If the Angles of a Triangle are in A.P., Then the Measure of One of the Angles in Radians is Question: If the angles of a triangle are in A.P., then the measure of one of the angles in radians is (a) \(\frac{\pi}{6}\) (b) \(\frac{\pi}{3}\) (c) \(\frac{\pi}{2}\) (d) \(\frac{2\pi}{3}\) Solution Let the three angles of […]
If D, G and R Denote Respectively the Number of Degrees, Grades and Radians in an Angle Then Question: If \(D\), \(G\) and \(R\) denote respectively the number of degrees, grades and radians in an angle, then (a) \(\frac{D}{100} = \frac{G}{90} = \frac{2R}{\pi}\) (b) \(\frac{D}{90} = \frac{G}{100} = \frac{R}{\pi}\) (c) \(\frac{D}{90} = \frac{G}{100} = \frac{2R}{\pi}\)
Find the degree measure of the angle subtended at the center of a circle of radius \(100\) cm by an arc of length \(22\) cm. \((\text{Use } \pi=\frac{22}{7})\) Solution: We know: \[ s=r\theta \] Given: \[ s=22 \text{ cm} \] \[ r=100 \text{ cm} \] Using, \[ \theta=\frac{s}{r} \] \[ \theta=\frac{22}{100} \] \[ \theta=\frac{11}{50} \text{
If the arcs of the same length in two circles subtend angles \(65^\circ\) and \(110^\circ\) at the center, find the ratio of their radii. Solution: We know: \[ s=r\theta \] Since the arc lengths are equal, \[ r_1\theta_1=r_2\theta_2 \] Therefore, \[ \frac{r_1}{r_2}=\frac{\theta_2}{\theta_1} \] Given: \[ \theta_1=65^\circ \] \[ \theta_2=110^\circ \] So, \[ \frac{r_1}{r_2}=\frac{110}{65} \] \[
Find the diameter of the sun in km supposing that it subtends an angle of \(32’\) at the eye of an observer. Given that the distance of the sun is \(91\times10^6\) km. Solution: Distance of the sun: \[ r=91\times10^6 \text{ km} \] Angular diameter: \[ 32′ \] Since, \[ 1^\circ=60′ \] \[ 32’=\frac{32}{60}^\circ \] Convert
Find the distance from the eye at which a coin of \(2\) cm diameter should be held so as to conceal the full moon whose angular diameter is \(31’\). Solution: Diameter of the coin: \[ s=2 \text{ cm} \] Angular diameter of the moon: \[ 31′ \] Since, \[ 1^\circ=60′ \] \[ 31’=\frac{31}{60}^\circ \] Convert
A railway train is travelling on a circular curve of \(1500\) metres radius at the rate of \(66\) km/hr. Through what angle has it turned in \(10\) seconds? Solution: Radius of the curve: \[ r=1500 \text{ m} \] Speed of the train: \[ 66 \text{ km/hr} \] Convert speed into m/s: \[ 66\times\frac{1000}{3600} \] \[
The radius of a circle is \(30\) cm. Find the length of an arc of this circle, if the length of the chord of the arc is \(30\) cm. Solution: Radius of the circle: \[ r=30 \text{ cm} \] Chord length: \[ c=30 \text{ cm} \] Using chord formula: \[ c=2r\sin\frac{\theta}{2} \] Substituting the values:
Find the angle in radians through which a pendulum swings if its length is \(75\) cm and the tip describes an arc of length \(21\) cm. Solution: We know: \[ s=r\theta \] Given: \[ s=21 \text{ cm} \] \[ r=75 \text{ cm} \] Using, \[ \theta=\frac{s}{r} \] \[ \theta=\frac{21}{75} \] \[ \theta=\frac{7}{25} \] Therefore, the
Find the angle in radians through which a pendulum swings if its length is \(75\) cm and the tip describes an arc of length \(15\) cm. Solution: We know: \[ s=r\theta \] Given: \[ s=15 \text{ cm} \] \[ r=75 \text{ cm} \] Using, \[ \theta=\frac{s}{r} \] \[ \theta=\frac{15}{75} \] \[ \theta=\frac{1}{5} \] Therefore, the