MCQ Questions on Circular Measurement of Angle for Competitive Exams

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

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

sec²θ + tan²θ = 7
1 + tan²θ + tan²θ = 7
2tan²θ = 7 - 1 = 6
tan²θ = 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°