The minimum number of points required to draw a line is 

1

(a)

1

(b)

2

(c)

3

(d)

4

In how many points two intersecting lines intersect?

1

(a)

4

(b)

3

(c)

2

(d)

1

The sum of two complimentary angles is 

1

(a)

\(180^{ 0 }\)

(b)

\(360^{ 0 }\)

(c)

\(90^{ 0 }\)

(d)

None of these

The angles whose sum is \(180^{ 0 }\) are called 

1

(a)

supplementary angles

(b)

complementary angles

(c)

alternate angles

(d)

corresponding angles

The sum of two supplementary angles  is

1

(a)

\(90^{ 0 }\)

(b)

\(180^{ 0 }\)

(c)

\(360^{ 0 }\)

(d)

None of these

In the figure below, AOB is a straight line. Calculate the measure of \(\angle COD\).

1

(a)

\(\because\) x+20^o+2x-20^o+60^o=180^o (\(\because\) Straight line makes an angle of 180^o)
\(\Rightarrow\) 3x=180^o-60^o=120^o
\(\Rightarrow\) x=40^o
Thus, \(\angle COD\)=2x-20^o=80^o-20^o=60^o

Write the angle which is one-fifth of its complement.

1

(a)

Let the angle be x, then
By given condition, \(x=\frac{1}{5}(90^o-x)\Rightarrow6x=90^0\)
\(\Rightarrow x=15^0\)

In the given fig., AB and CD are parallel to each other, then calculate the value of x/

1

(a)

Let, l||AB||CD (by construction)

then, \(\angle OEF+\angle BOE=180^o\)
\(\angle O'EF+\angle DOE=180^o\)[Corresponding interior angles]
\(\angle OEF+\angle O'EF+120^o+140^o\)
=360^o
\(\Rightarrow \angle OEO'=360^o-260^o\)
x=100^o

In the figure below, if x,y and z are exterior angles of \(\triangle ABC\), then calculate the value of x+y+z.

1

(a)

Since sum of all the exterior angles formed by producing the sides of a polygon is 360^o
x^o+y^o+z^o=360^o

In \(\triangle ABC,\angle A=\angle B/2=\angle C/6,\) then what will be the measurement of \(\angle A?\)

1

(a)

In \(\triangle ABC,\)
\(\angle A+\angle B+\angle C=180^o\)(By given conditions)
\(\Rightarrow \angle A+2\angle A+6\angle A=180^o\)
\(\Rightarrow 9\angle A=180^o\)
\(\Rightarrow \angle A=20^o\)