\(\because\) Exterior angle = 110°
One of the interior opposite angle = 40°
We know that, the sum of the interior opposite angles is equal to exterior angle.
x+ 40° = 110° \(\Rightarrow\) x = 110° - 40° = 70°

Yes, in a triangle (equilateral triangle), its median and altitude are same. In the adjoining figure. Hence, AL is an altitude as well as a median of \(\triangle ABC\).

(i) The side opposite to the vertex Q of \(\triangle PQR\) is \(\bar {PR}\).

(ii) The angle opposite to the side LM of  \(\triangle LMN\) is \(\angle MNL\).

(iii) The vertex opposite to the side RT of \(\triangle RST\) is S.

In the adjacent figure, we have \(\triangle PQR\) We know that, a median connects a vertex of a triangle to the mid-point of the opposite side. On joining Q and mid-point of PR, i.e. E.We get the required median QE.

(a) In the given figure,\(\bar { AC } =6cm,\bar { BC } =6cm\bar { ,AB } =4cm,\) \(\quad \because \bar { AC } =\bar { BC } \)
Hence, \(\triangle ABC\) is an isosceles triangle.
(b) In the given figure, \(\bar { AC } =4cm,\bar { AB } =4cm,\bar { BC } =4cm\)  \(\because \bar { AC } =\bar { AB } =\bar { BC } \)
Hence, \(\triangle ABC\) is an equilateral triangle.

(i) Consider a \(\triangle ABC\)

Side opposite to vertex B is AC.
(ii) Consider a \(\triangle XYZ,\)

Angle opposite to side XY is \(\angle Z\).
(iii) In \(\triangle PQR,\)

Vertex opposite to side PR is Q.

In the given figure,
\(\angle A=65^0,\angle B=x,\angle C=30^0\)
We know that, the sum of all angles in a triangle is equal to 180°.
\(\therefore\angle A+\angle B+\angle C=180^0\)
\(\Rightarrow\) 65° + x + 30° =180⁰ \(\Rightarrow\) x =180° - 65° - 30°
\(\Rightarrow\) x =180° - 95° = 85°

Given, \(\triangle ABC\) is right angled at C, in which

AB = 25 cm and AC = 7 cm
In \(\triangle ABC\), by using Pythagoras property,
AB² =AC² +BC² \(\Rightarrow\)BC² =AB² -AC²
\(\Rightarrow\) BC² = (25)² - (7)² \(\Rightarrow\) BC² = 625 - 49
\(\Rightarrow\) BC² = 576
\(\Rightarrow BC=\sqrt576=24\)
Hence, the length of BC is 24 cm.

Let given triangle be \(\triangle ABC\).

Then, we have AB = AC.
So,\(\triangle ABC\) is an isosceles triangle.
\(\therefore \angle ACB=\angle ABC=y^0\)
Also, \(\angle ACD\) is an exterior angle.
We know that, in a triangle an exterior angle and the interior adjacent angle form a linear pair.
\(\therefore\angle ACB+\angle ACD=180^0\)
\(\Rightarrow \angle ACD+120^0=180^0\Rightarrow\angle ACB=180^0-120^0=60^0\)
\(\therefore \angle ABC=\angle ACB=60^0\)
\(\triangle ABC\),\(\angle A+\angle B+\angle ACB=180^0\) [by angle sum property of a triangle]
\(\Rightarrow\) x+ y+ y=180⁰ \(\Rightarrow\) x+60⁰+60⁰=180⁰
\(\Rightarrow\) x = 180° -120° = 60°
Hence, the value of x is 60° and y is 60°.

Let the given triangle be \(\triangle ABC\),then we have AB = AC.
We know that, angle opposite to equal sides of an isosceles triangle are equal.
\(\therefore\) x=40⁰

Hence, the value of x is 40° .

(i) < (ii) < (iii) <

In the given figure, \(\angle BAC=80^0,\angle ABC=60^0,\angle ACE=x\ and \angle ECD=90^0\)
In \(\triangle ABC\), we know that, exterior angle is equal to the sum of interior opposite angles. 
\(\therefore \angle ACD=\angle CAB+\angle ABC\)
\(\Rightarrow \angle ACE+\angle ECD=80^0+60^0\)
\(\angle ACE+90^0=140^0\)
\(\Rightarrow \angle ACE=140^0-90^0\)
\(\therefore \angle ACE=50^0\)