Free Lines and Angles 02 Practice Test - 9th Grade 

From $△$ ABC;

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

From $△$ DEF;

$\angle D+\angle E+\angle F={180}^{\circ }$

So, adding these

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=1{80}^{\circ }+{180}^{\circ }={360}^{\circ }$

By angle sum property,
$x+y+z={180}^{\circ }$
$\angle BAF+\angle ACD+\angle CBE$
$=({180}^{\circ }-x)+({180}^{\circ }-y)+({180}^{\circ }-z)$
$={540}^{\circ }-(x+y+z)={540}^{\circ }-{180}^{\circ }={360}^{\circ }$

From the given figure,

$\angle ECD={180}^{\circ }-{150}^{\circ }={30}^{\circ }$ (sum of interior angles on the same side of the transversal is 180°)

$x=\angle BCD={25}^{\circ }+\angle ECD$ (alternate interior angles)
$={25}^{\circ }+{30}^{\circ }={55}^{\circ }$

$\frac{y}{x}=5$
$\Rightarrow y=5x$ and
$\frac{z}{x}=4$
$\Rightarrow z=4x$
$x^{\circ }+y^{\circ }+z^{\circ }={180}^{\circ }$
$x^{\circ }+5x^{\circ }+4x^{\circ }={180}^{\circ }$
$10x^{\circ }={180}^{\circ }$
$x={18}^{\circ }$

If two angles form a linear pair, either both of them are ${90}^{\circ }$ or one angle is acute and the other is obtuse. They cannot both be acute angles because in that case the sum would not be ${180}^{\circ }$ . 

x + 2x + 3x = 180${}^{\circ }$ [Angles on a straight line]

6x = 180${}^{\circ }$

x = 30${}^{\circ }$

Extend OP. Then, a triangle PQT will be formed; where T is the point at which OP cuts QR.

Now,

$\angle$OPQ + $\angle$QPT = 180${}^{\circ }$

$\Rightarrow \angle$QPT = 180${}^{\circ }$ - 110${}^{\circ }$ = 70${}^{\circ }$

Now, since if OP || RS,  $\angle$SRQ and $\angle$UTQ are corresponding angles hence, $\angle$UTQ = $\angle$SRQ = 130${}^{\circ }$

Therefore we have,

$\angle$UTQ + $\angle$PTQ = 180${}^{\circ }$

$\Rightarrow \angle$PTQ = 180${}^{\circ }$ - 130${}^{\circ }$ = 50${}^{\circ }$

In triangle PTQ,

$\angle$PTQ + $\angle$TQP + $\angle$QPT = 180${}^{\circ }$

$\Rightarrow {50}^{\circ }$ + $\angle$TQP + 70${}^{\circ }$ = 180${}^{\circ }$

$\Rightarrow \angle$TQP = 180${}^{\circ }$ - 120${}^{\circ }$ = 60${}^{\circ }$

$\Rightarrow \angle$TQP = $\angle$PQR = 60${}^{\circ }$

Parallel lines never intersect. Hence, they have no common point.

The angle on a straight line is ${180}^{\circ }$. Adjacent angles on a straight line add up to ${180}^{\circ }$.

Conversely, if adjacent angles add up to ${180}^{\circ }$ then the angles are on a straight line.

Hence, if x + y = ${180}^{\circ }$ , then A, B and C lie on a straight line.

Therefore, both the statements are true and statement 2 is the correct explanation of statement 1.

Since, the sum of co-interior angles (interior angles on the same side) = 180${}^{\circ }$

$\Rightarrow$x + 15${}^{\circ }$ + 6x - 10${}^{\circ }$ = 180${}^{\circ }$

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

$\Rightarrow$ 7x = 175${}^{\circ }$

$\Rightarrow$x = 25${}^{\circ }$