  Selina solutions

# Question 12

Q12) Find the unknown marked angles in the given figures : Solution:

We know that in a triangle, if one side of it is produced, then

Exterior angle = sum of its interior opposite angles

(i) In Fig. (i), 110° = x° + 30°

⇒ x° = 110° - 30° = 80°

(ii) In Fig. (ii), 120° = y° + 60°

⇒ y° = 120° - 60° = 60°

(iii) In Fig.(iii), 122° = k° + 35°

⇒ k° = 122° - 35° = 87°

(iv) In Fig. (iv), 135° = a° + 73°

⇒ a° = 135° - 73° = 62°

(v) In Fig. (v), 125° = a + c ….(i)

And 140° = a + b …..(ii)

a + c + a + b = 125° + 140°

⇒ a + a + b + c = 265°

But a + b + c = 180°

(Sum of angles of triangles)

∴ a + 180° = 265°

⇒ a = 265° - 180° = 85°

But a + b = 140°

⇒ 85° + b = 140°

⇒ b = 140° - 85° = 55°

And a + c = 125° ⇒ 85° + c = 125°

⇒ c = 125° - 85° = 40°

Hence, a = 85°, b = 55° = 40°

(vi) In Fig. (vi)

112° + x° = 180° (Linear pair)

⇒ x = 180° - 112° = 68°

And 112° = y + 63°

⇒ y = 112° - 63° = 49°

Hence x = 68°, y = 49°

(vii) In Fig. (vii),

120° = a + a ⇒ 2a = 120°

⇒ a = 120°/2 = 60°

∴ a = 60°

(viii) In Fig. (viii),

140° + a = 180° (Linear pair)

⇒ a = 180° - 140° = 40°

Now, 4m = 2m = a ⇒ 4m – 2m = a

⇒ 2m = 40° ⇒ m = 40°/2 = 20°

Hence, m = 20°

(ix) In Fig. (ix),

105° = b + b ⇒ 2b = 105°

⇒ b = 105°/2 = 52.5°

But a + 105° = 180° (Linear pair)

⇒ a = 180° - 105° = 75°

Hence a = 75° , b = 52.5°

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