Exam practice

90∘°

100ᵍ

200ᵍ

200ᵍ is two right angles: 180∘°.

$1^\circ = \left(\frac{10}{9}\right)^{\mathrm{g}}$

$1^{\mathrm{g}} = \left(\frac{9}{10}\right)^\circ$

35′ = (35 × 60)′′ = 2100′′.

(50 × 60 + 40)′′ = (3000 + 40)′′ = 3040′′.

(30 × 3600 + 40 × 60 + 50)′′ = (108000 + 2400 + 50)′′ = 110450′′.

(55 × 3600 + 30 × 60 + 10)′′ = (198000 + 1800 + 10)′′ = 199810′′.

(10 × 3600 + 25 × 60 + 48)′′ = (36000 + 1500 + 48)′′ = 37548′′.

(55 × 3600 + 56 × 60 + 28)′′ = (198000 + 3360 + 28)′′ = 201388′′.

In the sexagesimal system a complete turn is 360∘°, a right angle 90∘°, and 1∘° = 60′, 1′ = 60′′. It is the everyday system: school geometry, maps and GPS (latitude and longitude), navigation, astronomy.
In the centesimal system a complete turn is 400ᵍ, a right angle 100ᵍ, and 1ᵍ = 100′, 1′ = 100′′. It is used mainly by surveyors, especially in Europe, and is on most scientific calculators as GRAD.
Since 90∘° = 100ᵍ: degrees to grades, multiply by $\frac{10}{9}$, for example $45^\circ = \frac{10}{9} \times 45 = 50^{\mathrm{g}}$; grades to degrees, multiply by $\frac{9}{10}$, for example $50^{\mathrm{g}} = \frac{9}{10} \times 50 = 45^\circ$.

30ᵍ20′10′′ = 30.201ᵍ = ($\frac{9}{10}$ × 30.201)∘° = 27.1809∘°.
0.1809∘° = 10.854′, and 0.854′ = 51.24′′. So 27∘°10′51.24′′.

25.151ᵍ = (0.9 × 25.151)∘° = 22.6359∘°.
0.6359∘° = 38.154′, and 0.154′ = 9.24′′. So 22∘°38′9.24′′.

45.3525ᵍ = (0.9 × 45.3525)∘° = 40.81725∘°.
0.81725∘° = 49.035′, and 0.035′ = 2.1′′. So 40∘°49′2.1′′.

30.12ᵍ = (0.9 × 30.12)∘° = 27.108∘°.
0.108∘° = 6.48′, and 0.48′ = 28.8′′. So 27∘°6′28.8′′.

There are no minutes, so 26ᵍ0′15′′ = 26.0015ᵍ = (0.9 × 26.0015)∘° = 23.40135∘°.
0.40135∘° = 24.081′, and 0.081′ = 4.86′′. So 23∘°24′4.86′′.

47.4849ᵍ = (0.9 × 47.4849)∘° = 42.73641∘°.
0.73641∘° = 44.1846′, and 0.1846′ = 11.08′′. So 42∘°44′11.08′′.

25∘°45′30′′ = (25 + $\frac{45}{60}$ + $\frac{30}{3600}$)∘° = 25.758333∘°.
In grades: $\frac{10}{9}$ × 25.758333 = 28.620370ᵍ, so 62.0370′, which is 62′3.70′′. So 28ᵍ62′3.70′′.

30.254167∘° × $\frac{10}{9}$ = 33.615741ᵍ. So 33ᵍ61′57.41′′.

49.840278∘° × $\frac{10}{9}$ = 55.378086ᵍ. So 55ᵍ37′80.86′′.

44.590278∘° × $\frac{10}{9}$ = 49.544753ᵍ. So 49ᵍ54′47.53′′.

80.838889∘° × $\frac{10}{9}$ = 89.820988ᵍ, so 82.0988′, which is 82′9.88′′. So 89ᵍ82′9.88′′.

76.4425∘° × $\frac{10}{9}$ = 84.936111ᵍ. So 84ᵍ93′61.11′′.

$\frac{9}{10} \times 50 =$ 45∘°

$\frac{9}{10} \times 80 =$ 72∘°

$\frac{9}{10} \times 130 =$ 117∘°

$\frac{9}{10} \times 160 =$ 144∘°

$\frac{9}{10} \times 70 =$ 63∘°

$\frac{9}{10} \times 250 =$ 225∘°

$50 + \frac{40}{100} + \frac{8}{10000} =$ 50.4008ᵍ (about 50.40ᵍ).

40.3233ᵍ (about 40.32ᵍ).

56.855ᵍ (about 56.86ᵍ).

