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in Linear Equations by (41.5k points)
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Find ‘x’ in the following figures?

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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°

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