MCQ Questions on Circular Measurement of Angle for Competitive Exams

$$\eqalign{ & {180^ \circ } = \pi \,{\text{radian}} \cr & 1{\text{ radian}} = \frac{{{{180}^ \circ }}}{\pi } \cr & \frac{{{{180}^ \circ }}}{\pi } \Rightarrow \frac{{180 \times 7}}{{22}} \Rightarrow \frac{{630}}{{11}} \Rightarrow 57\frac{{{3^ \circ }}}{{11}} \cr & = \frac{1}{{\cot }} = {57^ \circ }\frac{{180'}}{{11}} \cr & = {57^ \circ }16'\frac{{4'}}{{11}} \cr & = {57^ \circ }16'\left( {\frac{4}{{11}} \times 60''} \right) \cr & = {57^ \circ }16'22'' \cr} $$

$$\eqalign{ & {\left[ {\frac{{3\pi }}{5}} \right]^c} \Rightarrow \pi = {180^ \circ } \cr & {\left( {\frac{{3\pi }}{5}} \right)^c} = {\left( {\frac{{3 \times 180}}{5}} \right)^ \circ } = {108^ \circ } \cr} $$

$$\eqalign{ & \sin \left( {\theta + 18} \right) = \frac{1}{2} \cr & \sin \left( {\theta + 18} \right) = \sin 30 \cr & \theta + 18 = 30 \cr & \therefore \theta = 12 \cr & {\text{We know that}} \cr & {180^ \circ } = \pi \cr & {12^ \circ } = \frac{\pi }{{180}} \times 12 = \frac{\pi }{{15}} \cr} $$

$$\eqalign{ & {\text{cosec X}} - \frac{2}{{\sqrt 3 }} \cr & = \frac{{\sqrt 2 x}}{x} - \frac{2}{{\sqrt 3 }} \cr & = \sqrt 2 - \frac{2}{{\sqrt 3 }} \cr & = \frac{{\sqrt 6 - 2}}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & PA = AR \cr & \angle APB = \angle ARB = {45^ \circ } \cr & {\text{If }}PR = \sqrt 2 x;\,\,PB = \frac{x}{{\sqrt 2 }} \cr & {\text{From }}\Delta APB \cr & \tan {45^ \circ } = \frac{{AB}}{{PB}} \cr & AB = PB = \frac{x}{{\sqrt 2 }} \cr & \therefore PA = \sqrt {\frac{{{x^2}}}{2} + \frac{{{x^2}}}{2}} = \sqrt {{x^2}} = x \cr & \therefore QA = PQ = PA = x \cr & \therefore \angle PAQ = {60^ \circ } \cr} $$

∠A + ∠B = 135° . . . . . . . . (i)
∠A - ∠B = $$\frac{\pi }{{12}}$$ = 15° . . . . . . . . (ii)
Adding both equation
2∠A = 150°
⇒ ∠A = 75°
∠B = 60°

$$\eqalign{ & \frac{{{\pi ^c}}}{4} = \frac{{{{180}^ \circ }}}{4} = {45^ \circ } \cr & \angle BAC = {180^ \circ } - {75^ \circ } - {45^ \circ } = {60^ \circ } \cr & {180^ \circ } \to \pi \cr & {1^ \circ } \to \frac{\pi }{{{{180}^ \circ }}} \cr & {60^ \circ } \to \frac{\pi }{{{{180}^ \circ }}} \times {60^ \circ } = \frac{\pi }{3}{\text{ radian}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \angle A = {30^ \circ } \cr & \therefore \angle C = {60^ \circ } \cr & \left( {\tan {{60}^ \circ } = \frac{{\sqrt 3 }}{1} = \frac{{AB}}{{BC}}} \right) \cr & AB = \sqrt 3 ,\,BC = 1 \cr & \therefore AC = \sqrt {A{B^2} + B{C^2}} \cr & = \sqrt {\left( {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2}} \right)} \cr & = 2\,{\text{unit}} \cr & 2\,{\text{unit}} = 10\,{\text{cm}} \cr & 1\,{\text{unit}} = 5{\text{ cm}} \cr & \boxed{AB = \sqrt 3 {\text{ unit}} = 5\sqrt 3 } \cr} $$

$$\eqalign{ & \tan U = \frac{{VW}}{{UV}} \cr & \frac{{12}}{5} = \frac{{VW}}{{10}} \cr & VW = \frac{{12}}{5} \times 10 = 24{\text{ cm}} \cr} $$

The hour hand of a watch traces an angle of 30° in an hour.
∴ Angle traced at 3 O' clock = 3 × 30° = 90°
$$\eqalign{ & \because {180^ \circ } = \pi {\text{ radian}} \cr & \therefore {90^ \circ } = \frac{\pi }{{180}} \times {90^ \circ } = \frac{\pi }{2}{\text{ radian}} \cr} $$