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The measure up of angles in degrees goes ago to antiquity. It may have emerged from the idea the there were roughly 360 work in a year, or it may have occurred from the Babylonian penchant because that base 60 numerals. In any kind of event, both the Greeks and also the Indians split the angle in a circle right into 360 equal parts, which us now call degrees. Castle further split each level into 60 equal parts referred to as minutes and divided every minute into 60 seconds. An example would be $$15^\circ22"16""$$. This means of measuring angle is an extremely inconvenient and also it to be realised in the 16th century (or even before) that it is far better to measure angle via arc length.

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We define one radian, created as $$1^c$$ (where the $$c$$ refers to circular measure), to be the edge subtended at the centre of a unit circle by a unit arc size on the circumference.

Since the complete circumference the a unit circle is $$2\pi$$ units, we have actually the counter formula

So one radian is equal to $$\dfrac180\pi$$ degrees, which is roughly $$57.3^\circ$$.

Since countless angles in degrees can be expressed as basic fractions that 180, we usage $$\pi$$ together a basic unit in radians and also often express angles as fractions of $$\pi$$. The commonly occurring angles $$30^\circ$$, $$45^\circ$$ and $$60^\circ$$ room expressed in radians respectively together $$\dfrac\pi6$$, $$\dfrac\pi4$$ and also $$\dfrac\pi3$$.

Example

$$135^\circ$$$$270^\circ$$$$100^\circ$$.Solution$$135^\circ = 3\times 45^\circ = \dfrac3\pi4^c$$$$270^\circ = 3\times 90^\circ = \dfrac3\pi2^c$$$$100^\circ = \dfrac100 \pi180^c = \dfrac5\pi9^c$$
Note. We will often leave turn off the $$c$$, specifically when the angle is express in terms of $$\pi$$.

Students should have a deal of exercise in recognize the trigonometric features of angle expressed in radians. Since students are much more familiar through degrees, the is often best to convert ago to degrees.

Example

Find

$$\cos \dfrac4\pi3$$$$\sin \dfrac7\pi6$$$$\tan \dfrac5\pi4$$.Solution$$\cos \dfrac4\pi3 = \cos 240^\circ = -\cos 60^\circ = -\dfrac12$$$$\sin \dfrac7\pi6 = \sin 210^\circ = -\sin 30^\circ = -\dfrac12$$$$\tan \dfrac5\pi4 = \tan 225^\circ = \tan 45^\circ = 1$$
Note. Friend can go into radians straight into your calculator to evaluate a trigonometric duty at an angle in radians, but you need to make sure your calculator is in radian mode. Students have to be reminded to inspect what setting their calculator is in when they space doing difficulties involving the trigonometric functions.Arc lengths, sectors and also segments

Measuring angles in radians enables us to create down quite simple formulas because that the arc size of part of a circle and also the area of a sector of a circle.

In any kind of circle of radius $$r$$, the proportion of the arc size $$\ell$$ come the circumference amounts to the ratio of the angle $$\theta$$ subtended through the arc at the centre and the edge in one revolution.

Thus, measure up the angle in radians,

\beginalignat*2 & & \dfrac\ell2\pi r &= \dfrac\theta2\pi\\ &\implies\ & \ell &= r\theta.\endalignat*

It have to be emphasize again that, to use this formula, we call for the angle to be in radians.

Example

In a one of radius 12 cm, discover the size of an arc subtending an edge of $$60^\circ$$ in ~ the centre.

Solution

With $$r=12$$ and $$\theta = 60^\circ = \dfrac\pi3$$, us have

\<\ell = 12\times \dfrac\pi3 = 4\pi \approx 12.57\text cm.\>

It is often ideal to leave her answer in regards to $$\pi$$ uneven otherwise stated.

We usage the same ratio idea to attain the area the a sector in a one of radius $$r$$ include an angle $$\theta$$ in ~ the centre. The ratio of the area $$A$$ the the ar to the full area of the circle equates to the ratio of the edge in the sector to one revolution.

Thus, v angles measured in radians,

\beginalignat*2& & \dfracA\pi r^2 &= \dfrac\theta2\pi\\&\implies\ & A &= \dfrac12 r^2\theta.\endalignat*

The arc length and sector area recipe given above should be cursed to memory.

Example

In a one of radius 36 cm, find the area that a sector subtending an angle of $$30^\circ$$ in ~ the centre.

Solution

With $$r=36$$ and $$\theta = 30^\circ = \dfrac\pi6$$, we have

\

As stated above, in problems such as these the is ideal to leave your answer in terms of $$\pi$$ unless otherwise stated.

The area $$A_s$$ the a segment of a circle is easily uncovered by taking the difference of 2 areas.

In a circle of radius $$r$$, take into consideration a segment that subtends an angle $$\theta$$ in ~ the centre. We can uncover the area that the segment by subtracting the area the the triangle (using $$\dfrac12ab\,\sin\theta$$) from the area that the sector. Thus

\
Example

Find the area of the segment shown.

Solution

With $$r=4$$ and also $$\theta = 18^\circ = \dfrac18\pi180 = \dfrac\pi10$$, we have

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Exercise 15

Around a circle of radius $$r$$, attract an inner and also outer hexagon as displayed in the diagram.

By considering the perimeters the the 2 hexagons, display that $$3 \leq \pi \leq 2\sqrt3$$.

Summary the arc, sector and also segment formulasLength the arcArea that sectorArea that segment
 $$\ell = r\theta$$ $$A=\dfrac12r^2\theta$$ $$A=\dfrac12r^2(\theta-\sin\theta)$$
Solve every equation because that $$0 \leq x \leq 2\pi$$:
$$\cos x = -\dfrac12$$$$\sin 3x = 1$$$$4\cos^2 x + 4\sin x =5$$.SolutionSince $$\cos 60^\circ = \dfrac12$$, the related angle is $$60^\circ$$. The angle $$x$$ can lie in the 2nd or third quadrant, therefore $$x = 180^\circ - 60^\circ$$ or $$x = 180^\circ + 60^\circ$$. Because of this $$x = 120^\circ$$ or $$x = 240^\circ$$. In radians, the remedies are $$x = \dfrac2\pi3, \dfrac4\pi3$$.Since $$\sin\dfrac\pi2 = 1$$, the related angle is $$\dfrac\pi2$$. Due to the fact that $$x$$ lies in between 0 and also $$2\pi$$, it adheres to that $$3x$$ lies in between 0 and $$6\pi$$. Thus\beginalign*3x &= \dfrac\pi2,\ 2\pi + \dfrac\pi2,\ 4\pi + \dfrac\pi2\\ &= \dfrac\pi2,\ \dfrac5\pi2,\ \dfrac9\pi2.\endalign*Hence $$x= \dfrac\pi6,\ \dfrac5\pi6,\ \dfrac3\pi2$$.In this case, we change $$\cos^2 x$$ through $$1-\sin^2 x$$ to achieve the quadratic\beginalignat*2 & & 4(1-\sin^2 x) + 4\sin x &= 5\\ &\implies\ & 4\sin^2 x - 4\sin x +1 &= 0.\endalignat*This factors to $$(2\sin x -1)^2=0$$ and also so $$\sin x = \dfrac12$$. In the provided range, this has actually solutions $$x= 30^\circ, 150^\circ$$. So, in radians, the solutions are $$x = \dfrac\pi6, \dfrac5\pi6$$.