Free Lines and Angles 03 Practice Test - 9th Grade 

The sum of all the three angles of a triangle = 180${}^{\circ }$

Since the angles are in the ratio 5:3:7, let x be a factor such that,

$5x+3x+7x$ =180${}^{\circ }$

$\Rightarrow$ $15x$ =180${}^{\circ }$

$\Rightarrow$ $x$ =12${}^{\circ }$

Hence the three angles of the triangle are:

$\Rightarrow$ $5x$ =$5\times {12}^{\circ }$ =${60}^{\circ }$

$\Rightarrow$ $3x$ =$3\times {12}^{\circ }$ =${36}^{\circ }$

$\Rightarrow$ $7x$ =$7\times {12}^{\circ }$ =${84}^{\circ }$

Since all the angles in the triangle are less than ${90}^{\circ }$, the triangle is an acute angled triangle.

Let the two angles be $x$ and $y$.

So, $x+y={95}^{\circ }$

$\Rightarrow x={95}^{\circ }-y$ $\cdot \cdot \cdot$ (i)

and, $x-y={25}^{\circ }$ $\cdot \cdot \cdot$ (ii)

Substitute (i) in (ii),

$\Rightarrow {95}^{\circ }-y-y={25}^{\circ }$

$\Rightarrow {95}^{\circ }-2y={25}^{\circ }$

$\Rightarrow y=\frac{{95}^{\circ }-{25}^{\circ }}{2}$

$\Rightarrow y={35}^{\circ }$

So, $x={95}^{\circ }-{35}^{\circ }$

$\Rightarrow$ $x={60}^{\circ }$

Third angle is ${180}^{\circ }-{95}^{\circ }={85}^{\circ }$

So, the angles are ${35}^{\circ }$, ${60}^{\circ }$ and ${85}^{\circ }$.

From the properties of triangles:

Exterior angle = Sum of interior opposite angles

Now let us take each interior opposite angle as x ( as both interior opposite angles are equal)

⇒ x + x = 105${}^{\circ }$

⇒ 2x = 105${}^{\circ }$

⇒ x = 52.5${}^{\circ }$

So the value of each of the opposite interior angle =  52.5${}^{\circ }$

In a triangle, if two sides are equal then it is an isosceles triangle. The angles opposite to the two equal sides are also equal.

Therefore, $\angle A=\angle B$

Since sum of all the angles in a triangle $={180}^{\circ }$

$\Rightarrow$ $\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow$ $2\times {40}^{\circ }+\angle C={180}^{\circ }$                                      ($\angle A=\angle B$)

$\Rightarrow$ $\angle C$$={180}^{\circ }-2\times {40}^{\circ }$

$\Rightarrow$ $\angle C$$={100}^{\circ }$

Since AB is a straight line, $\angle AOB={180}^{\circ }$

Therefore, $\angle AOC+\angle COB={180}^{\circ }$

$\Rightarrow 3x+{30}^{\circ }+2x-{60}^{\circ }={180}^{\circ }$

$\Rightarrow 5x-{30}^{\circ }={180}^{\circ }$

$\Rightarrow 5x={210}^{\circ }$

$\Rightarrow x={42}^{\circ }$

Therefore,

$\angle COB=2x-{60}^{\circ }={84}^{\circ }-{60}^{\circ }={24}^{\circ }$

$\angle AOC=3x+{30}^{\circ }={126}^{\circ }+{30}^{\circ }={156}^{\circ }$

The sum of two complementary angles is 90${}^{\circ }$. Hence, each angle will be less than 90${}^{\circ }$

Let $\angle$1 = 3x and $\angle$2 = 2x

Sum of angles on a straight line is 180${}^{\circ }$.

So, 3x + 2x = 180${}^{\circ }$

⇒ 5x = 180${}^{\circ }$

⇒ x = 36${}^{\circ }$

⇒$\angle$6 = $\angle$2       [$\because$∠6 and ∠2 are corresponding angles]

⇒ $\angle$6 = 2x = 2 × 36 = 72${}^{\circ }$ 

Hence, $\angle$6 = 72${}^{\circ }$

In the above figure, since the two lines are parallel and cut by a transversal, using the property of corresponding angles:

$\Rightarrow$ 2x - 60${}^{\circ }$ = x

$\Rightarrow$ x - 60${}^{\circ }$ = 0

$\Rightarrow$ x = 60${}^{\circ }$

From the given figure,
$\angle$BOC = 180${}^{\circ }-(\angle$2 + $\angle$4)

           = 180${}^{\circ }$ - ($\frac{\angle B}{2}$ + $\frac{\angle C}{2}$ )                   

           = 180${}^{\circ }$ - ($\frac{{180}^{\circ }-\angle A}{2}$)  
                    [$\because$ 180${}^{\circ }$ - $\angle$ A = $\angle$ B + $\angle$ C]

           = 180${}^{\circ }$ - 90${}^{\circ }$ + $\frac{\angle A}{2}$

           = 90${}^{\circ }$ + $\frac{\angle A}{2}$