triangle. In triangle ABC, AB = AC = 13 cm and BC = 10 cm. The altitude AD from the top meets BC at its midpoint, so BD = DC = 5 cm and $AD = \sqrt{169 - 25} = 12$ cm.

In right triangle ABD,

$$\sin B = \frac{12}{13} \qquad \cos B = \frac{5}{13} \qquad \tan B = \frac{12}{5}$$

The altitude also halves the top angle, so $\sin\angle BAD = \frac{BD}{AB} = \frac{5}{13}$ is the sine of half of $\angle A$.

Equilateral triangle. All three sides are equal, and the altitude again meets the base at its midpoint. For side 2a the altitude is $\sqrt{4a^2 - a^2} = a\sqrt{3}$. Its angles are 60° and 30°, whose values are worked out in Standard Angles and Complementary Angles. If a point D on the base is not the midpoint, drop the altitude AM anyway and work in triangle ADM, which has its right angle at M.

Any triangle with an altitude. In triangle ABC, AD is perpendicular to BC, AB = 15 cm, AC = 13 cm and AD = 12 cm.

The altitude splits the triangle into two right triangles that share the side AD.

• In triangle ABD: $BD = \sqrt{225 - 144} = 9$ cm, so $\sin B = \frac{12}{15} = \frac{4}{5}$, $\cos B = \frac{9}{15} = \frac{3}{5}$ and $\tan B = \frac{4}{3}$.
• In triangle ACD: $DC = \sqrt{169 - 144} = 5$ cm, so $\sin C = \frac{12}{13}$, $\cos C = \frac{5}{13}$ and $\tan C = \frac{12}{5}$.

The base BC is 9 + 5 = 14 cm. Read each angle's ratios inside the right triangle that contains it: B belongs to triangle ABD and C to triangle ACD. Using AB = 15 as if it were a hypotenuse for angle C mixes the two halves and gives nonsense.

How the Ratios Change from 0° to 90°

Place the right triangle inside a circle of radius 1, with the angle $\theta$ at the centre O and the hypotenuse along a radius. The hypotenuse is now 1, so

$$\sin\theta = \frac{\text{height}}{1} = \text{height of the point} \qquad \cos\theta = \frac{\text{base}}{1} = \text{distance across from O}$$

The figure shows this for 25° and 65°.

As $\theta$ grows from 0° to 90°, the point climbs the circle: its height grows from 0 to 1 while its distance across shrinks from 1 to 0. So

• $\sin\theta$ increases from 0 to 1, and $\cos\theta$ decreases from 1 to 0;
• $\tan\theta = \frac{\text{height}}{\text{base}}$ increases from 0 and grows without limit as $\theta$ nears 90°;
• the reciprocals move the other way: $\operatorname{cosec}\theta$ and $\cot\theta$ decrease, $\sec\theta$ increases.

Bounds. The hypotenuse is the longest side, so for an acute angle $\sin\theta$ and $\cos\theta$ lie between 0 and 1, and $\sec\theta$ and $\operatorname{cosec}\theta$ are greater than 1. $\tan\theta$ and $\cot\theta$ can be any positive number. Values such as $\sin\theta = \frac{5}{4}$ or $\sec\theta = \frac{1}{2}$ cannot happen.

Which is larger? The graphs of sin and cos cross at 45°.

• Below 45°: $\cos\theta > \sin\theta$ and $\tan\theta < 1$.
• At 45°: $\sin\theta = \cos\theta$ and $\tan\theta = 1$.
• Above 45°: $\sin\theta > \cos\theta$ and $\tan\theta > 1$.

So $\sin 50^\circ > \cos 50^\circ$ and $\cos 20^\circ > \sin 20^\circ$. Also $\tan\theta > \sin\theta$ for every acute angle, since $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and dividing by a number less than 1 makes it bigger.

Radians in a comparison. $\sin 1$ means the sine of 1 radian, which is about 57°. So $\sin 1 > \sin 1^\circ$, while $\cos 1 < \cos 1^\circ$.

[SSC Tip] To compare the sine of one angle with the cosine of another, turn the cosine into the sine of 90° minus the angle: $\cos 50^\circ = \sin 40^\circ$, which is less than $\sin 50^\circ$. The rule behind this is in Standard Angles and Complementary Angles.

Solved Examples

Question 1

What is the radian measure of 40°20′? (a) $\frac{121\pi}{540}$ (b) $\frac{2\pi}{9}$ (c) $\frac{121}{540}$ (d) $\frac{121\pi}{180}$

Answer. I change the minutes into degrees first: 20′ is $\frac{20}{60} = \frac{1}{3}$ of a degree, so the angle is $40\frac{1}{3} = \frac{121}{3}$ degrees. Now I multiply by $\frac{\pi}{180}$: $\frac{121}{3} \times \frac{\pi}{180} =$ (a) $\frac{121\pi}{540}$. $\frac{2\pi}{9}$ is 40° alone, with the 20′ thrown away.