No minutes: 45 + $\frac{35}{10000}$ = 45.0035ᵍ.

37.5ᵍ

98.4237ᵍ (about 98.42ᵍ).

$\frac{10}{9} \times 45 =$ 50ᵍ

$\frac{10}{9} \times 270 =$ 300ᵍ

$\frac{10}{9} \times 18 =$ 20ᵍ

$\frac{10}{9} \times 36 =$ 40ᵍ

$\frac{10}{9} \times 108 =$ 120ᵍ

$\frac{10}{9} \times 54 =$ 60ᵍ

The right angle, 90∘°.

The right angle is the largest (the other two add up to 90∘°), so 100ᵍ.

180∘° − 90∘° − 60∘° = 30∘°.

Let them be x, 2x, 3x. Then $6x = 180^\circ$, so $x = 30^\circ$.
The angles are 30∘°, 60∘° and 90∘°: a right-angled triangle.

$2x + 3x + 4x = 180^\circ$, so $9x = 180^\circ$ and $x = 20^\circ$. The angles are 40∘°, 60∘° and 80∘°.

In grades the sum is 200ᵍ: 9x = 200, x = 22.22. So $\left(\frac{400}{9}\right)^{\mathrm{g}}$, $\left(\frac{600}{9}\right)^{\mathrm{g}}$, $\left(\frac{800}{9}\right)^{\mathrm{g}}$, which are 44.44ᵍ, 66.67ᵍ and 88.89ᵍ. (Or convert a: $\frac{10}{9} \times 40 = 44.44$, and so on.)

$20x = 180^\circ$, $x = 9^\circ$. The angles are 45∘°, 63∘° and 72∘°.

$20x = 200^{\mathrm{g}}$, $x = 10^{\mathrm{g}}$. The angles are 50ᵍ, 70ᵍ and 80ᵍ.

A quadrilateral’s angles add up to 360∘°: $18x = 360^\circ$, $x = 20^\circ$. The angles are 60∘°, 80∘°, 100∘° and 120∘°.

$60^\circ = \frac{10}{9} \times 60 = \left(\frac{200}{3}\right)^{\mathrm{g}} \approx 66.67^{\mathrm{g}}$ and $120^\circ = \left(\frac{400}{3}\right)^{\mathrm{g}} \approx 133.33^{\mathrm{g}}$.
Difference: $\left(\frac{200}{3}\right)^{\mathrm{g}} = \left(\frac{600}{9}\right)^{\mathrm{g}} \approx$ 66.67ᵍ (which is 60∘°, as it should be).

They add up to 180∘° − 72∘° = 108∘°: $x + 3x = 108^\circ$, $x = 27^\circ$. The angles are 27∘° and 81∘°.

$27^\circ = 30^{\mathrm{g}}$, $81^\circ = 90^{\mathrm{g}}$, $72^\circ = 80^{\mathrm{g}}$. So 30ᵍ, 80ᵍ and 90ᵍ (they add up to 200ᵍ ✓).

$x + 5x = 108^\circ$, $x = 18^\circ$. The angles would be 18∘° and 90∘°: the triangle would be right-angled.

$\frac{9}{10} \times 60 =$ 54∘°.

The other two add up to 180∘° − 54∘° = 126∘°: $7x = 126^\circ$, $x = 18^\circ$. The angles are 54∘°, 72∘° and 54∘°.

Two angles are equal (54∘° and 54∘°), so it is an isosceles triangle (the sides opposite those angles are equal). All its angles are less than 90∘°, so it is also acute-angled.

90∘°

$\frac{3}{10} \times 90^\circ =$ 27∘°.

90∘° − 27∘° = 63∘° = $\frac{10}{9} \times 63 =$ 70ᵍ.

$1^\circ = \left(\frac{10}{9}\right)^{\mathrm{g}}$

20ᵍ = 18∘°. x + y = 100, x − y = 18: x = 59, y = 41. The angles are 59∘° and 41∘° (as in Example 7).

$59^\circ = \left(\frac{590}{9}\right)^{\mathrm{g}} \approx$ 65.56ᵍ and $41^\circ = \left(\frac{410}{9}\right)^{\mathrm{g}} \approx$ 45.56ᵍ. Check: their difference is $\frac{180}{9} = 20^{\mathrm{g}}$ ✓.

$1^{\mathrm{g}} = \left(\frac{9}{10}\right)^\circ$

30ᵍ = 27∘°. x + y = 45, x − y = 27: 2x = 72, x = 36, y = 9. The angles are 36∘° and 9∘°.

$36^\circ = 40^{\mathrm{g}}$ and $9^\circ = 10^{\mathrm{g}}$. So 40ᵍ and 10ᵍ (difference 30ᵍ ✓).

