CBSE Class 11 Mathematics Trigonometry Assignment Set 14

XI - Assignment No 7 - Trigonometric Functions

(Relation between degree and radian)

Question 1. Find the radian measure corresponding to \(-37^{\circ}30'\) and find the degree measure corresponding to \(\frac{5\pi}{3}\) radian.
Answer: To convert degrees to radians, use the relation:
Radian measure = \(\frac{\pi}{180^{\circ}} \times \text{Degree measure}\)
For \(-37^{\circ}30'\):
\(30' = \left(\frac{30}{60}\right)^{\circ} = 0.5^{\circ}\)
So, \(-37^{\circ}30' = -37.5^{\circ} = -\frac{75}{2}^{\circ}\)
Radian measure = \(\frac{\pi}{180} \times \left(-\frac{75}{2}\right) = -\frac{5\pi}{24}\)

To convert radians to degrees, use the relation:
Degree measure = \(\frac{180}{\pi} \times \text{Radian measure}\)
For \(\frac{5\pi}{3}\) radian:
Degree measure = \(\frac{180}{\pi} \times \frac{5\pi}{3} = \frac{180 \times 5}{3} = 300^{\circ}\) (Note: The provided answer key states \(210^{\circ}\) as a typo in the source paper, but the correct calculation for \(\frac{5\pi}{3}\) is \(300^{\circ}\)).
In simple words: Use the conversion formulas between degrees and radians to get the required values.

Exam Tip: Remember that \(1^{\circ} = 60'\) and \(1' = 60''\) when converting minutes to decimal degrees.

Question 2. The minute hand of a watch is \(1.5\text{ cm}\) long. How far does its tip move in \(50\) minutes? (Use \(\pi = 3.14\))
Answer: Length of the minute hand \(r = 1.5\text{ cm}\).
In 60 minutes, the minute hand completes 1 full revolution, turning an angle of \(2\pi\) radians.
Angle swept in 1 minute = \(\frac{2\pi}{60} = \frac{\pi}{30}\) radians.
Angle swept in 50 minutes (\(\theta\)) = \(50 \times \frac{\pi}{30} = \frac{5\pi}{3}\) radians.
Distance moved by the tip (\(l\)) = \(r \theta\)
\(l = 1.5 \times \frac{5 \times 3.14}{3} = 0.5 \times 5 \times 3.14 = 7.85\text{ cm}\).
In simple words: Find the total angle swept in radians and multiply by the radius using arc length formula \(l = r\theta\).

Exam Tip: Always ensure the angle is expressed in radians before applying the arc length formula \(l = r\theta\).

Question 3. If the arcs of the same length in two circles subtend angles \(75^{\circ}\) and \(120^{\circ}\) at the centre, find the ratio of their radii.
Answer: Let the lengths of the arcs be equal to \(l\).
Let the radii of the two circles be \(r_1\) and \(r_2\).
Let the angles subtended be \(\theta_1 = 75^{\circ}\) and \(\theta_2 = 120^{\circ}\).
Convert angles to radians:
\(\theta_1 = 75 \times \frac{\pi}{180} = \frac{5\pi}{12}\) radians
\(\theta_2 = 120 \times \frac{\pi}{180} = \frac{2\pi}{3}\) radians
Since \(l = r_1 \theta_1 = r_2 \theta_2\):
\(r_1 \left(\frac{5\pi}{12}\right) = r_2 \left(\frac{2\pi}{3}\right)\)
\(\frac{r_1}{r_2} = \frac{2\pi}{3} \times \frac{12}{5\pi} = \frac{8}{5}\)
Therefore, the ratio of their radii is \(8:5\).
In simple words: Since arc lengths are equal, equate \(r_1\theta_1 = r_2\theta_2\) to find the radius ratio.

Exam Tip: Ratios do not depend on whether angles are in degrees or radians as long as both are in the same unit, but radians are standard for arc length formulas.

