Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.2

Question 1. Express each of the following in radian measure.
(i) 30°
(ii) 135°
(iii) -205°
(iv) 150°
(v) 330°
Answer: We know that \( 180^\circ = \pi \) radians. So, to convert degrees to radians, we multiply by \( \frac{\pi}{180} \).
(i) For \( 30^\circ \):
\( 180^\circ = \pi \) radians
\( 1^\circ = \frac{\pi}{180} \) radians
\( 30^\circ = \frac{\pi}{180} \times 30 \) radians
\( \implies 30^\circ = \frac{\pi}{6} \) radians
(ii) For \( 135^\circ \):
\( 180^\circ = \pi \) radians
\( 1^\circ = \frac{\pi}{180} \) radians
\( 135^\circ = \frac{\pi}{180} \times 135 \) radians
\( \implies 135^\circ = \frac{3\pi}{4} \) radians
(iii) For \( -205^\circ \):
\( 180^\circ = \pi \) radians
\( 1^\circ = \frac{\pi}{180} \) radians
\( -205^\circ = \frac{\pi}{180} \times (-205) \) radians
\( \implies -205^\circ = -\frac{41\pi}{36} \) radians
(iv) For \( 150^\circ \):
\( 180^\circ = \pi \) radians
\( 1^\circ = \frac{\pi}{180} \) radians
\( 150^\circ = \frac{\pi}{180} \times 150 \) radians
\( \implies 150^\circ = \frac{5\pi}{6} \) radians
(v) For \( 330^\circ \):
\( 180^\circ = \pi \) radians
\( 1^\circ = \frac{\pi}{180} \) radians
\( 330^\circ = \frac{\pi}{180} \times 330 \) radians
\( \implies 330^\circ = \frac{11\pi}{6} \) radians
In simple words: To change an angle from degrees to radians, multiply the degree value by \( \frac{\pi}{180} \). Then, simplify the fraction to get the radian measure.

🎯 Exam Tip: Remember the conversion factor: multiply by \( \frac{\pi}{180^\circ} \) to go from degrees to radians, and by \( \frac{180^\circ}{\pi} \) to go from radians to degrees.

Question 2. Find the degree measure corresponding to the following radian measures.
(i) \( \frac{\pi}{3} \)
(ii) \( \frac{\pi}{9} \)
(iii) \( \frac{2 \pi}{5} \)
(iv) \( \frac{7 \pi}{3} \)
(v) \( \frac{10 \pi}{9} \)
Answer: We know that \( \pi \) radians \( = 180^\circ \). So, to convert radians to degrees, we multiply by \( \frac{180}{\pi} \).
(i) For \( \frac{\pi}{3} \) radians:
We know \( \pi \) radians \( = 180^\circ \)
So, \( 1 \) radian \( = (\frac{180}{\pi})^\circ \)
\( \implies \frac{\pi}{3} \) radians \( = (\frac{180}{\pi} \times \frac{\pi}{3})^\circ \)
\( \implies \frac{\pi}{3} \) radians \( = 60^\circ \)
(ii) For \( \frac{\pi}{9} \) radians:
We know \( \pi \) radians \( = 180^\circ \)
So, \( 1 \) radian \( = (\frac{180}{\pi})^\circ \)
\( \implies \frac{\pi}{9} \) radians \( = (\frac{180}{\pi} \times \frac{\pi}{9})^\circ \)
\( \implies \frac{\pi}{9} \) radians \( = 20^\circ \)
(iii) For \( \frac{2 \pi}{5} \) radians:
We know \( \pi \) radians \( = 180^\circ \)
So, \( 1 \) radian \( = (\frac{180}{\pi})^\circ \)
\( \implies \frac{2 \pi}{5} \) radians \( = (\frac{180}{\pi} \times \frac{2 \pi}{5})^\circ \)
\( \implies \frac{2 \pi}{5} \) radians \( = 72^\circ \)
(iv) For \( \frac{7 \pi}{3} \) radians:
We know \( \pi \) radians \( = 180^\circ \)
So, \( 1 \) radian \( = (\frac{180}{\pi})^\circ \)
\( \implies \frac{7 \pi}{3} \) radians \( = (\frac{180}{\pi} \times \frac{7 \pi}{3})^\circ \)
\( \implies \frac{7 \pi}{3} \) radians \( = 420^\circ \)
(v) For \( \frac{10 \pi}{9} \) radians:
We know \( \pi \) radians \( = 180^\circ \)
So, \( 1 \) radian \( = (\frac{180}{\pi})^\circ \)
\( \implies \frac{10 \pi}{9} \) radians \( = (\frac{180}{\pi} \times \frac{10 \pi}{9})^\circ \)
\( \implies \frac{10 \pi}{9} \) radians \( = 200^\circ \)
In simple words: To change an angle from radians to degrees, multiply the radian value by \( \frac{180}{\pi} \). This will convert the angle into its degree equivalent.

