Trigonometric Functions

11. Kindly go through the solution

10. Kindly go through the solution

9. Kindly go through the solution

8. Kindly go through the solution

7. Here,

r= length of pendulum.

r= 75 cm.

(i) Arc of length, l = 10 cm

Ø= $\frac{l}{r}$ = $\frac{10cm}{75cm}=\frac{2}{15}$ radian.

(ii) Arc of length, l = 15 cm.

So, Ø= $\frac{l}{r}$ = $\frac{15cm}{75cm}=\frac{1}{5}$ radian.

(iii) Arc for length, l= 21 cm.

So, Ø= $\frac{l}{r}=\frac{21cm}{75cm}=\frac{7}{25}$ radian.

6. Let r₁ and r₂ be the radii of two circles.

Then using relation

So, radius, r = $\frac{40}{2}$ cm = 20 cm

Length of chord (AB) = 20cm

In OAB

OA = OB=AB=20 cm

Hence, AOAB is equilateral triangle and end of the angle is 60°

:. Ø =60° = 60 *
$\frac{\mathrm{}\mathrm{\pi }}{180}$ radian =$\frac{\mathrm{\pi }}{3}$ radian

Hence, length of minor are of the chord, l=rØ.

l = 20 * $\frac{\mathrm{\pi }}{3}$ cm

l = $\frac{2\mathrm{0\pi }}{3}$ cm.

4. Here l = 22cm.

r =100cm.

Ø =?

Hence by r = $\frac{1}{\mathrm{Ø}}$

= Ø = $\frac{l}{r}$ = $\frac{22}{100}$ radian

= $\frac{22}{100}$ * $\mathrm{180°}$/π

$\frac{\mathrm{}\mathrm{}}{}$$\frac{}{}$= $\frac{22}{100}$ * 180° * $\frac{7}{22}$

= ${\left(\frac{63}{5}\right)}^0$

=12 $\frac{3°}{5}$ = 12° $\frac{3*60\text{'}}{5}=12°36\text{'}$

3. Given that a wheel makes 360 revolutions in one minute

Then, number of revolutions in one second = $\frac{360}{60}$ =6.

In 1 complete revolution the wheel turns 360°= 2π radian.

So, In 6 revolution, the wheel will turns 6*2π radian = 12π radian.

Hence, in one second the wheel will turn an angle of 12π radian.