From the given figure, find the three angles of the triangle ABC.

1.	From the given figure, find the three angles of the triangle ABC.
Answer» Here ∠DCE = 53° ∴∠ACB = 53° (Vertically opposite angle) In ∆ABC ∠ABC + ∠ACB = ∠CAF (By exterior angle property) ⇒ ∠ABC + 53°= 112° ⇒ ∠ABC = 112° – 53° ⇒ ∠ABC = 59° Now in ∆ABC ∠BAC + ∠ABC + ∠BCA = 180° (by angle sum property) ⇒ ∠BAC + 59° + 53° = 180° ⇒ ∠BAC = 68° Hence, ∠BAC = 68°, ∠ABC = 59° and ∠ACB = 53°.

Answer»

Here ∠DCE = 53°

∴∠ACB = 53°

(Vertically opposite angle)

In ∆ABC

∠ABC + ∠ACB = ∠CAF

(By exterior angle property)

⇒ ∠ABC + 53°= 112°

⇒ ∠ABC = 112° – 53°

⇒ ∠ABC = 59°

Now in ∆ABC

∠BAC + ∠ABC + ∠BCA = 180°

(by angle sum property)

⇒ ∠BAC + 59° + 53° = 180°

⇒ ∠BAC = 68°

Hence, ∠BAC = 68°, ∠ABC = 59° and ∠ACB = 53°.

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