🎯 Exam Tip: Always remember that \( \pi \) radians is equal to \( 180^\circ \). This fundamental relationship is key to all radian-degree conversions.

Question 3. What must be the radius of a circular running path, around which an athlete must run 5 times in order to describe 1 km?
Answer: Let the radius of the circular path be \( r \) meters.
The total length of the circular path covered, \( s = 1 \) km.
We convert 1 km to meters: \( s = 1000 \) m.
The athlete runs 5 times around the path to cover this distance.
The total angle covered in 5 runs is \( \theta \). Since one full circle is \( 360^\circ \) or \( 2\pi \) radians:
\( \theta = 5 \times 360^\circ \)
First, convert this angle to radians:
\( \theta = 5 \times 360^\circ \times \frac{\pi}{180^\circ} \) radians
\( \implies \theta = 5 \times 2 \pi \) radians
\( \implies \theta = 10\pi \) radians
We know the relationship between arc length \( s \), radius \( r \), and angle \( \theta \) (in radians):
\( s = r\theta \)
Substitute the known values:
\( 1000 = r \times 10\pi \)
Now, solve for \( r \):
\( r = \frac{1000}{10\pi} \)
\( \implies r = \frac{100}{\pi} \)
Using the approximation \( \pi \approx \frac{22}{7} \):
\( r = \frac{100}{22/7} \)
\( \implies r = \frac{100 \times 7}{22} \)
\( \implies r = \frac{700}{22} \)
\( \implies r = \frac{350}{11} \)
\( \implies r \approx 31.818 \) meters
The radius of the circular path is approximately \( 31.82 \) meters.
In simple words: The athlete runs a total distance of 1000 meters. Since they run 5 times around the circular path, the total angle covered is \( 5 \times 2\pi \) radians. We use the formula "distance = radius × angle" to find the radius of the path.

🎯 Exam Tip: Always ensure the angle is in radians when using the formula \( s = r\theta \). Carefully convert units, like kilometers to meters, before calculation.

Question 4. In a circle of diameter 40 cm a chord is of length 20 cm. Find the length of the minor arc of the chord.
Answer:Given the diameter \( AB = 40 \) cm.
Therefore, the radius \( r = \frac{40}{2} = 20 \) cm.
The length of the chord \( CD = 20 \) cm.
Let \( O \) be the center of the circle.
The lines \( OC \) and \( OD \) are radii, so \( OC = OD = r = 20 \) cm.
Since \( OC = OD = CD = 20 \) cm, the triangle \( OCD \) is an equilateral triangle.
In an equilateral triangle, all angles are \( 60^\circ \). So, the angle subtended by the chord at the center is \( \angle COD = 60^\circ \).
To find the length of the minor arc \( CD \), we first convert the angle from degrees to radians:
\( \theta = 60^\circ \times \frac{\pi}{180^\circ} \) radians
\( \implies \theta = \frac{\pi}{3} \) radians
Now, we use the formula for arc length \( s = r\theta \):
\( s = 20 \times \frac{\pi}{3} \) cm
\( s = \frac{20\pi}{3} \) cm
Using the approximation \( \pi \approx 3.14 \):
\( s = \frac{20 \times 3.14}{3} \)
\( s = \frac{62.8}{3} \)
\( s \approx 20.93 \) cm
The length of the minor arc is approximately \( 20.93 \) cm.
In simple words: We found that the triangle formed by the two radii and the chord is equilateral, which means the angle at the center is 60 degrees. We changed this angle to radians and then used the formula "arc length = radius × angle" to find the length of the arc.