πᶜ

πᶜ

180∘°

$\left(\frac{180}{\pi}\right)^\circ \approx$ 57.3∘° (57.27∘° with $\pi = \frac{22}{7}$).

$\left(\frac{\pi}{180}\right)^{\mathrm{c}}$

$\left(\frac{\pi}{200}\right)^{\mathrm{c}}$

One radian is the angle made at the center of a circle by an arc as long as the radius of the circle. It is a constant angle, $\frac{180^\circ}{\pi} \approx 57.3^\circ$.

The circular measure system measures an angle by the arc it cuts off on a circle, in radius-lengths. Its unit is the radian (ᶜ or rad); a complete turn is $2\pi$ radians.

l is the length of the arc and r is the radius of the circle (in the same unit); θ is the central angle the arc makes, in radians.

A complete turn is 360∘° = 400ᵍ = 2πᶜ, a right angle 90∘° = 100ᵍ = $\frac{\pi^{\mathrm{c}}}{2}$, two right angles 180∘° = 200ᵍ = πᶜ. So 180∘° = 200ᵍ = πᶜ, which converts any angle between the three systems.

Central angles are in the same ratio as their arcs: double the arc, double the angle. So the angle in radians is the arc divided by the radius, $\theta = \frac{l}{r}$.

$30 \times \frac{\pi}{180} =$ $\frac{\pi^{\mathrm{c}}}{6}$

$45 \times \frac{\pi}{180} =$ $\frac{\pi^{\mathrm{c}}}{4}$

$50 \times \frac{\pi}{200} =$ $\frac{\pi^{\mathrm{c}}}{4}$ (50ᵍ = 45∘°).

$70 \times \frac{\pi}{200} =$ $\frac{7\pi^{\mathrm{c}}}{20}$

$120 \times \frac{\pi}{180} =$ $\frac{2\pi^{\mathrm{c}}}{3}$

$150 \times \frac{\pi}{200} =$ $\frac{3\pi^{\mathrm{c}}}{4}$

$\frac{180^\circ}{2} =$ 90∘°

$\frac{3 \times 180^\circ}{2} =$ 270∘°

$\frac{7 \times 180^\circ}{50} =$ 25.2∘°

$\frac{3 \times 180^\circ}{4} =$ 135∘°

$\frac{2 \times 180^\circ}{3} =$ 120∘°

$\frac{5 \times 180^\circ}{6} =$ 150∘°

$\frac{4 \times 180^\circ}{9} =$ 80∘°

$\frac{5 \times 180^\circ}{12} =$ 75∘°

$\frac{180^\circ}{9} =$ 20∘°

Put 200ᵍ for $\pi$: $\frac{200}{5} =$ 40ᵍ

$\frac{3 \times 200}{10} =$ 60ᵍ

$\frac{4 \times 200}{25} =$ 32ᵍ

$\frac{200}{4} =$ 50ᵍ

$\frac{200}{8} =$ 25ᵍ

$\frac{3 \times 200}{2} =$ 300ᵍ

∠ABC is the right angle: 90∘°.

$\frac{3}{5} \times \frac{\pi}{2} =$ $\frac{3\pi^{\mathrm{c}}}{10}$ (54∘°).

$0.4 \times 90^\circ = 36^\circ = 36 \times \frac{\pi}{180} =$ $\frac{\pi^{\mathrm{c}}}{5}$.

$180^\circ = \pi^{\mathrm{c}}$

The right angle, 90∘° = $\frac{\pi^{\mathrm{c}}}{2}$, and 180∘° − 90∘° − 60∘° = 30∘° = $\frac{\pi^{\mathrm{c}}}{6}$.

60∘° − 30∘° = 30∘° = $\frac{\pi^{\mathrm{c}}}{6}$ (or $\frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$).

90∘° − 50∘° = 40∘° = $40 \times \frac{\pi}{180} =$ $\frac{2\pi^{\mathrm{c}}}{9}$.

50∘° − 40∘° = 10∘° = $\frac{\pi^{\mathrm{c}}}{18}$.

A straight angle is $\pi^{\mathrm{c}}$, so a third is $\frac{\pi^{\mathrm{c}}}{3}$ (60∘°).

$\frac{3\pi^{\mathrm{c}}}{10} = \frac{3 \times 200}{10} = 60^{\mathrm{g}}$. The acute angles add up to 100ᵍ: x + y = 100, x − y = 60, so x = 80, y = 20.
The angles are 100ᵍ, 80ᵍ and 20ᵍ.

All three are equal and add up to $\pi^{\mathrm{c}}$, so each is $\frac{\pi^{\mathrm{c}}}{3}$.

