Solution:
Circumcentre of a triangle lies outside then triangle is obtuse angled triangle.

Solution:
According to question,

Sides AB = x² - 1
          BC = 2x
          AC = x² + 1
By using Pythagoras theorem
AC² = AB² + BC²
(x² + 1)² = (x² - 1)² + (2x)²
x⁴ + 1 + 2x² = x² + 1 - 2x² + 4x²
(x² + 1)² = (x² + 1)²
∴ The triangle is right angle Δ

Solution:
According to question,

In right angle ΔBAC
∠A + ∠B + ∠C = 180°
∠B = 180° - 55° - 90°
∠B = 35°
In right angle ΔADB
∠ADB + ∠ABD + ∠BAD = 180°
∠BAD = 180° - 35° - 90°
∠BAD = 55°

Solution:
According to question,

Given : ∠A = 75°,       ∠B = 45°
∴ ∠ACD = ∠A + ∠B
x° = ∠ACD = 120°
Now, $$\frac{x}{3}$$% of 60° is
= $$\frac{{120}}{3}$$ % of 60°
= 40% of 60°
= $$\frac{{40}}{{100}}$$ × 60°
= 24°

Solution:
∵ ΔPQR ∼ ΔPMN
∵ ΔPQR is equilateral
∴ PQ = PR = QR

So, ΔPMN must be equilateral
So, MN = PN = 6 cm

Solution:

In Equilateral triangle
$$\eqalign{ AG:GD = 2:1 \cr AD = \frac{{\sqrt 3 }}{2}a \cr AD = \frac{{\sqrt 3 }}{2} \times 6 \cr AD = 3\sqrt 3 \cr 3 \to 3\sqrt 3 \cr 1 \to \sqrt 3 \cr AG = 2\,{\text{unit}} = 2\sqrt 3 \,{\text{cm}} \cr} $$

Solution:
According to question,
Given:

∴ ∠BOC = 90° + $$\frac{1}{2}$$ ∠A
∠BOC = 90° + $$\frac{1}{2}$$ × 30°
∠BOC = 90° + 15°
∠BOC = 105°