🎯 Exam Tip: When given a chord length equal to the radius, immediately recognize that an equilateral triangle is formed, leading to a central angle of \( 60^\circ \).

Question 5. Find the degree measure of the angle subtended at the centre of the circle of radius 100 cm by an arc of length 22 cm.
Answer: Given values:
Radius of the circle, \( r = 100 \) cm.
Length of the arc, \( s = 22 \) cm.
We need to find the angle \( \theta \) subtended by the arc at the center. The formula relating these is \( s = r\theta \), where \( \theta \) must be in radians.
Rearranging the formula to solve for \( \theta \):
\( \theta = \frac{s}{r} \)
Substitute the given values:
\( \theta = \frac{22}{100} \) radians
\( \implies \theta = \frac{11}{50} \) radians
Now, we convert this radian measure to degrees. We know that \( \pi \) radians \( = 180^\circ \), so \( 1 \) radian \( = (\frac{180}{\pi})^\circ \).
\( \theta = \frac{22}{100} \times \frac{180^\circ}{\pi} \)
Using the approximation \( \pi \approx \frac{22}{7} \):
\( \theta = \frac{22}{100} \times \frac{180^\circ}{(22/7)} \)
\( \implies \theta = \frac{22}{100} \times \frac{180 \times 7}{22} \)
\( \implies \theta = \frac{180 \times 7}{100} \)
\( \implies \theta = \frac{1260}{100} \)
\( \implies \theta = 12.6^\circ \)
To express this in degrees and minutes:
\( 0.6^\circ = 0.6 \times 60' = 36' \)
So, \( \theta = 12^\circ 36' \).
In simple words: We used the formula "angle = arc length / radius" to find the angle in radians first. Then, we changed this radian angle into degrees, knowing that pi radians is 180 degrees.

🎯 Exam Tip: Remember to express the final angle in the required units (degrees, minutes, or radians) and use the correct conversion factor for \( \pi \), usually \( \frac{22}{7} \) or \( 3.14 \).

Question 6. What is the length of the arc intercepted by a central angle of measure 41° in a circle of radius of 10 feet?
Answer: Given values:
Central angle, \( \theta = 41^\circ \).
Radius of the circle, \( r = 10 \) feet.
To find the length of the arc \( s \), we use the formula \( s = r\theta \). However, the angle \( \theta \) must be in radians.
First, convert the angle from degrees to radians:
\( \theta = 41^\circ \times \frac{\pi}{180^\circ} \) radians
\( \implies \theta = \frac{41\pi}{180} \) radians
Now, substitute \( r \) and \( \theta \) into the arc length formula:
\( s = 10 \times \frac{41\pi}{180} \)
\( \implies s = \frac{410\pi}{180} \)
\( \implies s = \frac{41\pi}{18} \)
Using the approximation \( \pi \approx \frac{22}{7} \):
\( s = \frac{41}{18} \times \frac{22}{7} \)
\( \implies s = \frac{41 \times 11}{9 \times 7} \) (after cancelling 2 from 18 and 22)
\( \implies s = \frac{451}{63} \)
\( s \approx 7.1587 \) feet
So, the length of the arc is approximately \( 7.16 \) feet.
In simple words: First, we changed the given angle from degrees into radians. Then, we used the formula "arc length = radius × angle" to calculate the length of the curved part of the circle.

