65°

80°

Since sides are :
10·2 cm, 5·8 cm, 4·5 cm
(a) 10·2 + 5·8 = 16 > 4·5
(b) 10·2 + 4·5 = 14·7 > 5·8
(c) 5·8 + 4·5 = 10·3 > 10·2
Since sum of any two sides is greater than third side.
Hence, there may be a triangle with these sides.

Let the equal angles be x
\(\therefore\) x + x + 110° = 180° (By angle sum property)
\(\Rightarrow\)2x + 110° = 180°
\(\Rightarrow\) 2x = 180°-110°
\(\Rightarrow\) 2x = 70°
\(\Rightarrow\)x=\(\frac{70^0}{2}\)
\(\Rightarrow\)x = 35°
Hence, equal angles are of 35°.

Other angles are 35°,35°.

No, we cannot have a triangle with two obtuse angles, because an obtuse angle has its measure more than 90°. Therefore, the sum of two obtuse angles is greater than 180°, which is not possible.

No, we cannot have a triangle with two right angles, because the sum of two right angles is 180°. On adding the measure of the third angle, the sum of three angles will be more than 180°, which is not true for a triangle.

By angle sum property of a triangle,
x + x + 50° = 180° \(\Rightarrow\)2x + 50° = 180°
\(\Rightarrow\)2x = 180° - 50° \(\Rightarrow\) 2x = 130° \(\Rightarrow\) \(x=\frac{130^0}{2}=65^0\)
Hence, the value of the unknown x is 65°.

\(\angle X=120^0\)

60°

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 \(\triangle ABC\).

Every triangle has three altitudes, one A from each vertex to the opposite side.
In the adjoining figure

\(AD\bot BC\),\(BE \bot AC\),and \(CF \bot AB\)
Therefore, line segments AD, BE and CF are three altitudes.

Yes,a median lieswholly in the interior of the triangle.

In the given figure, altitude can be drawn as below:

AL = Altitude from A to BC

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.