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The measure of angle in levels goes back to antiquity. It may have arisen from the idea the there were approximately 360 work in a year, or it might have emerged from the Babylonian penchant for base 60 numerals. In any event, both the Greeks and also the Indians split the edge in a circle right into 360 equal parts, which we now contact degrees. They further separated each degree into 60 same parts referred to as minutes and also divided every minute right into 60 seconds. An example would it is in $$15^\circ22"16""$$. This method of measuring angle is really inconvenient and also it to be realised in the 16th century (or even before) that it is better to measure angles via arc length.

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We specify one radian, composed as $$1^c$$ (where the $$c$$ refers to circular measure), to it is in the edge subtended at the centre of a unit one by a unit arc length on the circumference.

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

So one radian is equal to $$\dfrac180\pi$$ degrees, i m sorry is about $$57.3^\circ$$.

Since numerous angles in levels can be expressed as straightforward fractions of 180, we use $$\pi$$ as a simple unit in radians and often express angle as fractions of $$\pi$$. The commonly arising angles $$30^\circ$$, $$45^\circ$$ and $$60^\circ$$ are 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. Us will frequently leave turn off the $$c$$, particularly when the angle is expressed in regards to $$\pi$$.

Students should have a deal of practice in finding the trigonometric attributes of angles expressed in radians. Because students are an ext familiar through degrees, it is often ideal 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. Girlfriend can enter radians directly into her calculator to evaluate a trigonometric duty at an edge in radians, but you need to make certain your calculator is in radian mode. Students have to be reminded to check what setting their calculator is in when they room doing difficulties involving the trigonometric functions.Arc lengths, sectors and also segments

Measuring angles in radians allows us to create down quite simple formulas for the arc length of part of a circle and also the area that a ar of a circle.

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

Thus, measure the angles in radians,

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

It must be stressed again that, to use this formula, we call for the angle to it is in in radians.

Example

In a one of radius 12 cm, discover the size of an arc subtending an edge of $$60^\circ$$ at 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 finest to leave her answer in regards to $$\pi$$ uneven otherwise stated.

We use the same ratio idea to obtain the area the a ar in a circle of radius $$r$$ comprise an edge $$\theta$$ at the centre. The proportion of the area $$A$$ the the ar to the complete area of the circle equates to the proportion of the angle in the sector to one revolution.

Thus, with angles measure 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 circle of radius 36 cm, find the area that a ar subtending an edge of $$30^\circ$$ at the centre.

Solution

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

\

As mentioned above, in difficulties such together these the is ideal to leave her answer in regards to $$\pi$$ unless otherwise stated.

The area $$A_s$$ of a segment that a one is easily found by acquisition the distinction of two areas.

In a one of radius $$r$$, think about a segment that subtends an edge $$\theta$$ at the centre. We can discover the area the the segment by individually the area the the triangle (using $$\dfrac12ab\,\sin\theta$$) native the area the the sector. Thus

\
Example

Find the area the the segment shown.

Solution

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

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Exercise 15

Around a one of radius $$r$$, draw an inner and outer hexagon as displayed in the diagram.

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

Summary the arc, sector and also segment formulasLength of arcArea that sectorArea that segment
 $$\ell = r\theta$$ $$A=\dfrac12r^2\theta$$ $$A=\dfrac12r^2(\theta-\sin\theta)$$
Solve each 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 $$x$$ might lie in the 2nd or third quadrant, for this reason $$x = 180^\circ - 60^\circ$$ or $$x = 180^\circ + 60^\circ$$. Therefore $$x = 120^\circ$$ or $$x = 240^\circ$$. In radians, the solutions are $$x = \dfrac2\pi3, \dfrac4\pi3$$.Since $$\sin\dfrac\pi2 = 1$$, the connected angle is $$\dfrac\pi2$$. Due to the fact that $$x$$ lies between 0 and $$2\pi$$, it complies with 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 replace $$\cos^2 x$$ v $$1-\sin^2 x$$ to obtain 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 determinants to $$(2\sin x -1)^2=0$$ and so $$\sin x = \dfrac12$$. In the given range, this has actually solutions $$x= 30^\circ, 150^\circ$$. So, in radians, the remedies are $$x = \dfrac\pi6, \dfrac5\pi6$$.