Question 2

What is $\frac{5\pi}{24}$ radians in degrees and minutes? (a) 37°50′ (b) 75° (c) 37°30′ (d) 18°45′

Answer. I put 180° in place of $\pi$: $\frac{5 \times 180^\circ}{24} = 37.5^\circ$. Half a degree is 30 minutes, so the angle is (c) 37°30′. Writing 37°50′ reads the .5 as 50 minutes, but .5 of a degree is half of 60 minutes.

Question 3

Taking $\pi = \frac{22}{7}$, one radian is closest to (a) 57°27′ (b) 57°16′22″ (c) 57°30′ (d) 57°22′16″

Answer. 1 rad $= \frac{180^\circ}{\pi} = 180 \times \frac{7}{22} = \frac{630}{11} = 57\frac{3}{11}$ degrees. The leftover $\frac{3}{11}$ of a degree is $\frac{180}{11} = 16\frac{4}{11}$ minutes, and $\frac{4}{11}$ of a minute is $\frac{240}{11}$, about 21.8 seconds. So one radian is about (b) 57°16′22″. The decimal 57.27° is not 57°27′; its .27 has to be multiplied by 60.

Question 4

The angles of a triangle are in the ratio 3 : 4 : 5. What is the largest angle in radians? (a) $\frac{\pi}{3}$ (b) $\frac{5\pi}{6}$ (c) $\frac{5}{12}$ (d) $\frac{5\pi}{12}$

Answer. The parts add up to 3 + 4 + 5 = 12 and the angles share 180°, so one part is 15°. The largest angle is 5 × 15° = 75°, and $75 \times \frac{\pi}{180} =$ (d) $\frac{5\pi}{12}$. Sharing 360° by mistake gives 150°, which is $\frac{5\pi}{6}$.

Question 5

Two complementary angles differ by $\frac{\pi}{9}$ radians. What is the larger angle in degrees? (a) $55^\circ$ (b) $35^\circ$ (c) $65^\circ$ (d) $110^\circ$

Answer. Complementary angles add up to 90°. The difference $\frac{\pi}{9}$ is $\frac{180^\circ}{9} = 20^\circ$. So the larger angle is $\frac{90^\circ + 20^\circ}{2} =$ (a) $55^\circ$ and the smaller one is 35°.

Question 6

An arc of a circle of radius 35 cm makes an angle of 72° at the centre. How long is the arc? (Take $\pi = \frac{22}{7}$.) (a) 88 cm (b) 44 cm (c) 22 cm (d) 2520 cm

Answer. In radians, $72^\circ = 72 \times \frac{\pi}{180} = \frac{2\pi}{5}$. Then $l = r\theta = 35 \times \frac{2\pi}{5} = 14\pi = 14 \times \frac{22}{7} =$ (b) 44 cm. Multiplying 35 by 72 gives 2520 cm, longer than the whole circle (220 cm), because 72 is in degrees.

Question 7

An arc 22 cm long makes an angle at the centre of a circle of radius 28 cm. What is the angle in degrees? (Take $\pi = \frac{22}{7}$.) (a) $22.5^\circ$ (b) $90^\circ$ (c) $45^\circ$ (d) $\left(\frac{11}{14}\right)^\circ$

Answer. The angle in radians is $\frac{l}{r} = \frac{22}{28} = \frac{11}{14}$. To change it to degrees I multiply by $\frac{180}{\pi}$: $\frac{11}{14} \times 180 \times \frac{7}{22} =$ (c) $45^\circ$.

Question 8

A wheel makes 300 revolutions per minute. Through how many radians does it turn in one second? (a) $5\pi$ (b) $600\pi$ (c) 10 (d) $10\pi$

Answer. 300 revolutions a minute is 5 revolutions a second. Each revolution is $2\pi$ radians, so in one second the wheel turns $5 \times 2\pi =$ (d) $10\pi$. $600\pi$ is the angle for a whole minute.

Question 9

What is the angle between the hands of a clock at 5:20, in radians? (a) $\frac{\pi}{6}$ (b) $\frac{16\pi}{9}$ (c) $\frac{2\pi}{9}$ (d) $\frac{2}{9}$

Answer. At 5:20 the hour hand has moved past the 5: it is 30 × 5 + 0.5 × 20 = 160° from 12. The minute hand is 6 × 20 = 120° from 12. The gap is 40°, and $40 \times \frac{\pi}{180} =$ (c) $\frac{2\pi}{9}$. If the hour hand is kept exactly on the 5 (150°), the gap comes out as 30°, which is $\frac{\pi}{6}$.