The hands are 4 spaces apart: $4 \times 30^\circ = 120^\circ =$ $\frac{2\pi^{\mathrm{c}}}{3}$.

The minute hand is at 6 (180∘° from 12). The hour hand is half way from 1 to 2, at $1.5 \times 30^\circ = 45^\circ$. The angle is $180^\circ - 45^\circ = 135^\circ =$ $\frac{3\pi^{\mathrm{c}}}{4}$.

The minute hand is at 9 (270∘°). The hour hand is three quarters of the way from 6 to 7, at $6.75 \times 30^\circ = 202.5^\circ$. The angle is $270^\circ - 202.5^\circ = 67.5^\circ =$ $\frac{3\pi^{\mathrm{c}}}{8}$.

The minute hand is at 6 (180∘°); the hour hand is at $3.5 \times 30^\circ = 105^\circ$. The angle is $75^\circ =$ $\frac{5\pi^{\mathrm{c}}}{12}$.

$\theta = 60^\circ = \frac{\pi^{\mathrm{c}}}{3}$. From l = rθ: $r = \frac{l}{\theta} = \frac{44}{\pi/3} = \frac{132}{\pi} = 132 \times \frac{7}{22} =$ 42 cm.

$81^\circ = \frac{81\pi}{180} = \frac{9\pi^{\mathrm{c}}}{20}$. $r = \frac{18}{9\pi/20} = \frac{40}{\pi} = 40 \times \frac{7}{22} \approx$ 12.73 cm.

$r = \frac{15}{3\pi/4} = \frac{20}{\pi} = 20 \times \frac{7}{22} \approx$ 6.36 cm.

$\theta = \frac{11}{18}$ radians $= \frac{11}{18} \times \frac{180}{\pi} = \frac{110}{\pi} = 110 \times \frac{7}{22} =$ 35∘°.

$\theta = \frac{3.8}{7.254} \approx 0.52385$ radians $= 0.52385 \times \frac{180 \times 7}{22} \approx$ 30∘°. (With the calculator’s $\pi$ it is 30.01∘°: the numbers were chosen for $\pi = \frac{22}{7}$.)

In 20 minutes the minute hand turns $\frac{20}{60}$ of a turn: $\frac{1}{3} \times 2\pi = \frac{2\pi^{\mathrm{c}}}{3}$.
$l = r\theta = 3 \times \frac{2\pi}{3} = 2\pi = 2 \times \frac{22}{7} \approx$ 6.29 cm. (With $\pi = 3.14$, 6.28 cm.)

$\frac{7 \times 180^\circ}{18} =$ 70∘°.

$l = r\theta = 10 \times \frac{7\pi}{18} = 10 \times \frac{7 \times 22}{18 \times 7} = \frac{220}{18} \approx$ 12.22 m.

$70 \times \frac{\pi}{180} =$ $\frac{7\pi^{\mathrm{c}}}{18}$.

$l = 14 \times \frac{7\pi}{18} = 14 \times \frac{7 \times 22}{18 \times 7} = \frac{308}{18} \approx$ 17.11 m.

The rope is the same, so the angle doubles with the arc ($\theta = \frac{l}{r}$): 140∘°. (Check: $\frac{2 \times 17.11}{14} \approx 2.444$ radians $\approx 140^\circ$.)

$5^\circ = \frac{5\pi}{180} = \frac{\pi^{\mathrm{c}}}{36}$. The length is the radius: $r = \frac{l}{\theta} = \frac{20.5}{\pi/36} = \frac{738}{\pi} = 738 \times \frac{7}{22} \approx$ 234.82 cm, about 2.35 m.

36 seconds is $\frac{36}{60} = 0.6$ minute: 100 × 0.6 = 60 m.

$56^\circ = \frac{56\pi}{180} = \frac{14\pi^{\mathrm{c}}}{45}$. $r = \frac{60}{14\pi/45} = \frac{2700}{14\pi} = \frac{2700 \times 7}{14 \times 22} \approx$ 61.36 m.

$2\pi r = 2 \times \frac{22}{7} \times 61.36 \approx$ 385.71 m. (Or: 56∘° takes 60 m, so 360∘° takes $60 \times \frac{360}{56} \approx 385.71$ m.)

Two right angles are $180^\circ = 200^{\mathrm{g}} = \pi^{\mathrm{c}}$. An angle is the same fraction of two right angles in every system. In degrees that fraction is $\frac{D}{180}$, in grades $\frac{G}{200}$, in radians $\frac{C}{\pi}$. So $\frac{D}{180} = \frac{G}{200} = \frac{C}{\pi}$. Proved.
For example 90∘° = 100ᵍ = $\frac{\pi}{2}$: $\frac{90}{180} = \frac{100}{200} = \frac{\pi/2}{\pi} = \frac{1}{2}$.

