1) In a triangle the exterior angle is equal to the sum of its opposite interior angles.
∴ ∠ACD = ∠B + ∠A
⇒ 123°= x + 56°
⇒ x = 123°- 56° = 67°
∴ x = 67°
ii) Sum of three angles of a triangle = 180°
∠P + ∠Q +∠R = 180°
⇒ 45° + 3x + 16°+ 68° = 180°
⇒ 3x + 129° = 180°
3x = 180 – 129 = 51
∴ x = \(\frac{51}3\)
∴ x = 17°
iii) ∠A + ∠B + ∠C = 180°
⇒ 25° + x + 30° = 180°
x + 55°= 180°
x = 180 – 55 = 125°
∴ x = 125°
iv) In ΔXYZ,\(\overline {XY}= \overline {XZ}\) then ∠Y = ∠Z
∴ 2x + 7° = 45°
⇒ 2x = 45 – 7
⇒ 2x = 38
⇒ x = \(\frac{38}2\)
∴ x = 19°
v) From ΔBOA
\(\overline {AB}= \overline {OA}\)
⇒ ∠B = ∠O = 3x + 10° ………(1)
From ΔCOD
\(\overline {OC}= \overline {CD}\)
⇒ ∠O = ∠D …………………(2)
[∵ The angles which are opposite to the equal sides are equal].
from (1) & (2)
∠BOA = ∠COD
[∵ Vertically opposite angles are equal.]
But ∠COD = 90 – x (∵ 2x + ∠O + ∠D = 180 )
⇒ 2x + ∠O + ∠O = 180 (∵∠O = ∠D)
⇒ 2∠O = 180 – 2x
∠O = \(\frac{180\,-\,2x}{2}\) = 90 – x
∴ From ∠BOA = ∠COD
⇒ 3x + 10 = 90 – x
⇒ 3x + x = 90 – 10
⇒ 4x = 80
∴ x = 20°