Circular Measurement Of Angle MCQs with Explanations

$$\eqalign{ & 4\angle A = 3\angle B = 12\angle C \cr & A:B:C = \frac{1}{4}:\frac{1}{3}:\frac{1}{{12}} \cr & A:B:C = 3:4:1 \cr & {\text{Now}} \cr & 3x + 4x + x = 180 \cr & 8x = 180 \cr & x = \frac{{180}}{8} \cr & \angle A = 3x = \frac{{180}}{8} \times 3 = 67.5 \cr} $$

$$\eqalign{ & 5{\text{ unit}} \to 1\,{\text{cm}} \cr & {\text{1 unit}} \to \frac{1}{5}\,{\text{cm}} \cr & \cot R = \frac{{2.4}}{1} = 2.4\,{\text{cm}} \cr} $$

sec 2 θ + tan 2 θ = 7 1 + tan 2 θ + tan 2 θ = 7 2tan 2 θ = 7 - 1 = 6 tan 2 θ = 3 tanθ = √3 = tan 60° $$\eqalign{ & \because {180^ \circ } = \pi {\text{ radian}} \cr & \therefore {60^ \circ } = \frac{\pi }{{180}} \times 60 = \frac{\pi }{3}{\text{ radian}} \cr} $$

$$\eqalign{ & 5\,{\text{unit}} \to 1\,{\text{cm}} \cr & 1\,{\text{unit}} \to \frac{1}{5}\,{\text{cm}} \cr & {\text{EF}} = 2.4\,{\text{cm}} \cr} $$

8 unit → 16 cm 1 unit → 2 cm EF = 30 cm

$$\eqalign{ & {11^ \circ }15' = 11 + \frac{{15}}{{60}} = 11 + \frac{1}{4} = \frac{{{{45}^ \circ }}}{4} \cr & {\text{We know }}\pi \,{\text{radian}} = {180^ \circ } \cr & {1^ \circ } = \left( {\frac{\pi }{{180}}} \right){\text{ radian,}} \cr & \frac{{{{45}^ \circ }}}{4} = \frac{\pi }{{{{180}^ \circ }}} \times \frac{{{{45}^ \circ }}}{4} = \frac{{{\pi ^c}}}{{16}} \cr} $$

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

$$\eqalign{ & \Rightarrow {\text{cosec F}} \times \cot {\text{D}} \cr & = {\text{cosec 4}}{{\text{5}}^ \circ } \times \cot {45^ \circ } \cr & = \sqrt 2 \times 1 \cr & = \sqrt 2 \cr} $$

XY = 0.8 cm 8 unit → 0.8 cm 1 unit → $$\frac{1}{{10}}$$ XZ = 1.7 cm

The sum of a quadrilateral is 360° x + 2x + 4x + 5x = 360° 12x = 360° x = 30° Biggest angle = 5x = 5 × 30° = 150°

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

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

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

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

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

∠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° $$\eqalign{ & \because {180^ \circ } = \pi {\text{ radian}} \cr & \therefore {90^ \circ } = \frac{\pi }{{180}} \times {90^ \circ } = \frac{\pi }{2}{\text{ radian}} \cr} $$

$$\eqalign{ & \sec C + 2 \cr & \frac{{2x}}{{\sqrt 3 x}} + 2 \cr} $$ $$\eqalign{ & \frac{2}{{\sqrt 3 }} + 2 \cr & \frac{{2 + 2\sqrt 3 }}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & \theta = \frac{l}{r}{\text{ radian}} = \frac{{11}}{{14}}{\text{ radian}} \cr & \because \pi {\text{ radian}} = {180^ \circ } \cr & \therefore 1{\text{ radian}} = \frac{{{{180}^ \circ }}}{\pi } \cr & \therefore \frac{{11}}{{14}}{\text{ radian}} = \frac{{180}}{{\frac{{22}}{7}}} \times \frac{{11}}{{14}} \cr & = \frac{{180 \times 11 \times 7}}{{22 \times 14}} \cr & = {45^ \circ } \cr} $$

$$\eqalign{ & \tan {\text{P}} = \frac{4}{3} \cr & 3{\text{ unit}} \to 1.5{\text{ cm}} \cr & 1{\text{ unit}} \to \frac{{1.5}}{3}{\text{ cm}} = .5{\text{ cm}} \cr & {\text{PR}} = 2.5\,{\text{cm}} \cr} $$

XZ - YZ = 2 h - P = 2 . . . . . . . (i) h 2 = (2√6) 2 + P 2 h 2 - P 2 = (2√6) 2 (h - P)(h + P) = 4 × 6 (2)(h + P) = 24 h + P = 12 . . . . . . . (ii) Adding equation (i) and (ii) 2h = 14 h = 7, P = 5 $$\eqalign{ & \sec {\text{X}} + \tan {\text{X}} \cr & = \frac{{\text{h}}}{{{\text{XY}}}} + \frac{{\text{P}}}{{{\text{XY}}}} \cr & = \frac{7}{{2\sqrt 6 }} + \frac{5}{{2\sqrt 6 }} \cr & = \frac{{12}}{{2\sqrt 6 }} \cr & = \frac{6}{{\sqrt 6 }} \cr & = \sqrt 6 \cr} $$

$$\boxed{\tan L = \tan {{30}^ \circ } = \frac{1}{{\sqrt 3 }}}$$

$$\frac{{2\pi }}{9}$$ is the correct answer to “The circular measure of an angle of an isosceles triangle is $$\frac{{5\pi }}{9}$$, circular measure of one of the other angles must be:”. This question is standard in competitive examination syllabi for Circular Measurement Of Angle.