Question 10

The minute hand of a clock is 14 cm long. How far does its tip move in 45 minutes? (Take $\pi = \frac{22}{7}$.) (a) 66 cm (b) 88 cm (c) 22 cm (d) 33 cm

Answer. In 60 minutes the hand turns $2\pi$, so in 45 minutes it turns $\frac{45}{60} \times 2\pi = \frac{3\pi}{2}$ radians. The tip moves along an arc of radius 14 cm: $l = 14 \times \frac{3\pi}{2} = 21\pi = 21 \times \frac{22}{7} =$ (a) 66 cm.

Question 11

An angle of $\frac{3\pi}{10}$ radians is what percentage of a full turn? (a) 15% (b) 30% (c) 54% (d) 60%

Answer. In degrees, $\frac{3\pi}{10} = \frac{3 \times 180^\circ}{10} = 54^\circ$. A full turn is 360°, so the share is $\frac{54}{360} \times 100 =$ (a) 15%. Working in radians gives the same: $\frac{3\pi}{10} \div 2\pi = \frac{3}{20}$.

Question 12

A train runs at 66 km/h on a circular track of radius 1500 m. Through what angle (in degrees) does it turn in 10 seconds? (Take $\pi = \frac{22}{7}$.) (a) $14^\circ$ (b) $3.5^\circ$ (c) $22^\circ$ (d) $7^\circ$

Answer. 66 km/h is $66 \times \frac{5}{18} = \frac{55}{3}$ m per second, so in 10 seconds the train covers $\frac{550}{3}$ m of track. The angle at the centre is arc ÷ radius: $\frac{550}{3} \div 1500 = \frac{11}{90}$ radian. In degrees: $\frac{11}{90} \times 180 \times \frac{7}{22} =$ (d) $7^\circ$. Stopping before dividing by $\pi$ gives 22°.

Question 13

In triangle ABC, right-angled at B, AB = 24 cm and BC = 7 cm. What is $\sin A + \cos C$? (a) $\frac{31}{25}$ (b) $\frac{48}{25}$ (c) $\frac{14}{25}$ (d) 1

Answer. AC is the hypotenuse: $\sqrt{576 + 49} = \sqrt{625} = 25$ cm. For angle A the opposite side is BC, so $\sin A = \frac{7}{25}$. For angle C the adjacent side is BC again, so $\cos C = \frac{7}{25}$ too. The sum is (c) $\frac{14}{25}$.

Question 14

If $\tan\theta = \frac{20}{21}$, what is $\sin\theta + \cos\theta$? (a) $\frac{41}{21}$ (b) $\frac{41}{29}$ (c) $\frac{29}{41}$ (d) 1

Answer. Opposite 20 and adjacent 21, so the hypotenuse is $\sqrt{400 + 441} = \sqrt{841} = 29$. Then $\sin\theta = \frac{20}{29}$ and $\cos\theta = \frac{21}{29}$, and their sum is (b) $\frac{41}{29}$.

Question 15

If $\sec\theta = \frac{85}{36}$, what is $\operatorname{cosec}\theta + \cot\theta$? (a) $\frac{11}{7}$ (b) $\frac{7}{11}$ (c) $\frac{9}{2}$ (d) $\frac{121}{85}$

Answer. Secant is hypotenuse over adjacent, so the hypotenuse is 85 and the adjacent side 36. The opposite side is $\sqrt{7225 - 1296} = \sqrt{5929} = 77$. Then $\operatorname{cosec}\theta = \frac{85}{77}$ and $\cot\theta = \frac{36}{77}$, which add up to $\frac{121}{77} =$ (a) $\frac{11}{7}$. Adding sec and tan instead would give $\frac{85 + 77}{36} = \frac{9}{2}$.

Question 16

If $\tan\theta = \frac{4}{7}$, what is the value of $\frac{(1 + \cos\theta)(1 - \cos\theta)}{(1 + \sin\theta)(1 - \sin\theta)}$? (a) $\frac{49}{16}$ (b) $\frac{4}{7}$ (c) $\frac{65}{49}$ (d) $\frac{16}{49}$

Answer. I multiply out the brackets first. The top is $1 - \cos^2\theta = \sin^2\theta$ and the bottom is $1 - \sin^2\theta = \cos^2\theta$. So the whole fraction is $\frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta = \left(\frac{4}{7}\right)^2 =$ (d) $\frac{16}{49}$. I never needed the hypotenuse.

Question 17

If $\cos\theta = \frac{7}{25}$, what is $\tan\theta + \cot\theta$? (a) 2 (b) $\frac{625}{168}$ (c) $\frac{168}{625}$ (d) $\frac{527}{168}$

Answer. Adjacent 7 and hypotenuse 25, so the opposite side is $\sqrt{625 - 49} = 24$. Then $\tan\theta = \frac{24}{7}$ and $\cot\theta = \frac{7}{24}$. Over the common denominator 168 this is $\frac{576 + 49}{168} =$ (b) $\frac{625}{168}$.