🎯 Exam Tip: Always convert angles to radians before using the arc length formula \( s = r\theta \) or area of sector formula \( A = \frac{1}{2}r^2\theta \).

Question 7. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Answer: Let \( r_1 \) and \( r_2 \) be the radii of the two circles. Let \( l \) be the length of the arc, which is the same for both circles.
For the first circle:
Central angle \( \theta_1 = 60^\circ \)
Convert \( \theta_1 \) to radians: \( \theta_1 = 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3} \) radians.
The arc length \( l = r_1\theta_1 \implies l = r_1 \frac{\pi}{3} \) (Equation 1)
For the second circle:
Central angle \( \theta_2 = 75^\circ \)
Convert \( \theta_2 \) to radians: \( \theta_2 = 75^\circ \times \frac{\pi}{180^\circ} = \frac{5\pi}{12} \) radians.
The arc length \( l = r_2\theta_2 \implies l = r_2 \frac{5\pi}{12} \) (Equation 2)
Since the arc lengths are the same for both circles, we can equate Equation 1 and Equation 2:
\( r_1 \frac{\pi}{3} = r_2 \frac{5\pi}{12} \)
To find the ratio \( r_1 : r_2 \), we rearrange the equation:
\( \frac{r_1}{r_2} = \frac{5\pi/12}{\pi/3} \)
\( \implies \frac{r_1}{r_2} = \frac{5\pi}{12} \times \frac{3}{\pi} \)
\( \implies \frac{r_1}{r_2} = \frac{5 \times 3}{12} \)
\( \implies \frac{r_1}{r_2} = \frac{15}{12} \)
\( \implies \frac{r_1}{r_2} = \frac{5}{4} \)
So, the ratio of their radii is \( r_1 : r_2 = 5 : 4 \).
In simple words: We know the formula "arc length = radius × angle". Since the arc lengths are equal for both circles, we set their formulas equal to each other. After converting the angles to radians, we solved for the ratio of the two radii.

🎯 Exam Tip: When dealing with arc lengths and angles, always ensure angles are in radians. Equating expressions for common quantities (like arc length here) is a common strategy in such problems.

Question 8. The perimeter of a certain sector of a circle is equal to the length of the arc of a semi-circle having the same radius. Express the angle of the sector in degrees, minutes, and seconds.
Answer:Let the radius of the sector and the semi-circle be \( r \).
Let the angle of the sector be \( \theta \) radians.
The perimeter of a sector \( OAB \) is the sum of its two radii and the arc length \( AB \).
Perimeter of sector \( = OA + OB + \text{arc } AB \)
\( = r + r + r\theta \)
\( = 2r + r\theta \)
\( = r(2 + \theta) \) (Equation 1)
The length of the arc of a semi-circle is half the circumference of a full circle with the same radius.
Length of arc of semi-circle \( = \frac{1}{2} \times 2\pi r = \pi r \) (Equation 2)
According to the problem, the perimeter of the sector is equal to the length of the arc of the semi-circle:
\( r(2 + \theta) = \pi r \)
Since \( r \) is a radius, \( r \neq 0 \), so we can divide both sides by \( r \):
\( 2 + \theta = \pi \)
\( \implies \theta = \pi - 2 \) radians
Now, we need to convert this angle from radians to degrees, minutes, and seconds.
We use the conversion factor \( 1 \) radian \( = (\frac{180}{\pi})^\circ \).
\( \theta = (\pi - 2) \times \frac{180^\circ}{\pi} \)
\( \implies \theta = (1 - \frac{2}{\pi}) \times 180^\circ \)
Using the approximation \( \pi \approx \frac{22}{7} \):
\( \theta = (1 - \frac{2}{22/7}) \times 180^\circ \)
\( \implies \theta = (1 - \frac{14}{22}) \times 180^\circ \)
\( \implies \theta = (1 - \frac{7}{11}) \times 180^\circ \)
\( \implies \theta = (\frac{11 - 7}{11}) \times 180^\circ \)
\( \implies \theta = \frac{4}{11} \times 180^\circ \)
\( \implies \theta = \frac{720}{11}^\circ \)
Now, convert this fraction to degrees, minutes, and seconds:
\( \frac{720}{11} \approx 65.4545^\circ \)
First, find the whole degrees: \( 65^\circ \).
Remaining fraction: \( 0.4545^\circ \).
Convert to minutes: \( 0.4545 \times 60' \approx 27.27' \)
First, find the whole minutes: \( 27' \).
Remaining fraction: \( 0.27' \).
Convert to seconds: \( 0.27 \times 60'' \approx 16.2'' \)
So, \( \theta \approx 65^\circ 27' 16'' \).
In simple words: We set the perimeter of the sector equal to the arc length of a semi-circle. This helped us find the angle of the sector in radians. Then, we changed this radian angle into degrees, minutes, and seconds.

