\(\angle c=35^0\)

Let x cm be the length of the third side.
We know that, the sum of lengths of two sides of a triangle is greater than the length of third side.
∴ 12 cm + 15 cm > x cm \(\Rightarrow\) 27 cm> x cm or x < 27 cm
Also, the difference between the lengths of two sides of a triangle is less than the length of third side.
∴ 15 cm - 12 cm < x cm \(\Rightarrow\) 3 cm > x cm or x < 3 cm
Hence, the length of third side can be any length between 3 cm and 27 cm.

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

Then, we have AB = AC.
So, \(\triangle ABC\) is an isosceles triangle.
\(\therefore\angle ABC=\angle ACB\)
(since, angle opposite to equal sides are equal]
\(\Rightarrow\angle ACB=x\)
\(\angle BAC=92^0\) [vertically opposite angles]
In,\(\triangle ABC\), [by angle sum property of a traingle]
\(\angle ABC+\angle BAC+\angle ACB=180^0\)
\(\Rightarrow\) x + 92° + x = 180° 
\(\Rightarrow\) 2x + 92° = 180°
\(\Rightarrow x=\frac{88^0}{2}\Rightarrow x=44^0\)
Also, \(\angle ACD\) is an exterior angle of \(\triangle ABC\).
By exterior angle property,
Exterior angle = Sum of two interior opposite angles
\(\Rightarrow y= \angle BAC+\angle ABC\)
\(\Rightarrow\)y = 92° + x
\(\Rightarrow\) y = 92° + 44° = 136⁰
Hence, the value of x is 44° and y is 136°.

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
So,\(\triangle ABC\) is an isosceles triangle.
\(\therefore\angle ACB=\angle B\)

[since, the angle opposite to equal sides of an isosceles triangle are equal]
\(\angle ACB=x^0\)
We know that, linear pair of angles is supplementry.
\(\angle ACB+\angle ACD=180^0\)
\(\Rightarrow\) x + 120° = 180°[\(\therefore\angle ACD=120,given\) ]
\(\Rightarrow\) x = 180° -120° = 60°
Hence, the value of x is 60°.

Let the given triangle be \(\triangle ABC\).Then, we have AB=AC
So,\(\triangle ABC\) is an isosceles triangle.
\(\therefore\angle B=\angle C=x\)

[since, the angle opposite to equal sides of an isosceles triangle are equal]
Now, in \(\triangle ABC\) by angle sum property of a triangle,
\(\angle A+\angle B+\angle C=180^0\)
\(\Rightarrow\) 100° + x + x = 180⁰ \(\Rightarrow\) 100° + 2x = 180°
\(\Rightarrow\) 2x=180⁰-100 \(\Rightarrow\) 2x=80⁰ \(\Rightarrow=x=\frac{80^0}{2}=40^0\)
Hence, the value of x is 40°.

Let the given triangle be \(\triangle ABC\).
Then, we have AB = AC.
So, \(\triangle ABC\) is an isosceles triangle.
\(\therefore \angle A= \angle C=45^{0}\)

[since, the angle opposite to equal sides of an isosceles triangle are equal]
Now, in \(\triangle ABC\), by angle sum property of a triangle,
\(\angle A+\angle B+\angle C=180^0\Rightarrow45^0+x+45^0=180^0\)
\([\angle A=\angle B=45,because\ of\ equal\ side]\)
\(\Rightarrow\) x+90⁰=180°  \(\Rightarrow\) x=180⁰-90° =90°
Hence, the value of x is 90°.

\(\angle A+\angle B+\angle C=360^0\)

In \(\triangle OMP\),
\(\angle OMP=\angle OPM\)
\(\therefore\) OM=OP.................1
[Since, sides opposite to equal angles are equal]
Also, in \(\triangle OMN\),
\(\angle OMN=\angle ONM\Rightarrow OM=ON\Rightarrow OP=ON\)  [from Eq.1]
Hence, OM is the median of \(\triangle MPN\).

(a) In the following figure, we have \(\triangle ABC\).We know that, a median connects a vertex of a triangle to the mid-point of the opposite sides. On joining B and mid-point of AC i.e. E we get the required median BE.

(b) We know that, the perpendicular line segment from a vertex of a triangle to its opposite side is called an altitude of the triangle. In the following figure, we have \(\triangle PQR\) in which PQ and PR are the altitude drawn,from vertex Q to PR and from vertex R to PQ, respectively.

(c) In the adjoining figure, we have an obtuse angled triangle XYZ in which YL is an altitude drawn from Y to produced XZ such altitude lies in the exterior of the triangle.

20 cm

10m

26 cm

(i) 60°
(ii) 30°
(iii) 140°
(iv) 50°
(v) 35°
(vi) 60°

We know that, in a triangle the sum of two sides are always greater than the third side. Consider
\(\triangle POS,OP+OS>PS\)..................1
\(\triangle POQ,OP+OQ>PQ\)..................2
\(\triangle QOR,OQ+OR>QR\).............3
\(\triangle ROS,OR+OS>SR\)................4
On adding Eqs. (i), (ii), (iii) and (iv), we get
2 (OP + OS + OQ + OR) > PS + PQ+ QR + SR
\(\Rightarrow\) 2(PR+ QS) > PQ+ QR+RS+SP

(i) We know that, an equilateral triangle have equals sides and each angle equal to 60°.
i.e. 60° + 60° + 60° = 180°
Hence, answer of Ajay was wrong.
(ii) The value depicted by Ajay's answer is that he is not attentive in class, also do not understand the concept of equilateral triangle and their basic properties.