Radian is a Constant Angle

Let ABC be a circle whose center is O and radius is r. Let the length of arc AB of the circle be equal to r. Then by the definition of radian.

$ \displaystyle \angle AOB=1\text{ radian}$

Produce AO and let it cut the circle at C. Then AC is the diameter of the circle and arc ABC is equal to half the circumference of the circle and

$ \displaystyle \angle AOB=2\text{ right angle}={{180}^{\circ }}$

By geometry, we know that angles subtended at the center of a circle are proportional to the length of the arc which subtend them

$ \displaystyle {\therefore \dfrac{{\angle AOB}}{{\angle AOC}}=\dfrac{{\text{arc }AB}}{{\text{arc }ABC\text{ }}}}$

$ \displaystyle {\text{or, }\dfrac{{{{1}^{c}}}}{{{{{180}}^{c}}}}=\dfrac{r}{{\dfrac{{2\pi r}}{2}}}}$

$ \displaystyle {[\therefore \text{circumference of the circle}=2\pi r]}$

$ \displaystyle {\therefore 1\text{ radian}=\frac{{180}}{\pi }}$ $ \displaystyle {=\frac{{2\text{ right angle}}}{\pi }=\text{constant}}$

(Since a right angle and $ \displaystyle \pi $ $  \displaystyle {\text{are constant}}$)

To prove that $ \displaystyle \theta =\dfrac{l}{r}$

Let $ \displaystyle \theta $ be the angle in radian subtended by an arc of length $ \displaystyle l$ at the center of a circle of radius $ \displaystyle r$.

Let O be the center of a circle of radius $ \displaystyle r$.

Let angle $ \displaystyle \angle AOB=1\text{ radian}$, then arc $ \displaystyle AB=r$.

Let $ \displaystyle \angle AOC=\theta \text{ radian}$ and arc $ \displaystyle AC=l$

By geometry we know that angle subtended at the center of a circle are proportional to the lengths of the arc which subtend them

$ \displaystyle \begin{array}{l}\therefore \dfrac{{\angle AOC}}{{\angle AOB}}=\dfrac{{\text{arc }AC}}{{\text{arc }AB\text{ }}}\\\text{or}\\\dfrac{\theta }{1}=\dfrac{l}{r}\\\text{Thus, }\theta =\dfrac{l}{r}\end{array}$

Angle Subtended by a Very Small Arc

Let O be the center of a circle whose radius is r and AB be a very small arc of the circle. Let $ \displaystyle \angle AOB=\theta \text{ radian}$.

$ \displaystyle \begin{array}{l}\text{Then }\theta =\dfrac{{\text{arc }AB}}{r}\\=\dfrac{{\text{chord }AB}}{r}(\text{approximtely})\\\because \text{arc }AB\text{ is very small}\\\therefore \text{arc }AB\cong \text{chord }AB\end{array}$

Relation between degree, grades and radian

$ \displaystyle \begin{array}{l}1\text{ right angle}={{90}^{\circ }}={{100}^{g}}\\1\text{ radian}=\dfrac{{{{{180}}^{\circ }}}}{\pi }\text{ or, }{{180}^{\circ }}=\pi \text{ radian}\text{.}\\\therefore {{180}^{\circ }}={{200}^{g}}=\pi \text{ radian}\\\therefore {{1}^{\circ }}=\dfrac{{10}}{9}\text{grades; }{{\text{1}}^{g}}=\dfrac{9}{{10}}\text{degrees}\\{{1}^{\circ }}=\dfrac{\pi }{{180}}\text{ radians; }1\text{ radian}=\dfrac{{180}}{\pi }\text{degrees}\\{{1}^{g}}=\dfrac{\pi }{{200}}\text{ radians; }1\text{ radian}=\dfrac{{200}}{\pi }\text{grades}\end{array}$

Thus, if the measure of an angle in degrees, grades and radians be D, G and C respectively, then

$ \displaystyle \dfrac{D}{{180}}=\dfrac{G}{{200}}=\dfrac{C}{\pi }$

Relation between side and interior angles of a regular polygon

By geometry we know that the sum of interior angles of a triangle $ \displaystyle ={{180}^{\circ }}$

Thus, if n be the number of sides of a polygon and O be a poit inside the polygon then the polygon will be divided into n triangles by joining all the vertices to O.

