Selina Concise Class 7 Math Chapter 15 Triangles Exercise 15B Solution
EXERCISE 15B
(1) Find the unknown angles in the given figures:
(i) x = y (opposite to equal sides)
Then, x + x + 80° = 180°
⇒ 2x = 180° – 80° = 100°
⇒ x = 100°/2
⇒ x = 50°
So, y = 50°
(ii) a = b (opposite to equal sides)
Then, a + a + 40° = 180°
⇒ 2a = 180° – 40° = 140°
⇒ a = 140°/2
⇒ a = 70°
And, b = 70°
(iii) x = y (opposite to equal sides)
Then, x + x + 90° = 180°
⇒ 2x = 180° – 90°
⇒ x = 90°/2
⇒ x = 45°
y = 45°
(iv) a = b (opposite to equal angles)
Then, a + a + 80° = 180°
⇒ 2a = 180° – 80°
⇒ a = 100°/2
⇒ a = 50°
b = 50°
And, x + b = 180° (linear pair)
⇒ x = 180° – 50°
⇒ x = 130°
(v) Let each angle of an isosceles triangle be x.
Then, x + x = 86°
⇒ 2x = 86°
⇒ x = 86°/2
⇒ x = 43°
And, x + p = 180° (linear pair)
⇒ p = 180° – 43°
⇒ p = 137°
(vi) m = 35° (opposite to equal sides)
Now, m + n + (60° + 35°) = 180°
⇒ m + n = 180° – 95°
⇒ 35° + n = 85°
⇒ n = 85° – 35°
⇒ n = 50°
(vii) x = 60° (alternate angles)
Let each equal angle of an isosceles triangle.
Then, a + a + x = 180°
⇒ 2a = 180° – 60° = 120°
⇒ a = 120°/2
⇒ a = 60°
And, y + a = 180° (linear pair)
⇒ y = 180° – 60°
⇒ y = 120°
(2) Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures:
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