In the given figure- side BC of - ABC is produced to D-If - ACD-128-and- ABC-43-find- BAC-and - ACB-

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Standard IX

Mathematics

Exterior Angle of a Triangle = Sum of Opposite Internal Angles

In the given ...

Question

$In the given figure, side BC of Δ A B C is produced to D.If ∠ A C D = 128^{\circ} a n d ∠ A B C = 43^{\circ}, f i n d ∠ B A C a n d ∠ A C B .$

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Solution

Side BC of triangle ABC is produced to D.
∴∠ACD=∠A+∠B
[Exterior angle property]

⇒128°=∠A+43°⇒∠A=128-43°⇒∠A=85°⇒∠BAC=85°
Also, in triangle ABC,
∠BAC+∠ABC+∠ACB=180° (Sum of the angles of a triangle)

⇒85°+43°+∠ACB=180°⇒128°+∠ACB=180°⇒∠ACB=52°

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In the given figure, side BC of ΔABC is produced to D.If ∠ACD=128∘ and ∠ABC=43∘,find ∠BAC and∠ACB.

Side BC of triangle ABC is produced to D. ∴∠ACD=∠A+∠B [Exterior angle property] ⇒128°=∠A+43°⇒∠A=128-43°⇒∠A=85°⇒∠BAC=85° Also, in triangle ABC, ∠BAC+∠ABC+∠ACB=180° (Sum of the angles of a triangle) ⇒85°+43°+∠ACB=180°⇒128°+∠ACB=180°⇒∠ACB=52°

$In the given figure, side BC of Δ A B C is produced to D.If ∠ A C D = 128^{\circ} a n d ∠ A B C = 43^{\circ}, f i n d ∠ B A C a n d ∠ A C B .$

Side BC of triangle ABC is produced to D.
∴∠ACD=∠A+∠B
[Exterior angle property]

⇒128°=∠A+43°⇒∠A=128-43°⇒∠A=85°⇒∠BAC=85°
Also, in triangle ABC,
∠BAC+∠ABC+∠ACB=180° (Sum of the angles of a triangle)

⇒85°+43°+∠ACB=180°⇒128°+∠ACB=180°⇒∠ACB=52°