Given, ratio between three angles of a triangles is 1 : 2 : 1. Let three angles of a traingle be n, 2n and n, respectively.
Then, by angle sum property of a triangle,

n+ 2n+n= 180° \(\Rightarrow\) 4n= 180°
\(\Rightarrow n=\frac{180^0}{4}\Rightarrow n=45^0\)
Therefore, three angles of the given triangle are 45°, 2 X 45° and 45° i.e. 45, 90° and 45°.
We observe that, one angle of given triangle is 90° and each of the two other angles is of measure 45°.
Therefore, sides opposite of two equal angles are also equal.
Thus, we may classify the triangle in two different ways as follows:
(i) On the basis of angles triangle is a right angled triangle.
(ii) On the basis of sides triangle is an isosceles triangle.

We know that, the sum of all the angles (i.e. three angles) of a triangle is 180°.
Given, in \(\triangle ABC,\angle A=x,\angle B=50^0,\angle C=60^0\)
By angle sum property of a triangle,
\(\angle A+\angle B+\angle C=180^0\)
\(\Rightarrow\)x + 50° + 60° = 180°
\(\Rightarrow\) x + 110° = 180° \(\Rightarrow\) x = 180° - 110° = 70°
Hence, the value of the unknown x is 70°.

(i) When exterior angle is a right angle, then each of the interior opposite angles is complement to each other i.e. 90°.
(ii) When exterior angle is an obtuse angle, then both the interior opposite angles may be acute or one of the interior opposite angle is obtuse and other is acute.
(iii) When exterior angle is an acute angle, then both the interior opposite are acute angles.

Altitude QS

Rough sketches of altitudes from A to BC for the given triangle are as follows:

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. 
\(\therefore \angle ACD=\angle CAB+\angle ABC\)
\(\Rightarrow \angle ACE+\angle ECD=80^0+60^0\)
\(\angle ACE+90^0=140^0\)
\(\Rightarrow \angle ACE=140^0-90^0\)
\(\therefore \angle ACE=50^0\)

AC= 10 cm

(i) Yes, OP+OQ>PQ because on joining OP and OQ,we get a \(\triangle OPQ\) and in a triangle, sum of the lengths of any two sides is always greater than the third side.

(ii) Yes, OQ+OR>QR because on joining OQ and OR,we get a \(\triangle OQR\) and in a triangle, sum of the lengths of any two sides is always greater than the third side.
(iii) Yes, OR+OP>RP because on joining OR and OP, we get a \(\triangle OPR\) and in a triangle, sum of the lengths of any two sides is always greater than the third side.

(i)By exterior angle property of a triangle, Exterior angle = Sum of interior opposite angles
=> x = 50° + 70° = 120°
Hence, the exterior angle x is 120°.
(ii) By exterior angle property of a triangle,
Exterior angle = Sum of interior opposite angles
=> x=65°+45°=110°
Hence, the exterior angle x is 110°.
(iii) By exterior angle property of a triangle,
Exterior angle = Sum of interior opposite angles
=> x = 30° + 40° = 70°
Hence, the exterior angle x is 70°.
(iv) By exterior angle property of a triangle,
Exterior angle = Sum of interior opposite angles
=> x=60° +60° =120°
Hence, the exterior angle x is 120°.
(v) By exterior angle property of a triangle,
Exterior angle = Sum of interior opposite angles
\(\therefore\) x = 50° + 50° = 100°
Hence, the exterior angle x is 100°.
(vi) By exterior angle property of a triangle,
Exterior angle = Sum of interior opposite angles
=> x = 30° + 60° = 90°
Hence, the exterior angle x is 90°.

13cm

40°, 60°, 80°

In \(\Delta\)ABD, \(\angle\)D = 90°   (Median of isosceles triangle)
Hence, \(\Delta\)ABC is right angled triangle
\(\therefore\)  AB² = BD² + AD²
\(\Rightarrow\) (37)² = (12)² + (x/2)²
\(\Rightarrow\)1369 = 144 + (x/2)²
\(\Rightarrow\)\(\frac{x^2}{4}\)= 1369 -144 = 1225
\(\Rightarrow\)x² = 4 x 1225
\(\Rightarrow\)x = \(\sqrt{4\times1225}\)
\(\therefore\) x = 2\(\sqrt{5\times5\times7\times7}\)
= 2 x 5 x 7 = 70

(i) x + y = 1200                                  ...(1)
The exterior angle of a triangle is equal to the sum of its two interior opposite angles
x + y + y = 1800
Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of a triangle is 180⁰
\(\Rightarrow\)x + 2y = 1800                                  ... (2)
Subtracting equation (1) from equation (2),
y = 60°
Put y = 60° in equation (1),
x + 60° = 120°
\(\Rightarrow\)x = 120° - 60°
\(\Rightarrow\)x = 60°