Ravi Kant Kumar

The angles of a triangle are in A.P. and the number of degrees in the least angle is to the number of degrees in the mean angle as \(1:2\). Find the angles in radians. Solution: Let the three angles in A.P. be: \[ a-d,\ a,\ a+d \] Given, \[ \frac{a-d}{a}=\frac{1}{2} \] \[ 2a-2d=a \] \[

The angles of a quadrilateral are in A.P. and the greatest angle is \(120^\circ\). Express the angles in radians. Solution: Let the four angles in A.P. be: \[ a-3d,\ a-d,\ a+d,\ a+3d \] Given greatest angle: \[ a+3d=120^\circ \] Sum of angles of a quadrilateral: \[ 360^\circ \] So, \[ (a-3d)+(a-d)+(a+d)+(a+3d)=360^\circ \] \[ 4a=360^\circ \]

Find the magnitude, in radians and degrees, of the interior angle of a regular duodecagon. Solution: A regular duodecagon has \(12\) sides. Interior angle of a regular polygon: \[ \frac{(n-2)\times180^\circ}{n} \] Substituting \(n=12\): \[ \frac{(12-2)\times180^\circ}{12} \] \[ \frac{10\times180^\circ}{12} \] \[ 150^\circ \] Now convert into radians: \[ 150^\circ \times \frac{\pi}{180^\circ} \] \[ \frac{5\pi}{6} \] Therefore,

Find the magnitude, in radians and degrees, of the interior angle of a regular heptagon. Solution: A regular heptagon has \(7\) sides. Interior angle of a regular polygon: \[ \frac{(n-2)\times180^\circ}{n} \] Substituting \(n=7\): \[ \frac{(7-2)\times180^\circ}{7} \] \[ \frac{900^\circ}{7} \] \[ 128\frac{4}{7}^\circ \] Now convert into radians: \[ \frac{900^\circ}{7}\times\frac{\pi}{180^\circ} \] \[ \frac{5\pi}{7} \] Therefore, the interior

Find the magnitude, in radians and degrees, of the interior angle of a regular octagon. Solution: A regular octagon has \(8\) sides. Interior angle of a regular polygon: \[ \frac{(n-2)\times180^\circ}{n} \] Substituting \(n=8\): \[ \frac{(8-2)\times180^\circ}{8} \] \[ \frac{6\times180^\circ}{8} \] \[ 135^\circ \] Now convert into radians: \[ 135^\circ \times \frac{\pi}{180^\circ} \] \[ \frac{3\pi}{4} \] Therefore,

Find the magnitude, in radians and degrees, of the interior angle of a regular pentagon. Solution: A regular pentagon has \(5\) sides. Interior angle of a regular polygon: \[ \frac{(n-2)\times180^\circ}{n} \] Substituting \(n=5\): \[ \frac{(5-2)\times180^\circ}{5} \] \[ \frac{3\times180^\circ}{5} \] \[ 108^\circ \] Now convert into radians: \[ 108^\circ \times \frac{\pi}{180^\circ} \] \[ \frac{3\pi}{5} \] Therefore,

One angle of a triangle is \( \frac{2x}{3} \) grade and another is \( \frac{3x}{2}^\circ \) while the third is \( \frac{\pi x}{75} \) radians. Express all the angles in degrees. Solution: We know: \[ 1 \text{ grade} = \frac{9^\circ}{10} \] and \[ 1 \text{ radian} = \frac{180^\circ}{\pi} \] First angle: \[ \frac{2x}{3} \text{ grade}

The difference between the two acute angles of a right-angled triangle is \( \frac{2\pi}{5} \) radians. Express the angles in degrees. Solution: In a right-angled triangle, the sum of the two acute angles is \(90^\circ\). Let the acute angles be \(A\) and \(B\). \[ A + B = 90^\circ \] Given, \[ A – B

Find the Radian Measure Corresponding to the Following Degree Measure : \( -47^\circ 30′ \) First convert minutes into degrees: \[ 30′ = \frac{30}{60}^\circ = \frac{1}{2}^\circ \] Therefore, \[ -47^\circ 30′ = -47.5^\circ = -\frac{95}{2}^\circ \] To convert a degree measure into radians, we use the formula: \[ \text{Radian} = \text{Degree} \times \frac{\pi}{180^\circ} \] Conversion

Find the Radian Measure Corresponding to the Following Degree Measure : \( 125^\circ 30′ \) First convert minutes into degrees: \[ 30′ = \frac{30}{60}^\circ = \frac{1}{2}^\circ \] Therefore, \[ 125^\circ 30′ = 125.5^\circ = \frac{251}{2}^\circ \] To convert a degree measure into radians, we use the formula: \[ \text{Radian} = \text{Degree} \times \frac{\pi}{180^\circ} \] Conversion