$\operatorname{cosec}\theta = \frac{1}{\sin\theta}$

$\tan\theta = \frac{1}{\cot\theta}$

$\cos\theta = \frac{1}{\sec\theta}$

$\cot\theta = \frac{1}{\tan\theta}$

$\sin\theta = \frac{1}{\operatorname{cosec}\theta}$

$\sec\theta = \frac{1}{\cos\theta}$

$\cot\theta = \frac{\cos\theta}{\sin\theta}$

$\tan\theta = \frac{\sin\theta}{\cos\theta}$

$\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\cot\theta = \frac{\cos\theta}{\sin\theta}$.

For example: $\sin\theta \times \operatorname{cosec}\theta = 1$; $\tan\theta = \frac{\sin\theta}{\cos\theta}$; $\sin^2\theta + \cos^2\theta = 1$. (Any three of the reciprocal, quotient and Pythagoras relations.)

$\sin^2\theta + \cos^2\theta = 1$, $\sin^2\theta = 1 - \cos^2\theta$, $\cos^2\theta = 1 - \sin^2\theta$;
$\sec^2\theta - \tan^2\theta = 1$, $\sec^2\theta = 1 + \tan^2\theta$, $\tan^2\theta = \sec^2\theta - 1$;
$\operatorname{cosec}^2\theta - \cot^2\theta = 1$, $\operatorname{cosec}^2\theta = 1 + \cot^2\theta$, $\cot^2\theta = \operatorname{cosec}^2\theta - 1$.

4 tan θ

$12\cot A$

$11\sec^2 A$

$6\sin^3 x$

−3 sin θ

5 tan A

$-9\sec^2 A$

$\sin^3 x$

17 sin θ

18 tan A

$20\cos^2 A$

$14\operatorname{cosec}^3 x$

$\sin^2 A - \sin^2 B$

$1 - \cos^2\theta$, which is $\sin^2\theta$.

$1 - \sin^2\theta$, which is $\cos^2\theta$.

$1 + 2\cot^2 A + \cot^4 A$ (which is $\operatorname{cosec}^4 A$).

$(1 - \sin^2\theta)(1 + \sin^2\theta) = 1 - \sin^4\theta$

$(1 - \tan^2\theta)(1 + \tan^2\theta) = 1 - \tan^4\theta$

( cos A +  sin A)( cos A −  sin A)

$(\sec A + \operatorname{cosec} A)(\sec A - \operatorname{cosec} A)$

$\cos^2 A(1 + \sin^2 A)$

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $(\tan\theta - \cot\theta)(\tan^2\theta + \tan\theta\cot\theta + \cot^2\theta)$, and since $\tan\theta\cot\theta = 1$, $= (\tan\theta - \cot\theta)(\tan^2\theta + 1 + \cot^2\theta)$.

$(\sec^2\theta + \operatorname{cosec}^2\theta)(\sec^2\theta - \operatorname{cosec}^2\theta) = (\sec^2\theta + \operatorname{cosec}^2\theta)(\sec\theta + \operatorname{cosec}\theta)(\sec\theta - \operatorname{cosec}\theta)$

Two numbers with product 2 and sum 3: 1 and 2. ( sin x + 1)( sin x + 2).

L.H.S. $= \frac{\cos A}{\sin A} \times \sin A = \cos A =$ R.H.S.

L.H.S. $= \cos A \times \frac{1}{\sin A} = \frac{\cos A}{\sin A} = \cot A =$ R.H.S.

L.H.S. $= \frac{1}{\cos\theta} \times \sin\theta \times \frac{\cos\theta}{\sin\theta} = 1 =$ R.H.S.

L.H.S. $= \frac{\sin\theta}{\cos\theta} \times \cos\theta = \sin\theta =$ R.H.S.

L.H.S. $= \frac{1}{\sec\theta} = \cos\theta =$ R.H.S., since $\sin\theta\operatorname{cosec}\theta = 1$.

L.H.S. $= \frac{1}{\sec\theta\operatorname{cosec}\theta} = \cos\theta \times \sin\theta =$ R.H.S.

L.H.S. $= \sin^2\theta \times \sec^2\theta = \sin^2\theta \times \frac{1}{\cos^2\theta} = \tan^2\theta =$ R.H.S.

L.H.S. $= \operatorname{cosec}^2 A \times \cos^2 A = \frac{\cos^2 A}{\sin^2 A} = \cot^2 A =$ R.H.S.