In triangle ABC ,write tan A+B/2 in terms of angle C.
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Solution:-
Given. A, B and C are three angle of a triangle.
In Δ ABC
∠ A + ∠ B + ∠ C = 180° (Angle sum property)
⇒ ∠ A + ∠ B = π - ∠ C
⇒ (∠ A + ∠ B)/2 = (π - ∠ C)/2
⇒ (∠ A + ∠ B)/2 = π/2 - ∠ C/2
∴ tan {(∠ A + ∠ B)/2} = tan (π/2 - ∠ C/2)
⇒ tan {(∠ A + ∠ B)/2} = cot C/2.
Answer.
Given. A, B and C are three angle of a triangle.
In Δ ABC
∠ A + ∠ B + ∠ C = 180° (Angle sum property)
⇒ ∠ A + ∠ B = π - ∠ C
⇒ (∠ A + ∠ B)/2 = (π - ∠ C)/2
⇒ (∠ A + ∠ B)/2 = π/2 - ∠ C/2
∴ tan {(∠ A + ∠ B)/2} = tan (π/2 - ∠ C/2)
⇒ tan {(∠ A + ∠ B)/2} = cot C/2.
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