**Find the radian measures corresponding to the following degree**

**(i) 25˚**

**(ii) -47˚30’**

**(iii) 240˚**

**(iv) 520˚**

Solution:

(i) 25˚

We know that 180˚ = π radian

25˚ = ^{π}/_{180 }x 25radian = ^{5π}/_{36} radians

(ii) -47˚30’

-47˚30’ = -47^{1}/_{2} = ^{-95}/_{2} degree

^{-95}/_{2} degree = ^{π}/_{180 }x ^{-95}/_{2} radians = (^{-19}/_{36×2}) π radian = ^{-19}/_{72} π radians

Therefore -47˚30` = ^{-19}/_{72} π radian

(iii) 240˚

We know that 180˚ = π radians

Therefore 240˚ = ^{π}/_{180} x 240 radian = ^{4}/_{3} π radian

(iv) 520˚

We know that 180˚ = π radian

Therefore 520˚ = ^{π}/_{180} x 520 radian = ^{26π}/_{9} radian

**Find the degree measures corresponding to the following radian measures**

**(use π = ^{22}/_{7})**

**(i) ^{11}/_{16}**

**(ii) -4**

**(iii) ^{5π}/_{3}**

**(iv) ^{7π}/_{6}**

Solution:

(i) ^{11}/_{16}

We know that π radian = 180˚

^{11}/_{16} radian = ^{180}/_{π} x ^{11}/_{16 }degree = ^{54×11}/_{πx4} degree

= ^{54x11x7}/_{22×4} degree = ^{315}/_{8} degree

= 39^{ 3}/_{8} degree

= 39˚ + 22’ + ^{1}/_{2} minutes

= 39˚22’30”

(ii) -4

We know that π radian = 180˚

-4 radian = ^{180}/_{π} x (-4) degree = ^{180×7(-4)}/_{22} degree

= ^{-2520}/_{11} degree = -229^{1}/_{11} degree

=-229˚+ ^{1×60}/_{11} minutes [1˚ = 60’]

= -229˚+5’+^{5}/_{11} minutes

= -229˚5’27”

(iii)^{5π}/_{3}

We know that π radian = 180˚

^{5π}/_{3} radian = ^{180}/_{π} x ^{5π}/_{3} degree = 300˚

(iv) ^{7π}/_{6}

We know that π radian = 180˚

Therefore 7π/6 radian = ^{180}/_{π} x ^{7π}/_{6} = 210˚

**A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?**

Solution:

Number of revolutions made by the wheel in 1 minute = 360

Number of revolutions made by the wheel in 1 second = ^{360}/_{60} = 6

In one complete revolution, the wheel turns an angle of 2π radian.

Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., 12 π radian

Thus, in one second, the wheel turns an angle of 12π radian.

4**: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm.**

**(π = ^{22}/_{7})**

Solution:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then

Therefore, r = 100 cm, l = 22 cm,

we have

θ = ^{22}/_{100} radian = ^{180}/_{π} x ^{22}/_{100} degree = ^{180x7x22}/_{22×100} degree

= ^{126}/_{10 }degree = 12^{3}/_{5} degree = 12˚36’

Thus, the required angle is 12°36′.

**In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.**

Solution:

Diameter of the circle = 40 cm

Radius (r) of the circle = ^{40}/_{2} cm = 20 cm

Let AB be a chord (length = 20 cm) of the circle.

In ∆OAB,

OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, ∆OAB is an equilateral triangle.

θ = 60° = ^{π}/_{3} radians.

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

θ = ^{1}/_{r}

^{π}/_{3} = ^{AB}/_{20}

thus, AB = ^{20 π}/_{3} cm

Thus, the length of the minor arc of the chord is ^{20 π}/_{3} cm

**If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.**

Solution:

Let the radii of the two circles be r_{1} and r_{2}.

Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r_{1}, while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r_{2}.

Now, 60°= ^{π}/_{3 }radian and 75° = ^{5}^{π}/_{12} radian

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

Therefore, *l = ^{r1}*

^{π}/

_{3}and l = (

^{r}

_{2}

^{5}

^{π})/

_{12}

⇒ ^{r1}^{π}/_{3} = (^{r}_{2}^{5} ^{π})/_{12}

⇒ r_{1}= (^{r}_{2}^{5} ^{π})/_{4}

⇒ ^{r1}/_{r2} = ^{5}/_{4}

Thus the ratio of the radii is 5:4

**Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length**

**(i) 10 cm**

**(ii) 15 cm**

**(iii) 21 cm**

Solution:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = ^{1}/_{r}

It is given that r = 75 cm

(i) Here, l = 10 cm

θ = ^{10}/_{75} radian = ^{2}/_{15} radian

(ii) Here, l = 15 cm

θ = ^{15}/_{75} radian = ^{2}/_{5} radian

(iii) Here, l = 21 cm

θ = ^{21}/_{75} radian = ^{7}/_{25} radian

## 1 thought on “Trigonometry Functions – Class XI -Exercise 3.1”

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