Question

find x in the following figure class 8 exercise 2.2 solutions ​

Answer 1

x + 56° = 123°

x = 123 - 56

x = 67°

Hope it helps

Answer 2

• (i) x° = 67° .
• (ii) x° = 17° .
• (iii) x° = 125°
• (iv) x° = 19° .

To Find :-

• Value of x ?

Concept used :-

• Sum of all three angles of a triangle is equal to 180° .
• Angle opposite to equal sides of a triangle are equal in measure .
• Exterior angle of a traingle is equal to sum of opposite interior angles .

Solution :-

(i)

In ∆ABC we have,

→ ∠ACD = ∠ABC + ∠BAC { Exterior angle is equal to sum of opposite interior angles }

→ 123° = 56° + x°

→ x° = 123° - 56°

→ x° = 67° (Ans.)

(i)

In ∆PQR we have,

→ ∠PQR + ∠QPR + ∠PRQ = 180° { Angle sum property of a ∆ }

→ (3x + 16)° + 45° + 68° = 180°

→ 3x° + 16° + 45° + 68° = 180°

→ 3x° + 129° = 180°

→ 3x° = 180° - 129°

→ 3x° = 51°

→ 3x° = 3 × 17°

dividing both sides by 3,

→ x° = 17° (Ans.)

(iii)

In ∆ABC we have,

→ ∠ABC + ∠BAC + ∠BCA = 180° { Angle sum property of a ∆ }

→ x° + 25° + 30° = 180°

→ x° + 55° = 180°

→ x° = 180° - 55°

→ x° = 125° (Ans.)

(iv)

In ∆XYZ we have,

→ XY = XZ

So,

→ ∠XYZ = ∠XZY { Angle opposite to equal sides are equal in measure }

then,

→ (2x + 7)° = 45°

→ 2x° + 7° = 45°

→ 2x° = 45° - 7°

→ 2x° = 38°

→ 2x° = 2 × 19°

dividing both sides by 2,

→ x° = 19° (Ans.)

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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