🎯 Exam Tip: Be careful with unit conversions from radians to degrees, minutes, and seconds. Remember that \( 1^\circ = 60' \) and \( 1' = 60'' \).

Question 9. An airplane propeller rotates 1000 times per minute. Find the number of degrees that a point on the edge of the propeller will rotate in 1 second.
Answer: Given that the airplane propeller rotates 1000 times per minute.
A point on the edge of the propeller also completes 1000 rotations in 1 minute.
One full rotation corresponds to an angle of \( 2\pi \) radians or \( 360^\circ \).
Angle rotated in 1 minute (60 seconds):
Total angle \( = 1000 \times 2\pi \) radians \( = 2000\pi \) radians.
Now, we need to find the angle rotated in 1 second:
Angle in 1 second \( = \frac{2000\pi}{60} \) radians
\( \implies = \frac{100\pi}{3} \) radians
Now, convert this angle from radians to degrees. We know that \( 1 \) radian \( = (\frac{180}{\pi})^\circ \).
Angle in degrees \( = \frac{100\pi}{3} \times \frac{180^\circ}{\pi} \)
\( \implies = \frac{100 \times 180}{3}^\circ \)
\( \implies = 100 \times 60^\circ \)
\( \implies = 6000^\circ \)
A point on the edge of the propeller will rotate \( 6000^\circ \) in 1 second.
In simple words: First, we found the total angle the propeller turns in one minute by multiplying the number of rotations by \( 2\pi \) radians. Then, we divided this by 60 to find the angle turned in one second. Finally, we changed this angle from radians to degrees.

🎯 Exam Tip: Always pay attention to the units given (e.g., rotations per minute) and the required units for the answer (e.g., degrees per second). Unit conversion is a crucial first step.

