As per the given information in the question, we can draw the following figure:
Exterior angle = \(\angle DCA\),interior angles,\(\angle A=40^0\) and \(\angle B=40^0\)
We know that, the sum of interior opposite angles is equal to the exterior angle.
\(\therefore\angle A+\angle B=\angle DCA\Rightarrow40^0+40^0=\angle DCA\Rightarrow\angle DCA=80^0\)
Hence, the exterior angle is 80°.

An altitude does not alwayslie in the interior of a triangle.\(\triangle ABC\) is an obtuse angled triangle such that \(\angle C\) is obtuse and AI is the altitude fromA.
Clearly, L is not a point on BC but on BC produced.
So, altitude AI lie in the exterior of the triangle.

The six elements i.e. the three sides and the three angles of \(\triangle ABC\) are as follows:
sides \(\bar{AB},\bar{BC},\bar{CA}\)
Angles \(\angle ABC,\angle BAC,\angle BCA\)

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.

According to the given information, we have a \(\triangle OAB\) right angled at A.

Now, least distance from 0 to B is the hypotenuse OB of \(\triangle OAB\).
OB² = OA²+ AB²[by Pythagoras theorem]
\(\Rightarrow\) OB² = 7² + 15²= 49 + 225 \(\Rightarrow\) OB² = 289
\(\Rightarrow\) OB= 17 km
Hence, the least distance from O to B is 17 km.

Since, \(\angle Y\) and 45⁰ form a linear pair.
So, \(\angle y+45°=180°\)
\(\Rightarrow\angle y=180°-45°\)
\(\Rightarrow\angle y=135°\)
The sum of all angles in a triangle is equal to 180°.
So, \(45^0+60^0+\angle x=180°\)
\(\Rightarrow 105^0+\angle x=180°\)
\(\Rightarrow\angle x=180°-105°\)
=75°

Let ABCD be a rectangle, whose length, AB = 40 cm  and diagonal AC = 41 cm
In right angled \(\triangle ABC\), by using Pythagoras property,

AC² = AB² + BC² =>BC² = AC² - AB²
\(\Rightarrow\) BC² = (41)² -(40)²
\(\Rightarrow\) BC² =1681-1600 => BC² =81
\(\Rightarrow BC=\sqrt81=9\)
Now, perimeter of the rectangle = 2 (AB + BC) [∵  perimeter of a rectangle = 2 (l + b)]
= 2 (40 + 9) = 2 \(\times\) 49 =98
Hence, the perimeter of the rectangle is 98 cm.

Let given triangle be \(\triangle ABC\) and \(\triangle ABC\) a right angled at B.
Again, let AC = x cm
Given, AB = 8 cm, BC = 15 cm

In \(\triangle ABC\), by using Pythagoras property,
AC² = AB² + BC²
\(\Rightarrow\) x² = (8)² + (15)² = 64 + 225
\(\Rightarrow\) x²=289 \(\Rightarrow x=\sqrt 289=17\)
Hence, the unknown length of x is 17 cm.
Hence, the value of unknown length of x is 25.

Draw a line segment Be. By paper folding locate the perpendicular bisector of BC.The folded crease meets BC at D, its mid-point.
Take any point A on this perpendicular bisector. Join AB and Ae. Thus, the triangle obtained is an isoscelesMBC in which AB = AC.
Since, D is the mid-point of BC, so, AD is its median. Also, AD is perpendicular bisector of Be. So, AD is the altitude of \(\triangle ABC\).
Thus, it is verified that the median and altitude of an isoscelestriangle are same.

As per the given information in the question, one angle of a triangle is 60°. Let the other two angles be x and 3x.
We know that, the sum of all angles in a triangle is equal to 180°.
So, x+3x+60⁰=180°
\(\Rightarrow\) 4x + 60° =180°
\(\Rightarrow\) 4x =180° - 60°
\(\Rightarrow\) 4x =120°
\(\Rightarrow x=\frac{120^0}{4}=30^0\)
x=30⁰
So, angles will be x = 30° and 3x = 3 x 30° = 90°
Hence, the two angles are 90° and 30°.

The two interior opposite angles are 48⁰ and 64⁰.