Question 4. In a circle of diameter \(40\text{ cm}\), the length of a chord is \(20\text{ cm}\). Find the length of minor arc of the chord.
Answer: Diameter = \(40\text{ cm}\), so radius \(r = 20\text{ cm}\).
Length of chord \(AB = 20\text{ cm}\).
In triangle \(OAB\) (where \(O\) is the centre), \(OA = OB = r = 20\text{ cm}\) and \(AB = 20\text{ cm}\).
Since all three sides are equal, triangle \(OAB\) is an equilateral triangle.
Therefore, the angle subtended by the chord at the centre is \(\theta = 60^{\circ} = \frac{\pi}{3}\) radians.
Length of minor arc \(l = r \theta\)
\(l = 20 \times \frac{\pi}{3} = \frac{20\pi}{3}\text{ cm}\).
In simple words: Form an equilateral triangle with the radii and chord to find the central angle, then use \(l = r\theta\).

Exam Tip: Recognize standard geometric shapes (like equilateral triangles) inside circles to easily determine central angles.

Question 5. Express in circular measure and also in degrees the angle of a regular polygon of \(40\) sides.
Answer: Hint given: Angle of a polygon = \(180^{\circ} - \frac{360^{\circ}}{\text{no. of sides}}\)
Number of sides \(n = 40\)
Interior angle in degrees = \(180^{\circ} - \frac{360^{\circ}}{40} = 180^{\circ} - 9^{\circ} = 171^{\circ}\)
To convert to circular measure (radians):
Radian measure = \(171 \times \frac{\pi}{180} = \frac{19\pi}{20}\) radians.
In simple words: Calculate the interior angle in degrees using the formula, then convert it to radians.

Exam Tip: Memorize the interior angle formula for regular polygons: \(180^{\circ} \cdot (n-2)/n\) or the exterior angle method \(360^{\circ}/n\).

Question 6. Express in radians the fourth angle of a quadrilateral which has three angles of \(46^{\circ}30'10''\), \(75^{\circ}44'45''\), \(123^{\circ}9'35''\) respectively.
Answer: Sum of all four angles of a quadrilateral = \(360^{\circ}\).
First angle = \(46^{\circ} 30' 10''\)
Second angle = \(75^{\circ} 44' 45''\)
Third angle = \(123^{\circ} 9' 35''\)
Let us add the seconds first:
\(10'' + 45'' + 35'' = 90'' = 1' 30''\)
Add the minutes including the carry over \(1'\):
\(30' + 44' + 9' + 1' = 84' = 1^{\circ} 24'\)
Add the degrees including the carry over \(1^{\circ}\):
\(46^{\circ} + 75^{\circ} + 123^{\circ} + 1^{\circ} = 245^{\circ}\)
So the sum of the three angles = \(245^{\circ} 24' 30''\).
Fourth angle = \(360^{\circ} - 245^{\circ} 24' 30''\)
\(= 359^{\circ} 59' 60'' - 245^{\circ} 24' 30''\)
\(= (359 - 245)^{\circ} + (59 - 24)' + (60 - 30)''\)
\(= 114^{\circ} 35' 30''\)
Convert to decimal degrees:
\(30'' = \left(\frac{30}{60}\right)' = 0.5'\)
\(35.5' = \left(\frac{35.5}{60}\right)^{\circ} \approx 0.59167^{\circ}\)
Degree measure = \(114.59167^{\circ}\)
In radians = \(114.59167 \times \frac{\pi}{180} \approx \frac{13751\pi}{21600}\) radians (or matching answer key form \(\frac{13751}{...} \pi\)).
In simple words: Subtract the sum of the three given angles from \(360^{\circ}\), then convert the resulting degree measure into radians.

Exam Tip: When subtracting angles with degrees, minutes, and seconds, borrow \(1^{\circ}\) as \(60'\) and \(1'\) as \(60''\).

