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${}^{\circ }$

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.

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

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${}^{\circ }$

Let the two equal angles be x.

Then the third angle y = 2x

Also we know,
y + 2x = 180°  (Sum of the angles of a triangle is 180°)

2x + 2x = 180°                              

4x = 180°

x = 45°

Hence the two equal angles are 45° each and the other angle is 90° which makes this triangle a right angled triangle.

Let $\angle A$ and $\angle B$ be adjacent supplementary angles as shown in the figure.

Then, $\angle A+\angle B={180}^{\circ }$
The bisectors of the angles are drawn. We have to find $\angle 1+\angle 2$.
$\angle 1+\angle 2=(\frac{\angle A}{2}+\frac{\angle B}{2})$
$=\frac{180}{2}={90}^{\circ }$
Hence, the angle between the bisectors of adjacent supplementary angles is ${90}^{\circ }$.

The sum of the three angles of a triangle should be ${180}^{\circ }$. Hence it cannot have two right angles. If it has all angles less than ${60}^{\circ }$ then they cannot add up to form ${180}^{\circ }$. A triangle can have two or three acute angles. It cannot have two obtuse angles because then the sum of the three angles would be greater than ${180}^{\circ }$.

Given that,

 2a - 3b = 60${}^{\circ }$

Also if they form a linear pair, then

2 × (180${}^{\circ }$ - b) - 3b = 60${}^{\circ }$      ( since  a = 180${}^{\circ }$ - b)

⇒ 360${}^{\circ }$ - 2b - 3b = 60${}^{\circ }$

⇒ 5b =300${}^{\circ }$

The sum of two complementary angles is 90${}^{\circ }$.
If one angle is x, then another angle will be ${90}^{\circ }-x$.
$\Rightarrow x-({90}^{\circ }-x)={14}^{\circ }$
$\Rightarrow 2x={104}^{\circ }$
$\Rightarrow x={52}^{\circ }$

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

Since the angles are in the ratio 2 : 4 : 3, let x be a factor such that,

2x + 4x + 3x=180${}^{\circ }$

=> 9x = 180${}^{\circ }$

=> x = 20${}^{\circ }$

Hence the three angles of the triangle are:

=> 2x = 40${}^{\circ }$

=> 4x = 80${}^{\circ }$

=> 3x = 60${}^{\circ }$

So the smallest angle in the given triangle is 40${}^{\circ }$.