Question 10. A train is moving on a circular track of a 1500 m radius at the rate of 66 km/hr. What angle will it turn in 20 seconds?
Answer: Given values:
Radius of the circular track, \( r = 1500 \) m.
Speed of the train, \( v = 66 \) km/hr.
Time, \( t = 20 \) seconds.
First, convert the speed from km/hr to m/s:
\( v = 66 \text{ km/hr} = 66 \times \frac{1000 \text{ m}}{3600 \text{ s}} \)
\( \implies v = 66 \times \frac{5}{18} \text{ m/s} \)
\( \implies v = \frac{11 \times 5}{3} \text{ m/s} = \frac{55}{3} \text{ m/s} \)
Next, calculate the distance \( s \) the train travels in 20 seconds:
\( s = v \times t \)
\( \implies s = \frac{55}{3} \text{ m/s} \times 20 \text{ s} \)
\( \implies s = \frac{1100}{3} \) m
Now, we use the arc length formula \( s = r\theta \) to find the angle \( \theta \) (in radians) turned by the train.
\( \theta = \frac{s}{r} \)
\( \implies \theta = \frac{1100/3}{1500} \)
\( \implies \theta = \frac{1100}{3 \times 1500} \)
\( \implies \theta = \frac{11}{3 \times 15} \)
\( \implies \theta = \frac{11}{45} \) radians
If the question requires the answer in degrees (which it implies by "What angle"), we convert radians to degrees:
\( \theta = \frac{11}{45} \times \frac{180^\circ}{\pi} \)
Using \( \pi \approx \frac{22}{7} \):
\( \theta = \frac{11}{45} \times \frac{180 \times 7}{22}^\circ \)
\( \implies \theta = \frac{11 \times 180 \times 7}{45 \times 22}^\circ \)
\( \implies \theta = \frac{1 \times 4 \times 7}{1 \times 2}^\circ \) (after cancelling 11 from 11 and 22, and 45 from 45 and 180)
\( \implies \theta = \frac{28}{2}^\circ \)
\( \implies \theta = 14^\circ \)
The train will turn an angle of \( 14^\circ \) in 20 seconds.
In simple words: First, we converted the train's speed to meters per second and calculated the distance it travels in 20 seconds. Then, using the distance and the track's radius, we found the angle turned in radians, and finally converted it to degrees.

🎯 Exam Tip: Be very careful with units! Ensure all measurements (distance, speed, time, radius) are in consistent units (e.g., meters and seconds) before applying formulas. Convert the final angle to the required unit.

Question 11. A circular metallic plate of radius 8 cm and thickness 6 nuns is melted and molded into a pie (a sector of the circle with thickness) of radius 16 cm and thickness 4 mm. Find the angle of the sector.
Answer: This problem involves equating volumes before and after melting and molding.
**Original Metallic Plate (Cylinder):**
Radius \( r_1 = 8 \) cm
Thickness \( h_1 = 6 \) mm \( = 0.6 \) cm (since 1 cm = 10 mm)
Volume of a cylinder \( V_1 = \pi r_1^2 h_1 \)
\( V_1 = \pi (8)^2 (0.6) = \pi \times 64 \times 0.6 = 38.4\pi \) cubic cm.

**Molded Pie (Sector with thickness):**
Radius \( r_2 = 16 \) cm
Thickness \( h_2 = 4 \) mm \( = 0.4 \) cm
Let the angle of the sector be \( \theta \) degrees. The volume of a sector with thickness is given by \( V_2 = \frac{\theta}{360^\circ} \times \pi r_2^2 h_2 \).
Substitute the known values:
\( V_2 = \frac{\theta}{360} \times \pi (16)^2 (0.4) \)
\( V_2 = \frac{\theta}{360} \times \pi \times 256 \times 0.4 \)
\( V_2 = \frac{\theta}{360} \times \pi \times 102.4 \)
Since the plate is melted and molded, the volume remains the same:
\( V_1 = V_2 \)
\( 38.4\pi = \frac{\theta}{360} \times 102.4\pi \)
We can cancel \( \pi \) from both sides:
\( 38.4 = \frac{\theta}{360} \times 102.4 \)
Now, solve for \( \theta \):
\( \theta = \frac{38.4 \times 360}{102.4} \)
\( \implies \theta = \frac{13824}{102.4} \)
\( \implies \theta = 135^\circ \)
The angle of the sector is \( 135^\circ \).
Alternatively, if the solution converts to radians for the angle: \( \theta \) in radians is \( 135 \times \frac{\pi}{180} = \frac{3\pi}{4} \) radians.
In simple words: We calculated the volume of the original circular plate (which is like a flat cylinder) and the volume of the new pie shape (which is a sector with thickness). Since melting doesn't change the amount of material, these two volumes must be equal. By setting them equal, we found the angle of the pie sector.

🎯 Exam Tip: When dealing with melting and recasting problems, the key principle is the conservation of volume. Always ensure all dimensions are in consistent units (e.g., all in cm) before calculating volumes.