Question 7. The moon's distance from the Earth is \(240,000\text{ km}\) and its diameter subtends an angle of \(31'\) at the eye of the observer. Find the diameter of the moon. [use \(\theta = \frac{l}{r}\), given \(r, \theta\), find \(l\)]
Answer: Distance \(r = 240,000\text{ km}\).
Angle \(\theta = 31' = \frac{31}{60}^{\circ}\).
Convert angle to radians:
\(\theta = \frac{31}{60} \times \frac{\pi}{180} = \frac{31\pi}{10800}\) radians.
Using \(l = r \theta\):
Diameter \(l = 240,000 \times \frac{31\pi}{10800}\)
\(l = \frac{2400 \times 31 \times \frac{22}{7}}{108} = \frac{2400 \times 682}{756} = \frac{1636800}{756} \approx 2165 \frac{5}{63}\text{ km}\).
In simple words: Use the arc length formula \(l = r\theta\) by treating the moon's diameter as arc length \(l\) and distance as radius \(r\).

Exam Tip: Astronomical problems use the formula \(l = r\theta\) where the chord/diameter is approximated as the arc length because the angle is very small.

Question 8. At what distance does a man, whose height is \(6\text{ feet}\) subtend an angle of \(10'\)? [Find \(r\), given \(l = 6\text{ feet}, \theta = 10'\)]
Answer: Height \(l = 6\text{ feet}\).
Angle \(\theta = 10' = \left(\frac{10}{60}\right)^{\circ} = \left(\frac{1}{6}\right)^{\circ}\).
Convert angle to radians:
\(\theta = \frac{1}{6} \times \frac{\pi}{180} = \frac{\pi}{1080}\) radians.
Using \(l = r \theta\), we need to find distance \(r\):
\(r = \frac{l}{\theta} = \frac{6}{\frac{\pi}{1080}} = \frac{6 \times 1080}{\pi} = \frac{6480}{\frac{22}{7}} = \frac{45360}{22} = \frac{22680}{11} \approx 2061 \frac{9}{11}\text{ feet}\).
In simple words: Rearrange \(l = r\theta\) to \(r = \frac{l}{\theta}\) after converting the angle into radians.

Exam Tip: Keep units consistent; since height is in feet, the calculated distance will also be in feet.

Question 9. A horse is tethered to a stake by a rope \(5\text{ m}\) long. If the horse moves along the circumference of a circle always keeping the rope tight, how far it will have gone when the rope has traced out an angle of \(70^{\circ}\)? [Find \(l\), given \(r = 5\text{ m}, \theta = 70^{\circ}\)]
Answer: Radius \(r = 5\text{ m}\).
Angle \(\theta = 70^{\circ} = 70 \times \frac{\pi}{180} = \frac{7\pi}{18}\) radians.
Distance moved \(l = r \theta\)
\(l = 5 \times \frac{7 \times \frac{22}{7}}{18} = \frac{5 \times 22}{18} = \frac{110}{18} = \frac{55}{9} \approx 6.11\text{ m}\).
In simple words: Multiply the rope length (radius) by the angle in radians to find the distance traveled along the circular path.

Exam Tip: This is a direct application of the arc length formula where the rope length acts as the radius.

Question 10. Show that the minute hand of a watch gains \(5^{\circ}30'\) on the hour hand in a minute.
Answer: Hint: Minute hand covers \(6^{\circ}\) in one minute. Hour hand covers \(\frac{1}{2}^{\circ}\) in one minute.
Angle covered by minute hand in 1 minute = \(\frac{360^{\circ}}{60} = 6^{\circ}\).
Angle covered by hour hand in 1 minute = \(\frac{360^{\circ}}{12 \times 60} = \frac{360^{\circ}}{720} = \left(\frac{1}{2}\right)^{\circ} = 0.5^{\circ}\).
Gain of minute hand over hour hand in 1 minute = \(6^{\circ} - \left(\frac{1}{2}\right)^{\circ} = 5.5^{\circ} = 5^{\circ}30'\).
Hence proved.
In simple words: Subtract the speed of the hour hand per minute from the speed of the minute hand per minute.

Exam Tip: Relative speed concepts in clock problems apply directly to angular speeds in trigonometry.