Question 18

If $\cos\theta = \frac{2}{3}$ and $\theta$ is acute, what is $\tan\theta$? (a) $\frac{2}{\sqrt{5}}$ (b) $\frac{3}{2}$ (c) $\frac{\sqrt{5}}{2}$ (d) $\frac{\sqrt{5}}{3}$

Answer. Adjacent 2 and hypotenuse 3, so the opposite side is $\sqrt{9 - 4} = \sqrt{5}$. Then $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} =$ (c) $\frac{\sqrt{5}}{2}$.

Question 19

In triangle ABC, right-angled at B, AB = 5 cm and AC + BC = 25 cm. What is $\sin C$? (a) $\frac{12}{13}$ (b) $\frac{5}{13}$ (c) $\frac{5}{12}$ (d) $\frac{12}{5}$

Answer. AC is the hypotenuse, so $AC^2 - BC^2 = AB^2 = 25$. Factorising, $(AC - BC)(AC + BC) = 25$, and since AC + BC = 25, AC − BC = 1. That gives AC = 13 cm and BC = 12 cm. For angle C the opposite side is AB, so $\sin C = \frac{AB}{AC} =$ (b) $\frac{5}{13}$. The value $\frac{12}{13}$ would be $\sin A$.

Question 20

In triangle ABC, AB = AC = 25 cm and BC = 14 cm. What is $\tan B$? (a) $\frac{7}{24}$ (b) $\frac{24}{25}$ (c) $\frac{25}{7}$ (d) $\frac{24}{7}$

Answer. The triangle is isosceles, so the altitude AD from A meets BC at its midpoint and BD = 7 cm. Then $AD = \sqrt{625 - 49} = 24$ cm. In right triangle ABD, $\tan B = \frac{AD}{BD} =$ (d) $\frac{24}{7}$.

Question 21

In triangle ABC, AD is perpendicular to BC, AB = 20 cm, BD = 16 cm and DC = 5 cm. What is $\sin B + \cos C$? (a) $\frac{64}{65}$ (b) $\frac{77}{65}$ (c) $\frac{99}{65}$ (d) 1

Answer. From triangle ABD, $AD = \sqrt{400 - 256} = 12$ cm. From triangle ADC, $AC = \sqrt{144 + 25} = 13$ cm. So $\sin B = \frac{AD}{AB} = \frac{12}{20} = \frac{3}{5}$ and $\cos C = \frac{DC}{AC} = \frac{5}{13}$. Adding: $\frac{39 + 25}{65} =$ (a) $\frac{64}{65}$.

Question 22

Which of these is true? (a) $\sin 50^\circ < \cos 50^\circ$ (b) $\sin 50^\circ = \cos 50^\circ$ (c) $\sin 50^\circ > \cos 50^\circ$ (d) $\tan 50^\circ < 1$

Answer. 50° is more than 45°, and past 45° the sine is the larger of the two. Another way to see it: $\cos 50^\circ = \sin 40^\circ$, and sine increases from 0° to 90°, so $\sin 50^\circ$ is bigger. For the same reason $\tan 50^\circ = \frac{\sin 50^\circ}{\cos 50^\circ}$ is more than 1. The true statement is (c) $\sin 50^\circ > \cos 50^\circ$.

Question 23

$\theta$ is an acute angle. Which of these values is impossible? (a) $\sin\theta = 0.95$ (b) $\sec\theta = 0.95$ (c) $\cot\theta = 0.2$ (d) $\tan\theta = 12$

Answer. For an acute angle the hypotenuse is the longest side. So sin and cos are less than 1, while sec and cosec, which have the hypotenuse on top, are more than 1. But tan and cot can be any positive number. A secant below 1 cannot happen, so the impossible one is (b) $\sec\theta = 0.95$.

Question 24

ABC is an equilateral triangle of side 6 cm, and D is the point on BC with BD = 2 cm. What is $\tan\angle ADC$? (a) $\sqrt{3}$ (b) $\frac{3\sqrt{3}}{2}$ (c) $\frac{1}{3\sqrt{3}}$ (d) $3\sqrt{3}$

Answer. I drop the altitude AM. It meets BC at its midpoint, so BM = 3 cm and DM = 3 − 2 = 1 cm, with M between D and C. The altitude is $AM = \sqrt{36 - 9} = 3\sqrt{3}$ cm. Triangle ADM has its right angle at M, and $\angle ADC$ is its angle at D, so $\tan\angle ADC = \frac{AM}{DM} = \frac{3\sqrt{3}}{1} =$ (d) $3\sqrt{3}$. Dividing by BD = 2 instead of DM = 1 gives $\frac{3\sqrt{3}}{2}$.