Now, sum of all the angles of the triangles thus formed

$ \displaystyle =n\times {{180}^{\circ }}……….(i)$

and from figure it is clear that sum of all the angles of the triangles

$ \displaystyle =n\times {{360}^{{}^\circ }}$ $ \displaystyle +$ sum of all the interior angles of the polygon …(ii)

From (i) and (ii), we get

Sum of interior angles of the polygon $ \displaystyle +{{360}^{\circ }}={{180}^{\circ }}\times n$

$ \displaystyle \therefore $ Sum of all the interior angles $ \displaystyle =\left( {2n-4} \right)\times {{90}^{{}^\circ }}$

If the polygon is regular, each interior angle of the polygon

$ \displaystyle =\dfrac{{2n-4}}{n}\times {{90}^{\circ }}$

Example 1:

Express $ \displaystyle {{45}^{\circ }}2{0}’1{0}”$ in

(i) Centesimal measure

(ii) radian measure ($ \displaystyle {{45}^{\circ }}2{0}’1{0}”$)

Solution:

$ \displaystyle \begin{array}{*{20}{l}} \begin{array}{l}1{0}”=\dfrac{{10}}{{60}}\text{minutes }=\dfrac{{10}}{{60\times 60}}\text{degrees }\\=\dfrac{1}{{360}}\text{degrees}\end{array} \\ {2{0}’=\dfrac{{20}}{{60}}\text{degrees }=\dfrac{1}{3}\text{degrees}} \\ {\therefore {{{45}}^{{}^\circ }}2{0}’1{0}”=\left( {45+\frac{1}{{360}}+\dfrac{1}{3}} \right)\text{degrees}} \\ {=\dfrac{{16200+1+120}}{{360}}=\dfrac{{16321}}{{360}}} \end{array}$

(i)

$ \displaystyle \begin{array}{*{20}{l}} \begin{array}{l}{{\left( {\dfrac{{16321}}{{360}}} \right)}^{{}^\circ }}=\dfrac{{16321}}{{360}}\times \dfrac{{10}}{9}\\=\dfrac{{16321}}{{324}}\text{grades}\end{array} \\ \begin{array}{l}=50.3734\text{ grades }\\={{50}^{g}}(.3734\times 100{)}’\end{array} \\ \begin{array}{l}={{50}^{g}}37{{\left( {.34\times 100} \right)}^{{\prime \prime }}}\\={{50}^{g}}3{7}’3{4}”\end{array} \end{array}$

(ii)

$ \displaystyle \begin{array}{*{20}{l}} {{{{\left( {\dfrac{{16321}}{{360}}} \right)}}^{{}^\circ }}=\dfrac{{16321}}{{360}}\times \dfrac{\pi }{{180}}\text{radian}} \\ \begin{array}{l}=\dfrac{{16321}}{{360}}\times \dfrac{{3.1415}}{{180}}\\=\dfrac{{51274.054}}{{64800}}=.79\text{radian}\end{array} \end{array}$

Example 2:

Express 1.2radian in degree measure.

Solution:

$ \displaystyle \begin{array}{l}{{\left( {1.2} \right)}^{\circ }}=1.2\times \dfrac{{180}}{\pi }\text{degrees}\\=1.2\times \dfrac{{180}}{{22}}\left[ {\therefore \pi =\dfrac{{27}}{7}\text{(approx)}} \right]\\=\dfrac{{1.2\times 180\times 7}}{{22}}=68.7272\\={{68}^{\circ }}{{\left( {.7272\times 60} \right)}^{\prime }}={{68}^{\circ }}{{\left( {43.63} \right)}^{\prime }}\\={{68}^{\circ }}4{3}'(.63\times 60{)}”={{68}^{\circ }}4{3}’37.{8}”\end{array}$

Example 3:

The angles of a triangle are in the ratio 3:4:5, find the smallest angle in degree and the greatest angle in radians.

Solution:

Let the three angles be $ \displaystyle 3x,4x\text{ and }5x$ degrees, then

$ \displaystyle \begin{array}{*{20}{l}} {3x+4x+5x={{{180}}^{{}^\circ }}} \\ {x={{{15}}^{{}^\circ }}} \\ {\therefore \text{smallest angle }={{{45}}^{{}^\circ }}} \\ \begin{array}{l}\text{greatest angle }=5x={{75}^{{}^\circ }}\\=75\times \dfrac{\pi }{{{{{180}}^{{}^\circ }}}}=\dfrac{{5\pi }}{{12}}\end{array} \end{array}$

Example 4:

Assuming the distance of the earth from the moon to be 38400 km. and the angle subtended by the moon at the eye of a person on the earth to be 31′, find the diameter of the moon.

Solution:

Let AB be the diameter of the moon and O be the observer.

Given $ \displaystyle \angle AOB=3{1}’=\dfrac{{31}}{{60}}\times \dfrac{\pi }{{180}}\text{radians}$

Since, angle subtended by the moon is very small, therefore, its diameter will be approximately equal to a small arc of a circle whose center is the eye of the observer and radius id the distance of the earth from the moon. Also the moon subtends an angle of 31′ at the center of this circle.

$ \displaystyle \begin{array}{l}\because \theta =\dfrac{l}{r}\\\therefore \dfrac{{31}}{{60}}\times \dfrac{\pi }{{180}}=\dfrac{{AB}}{{38400}}\\\therefore AB=\dfrac{{31}}{{60}}\times \dfrac{{22}}{{7\times 180}}\times 38400\\=3464\dfrac{8}{{63}}\text{ km}.\end{array}$