What is the measure of the angle A in triangle ABC if tan^2 B/2=(1-sinC)/(1+sinC)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll recall that the tangent function is the ratio:

tan x = sin x/cos x

tan (B/2) = sin (B/2)/cos (B/2)

We'll raise to square both sides:

[tan (B/2)]^2 = [sin (B/2)]^2/[cos (B/2)]^2

We'll use the half angle identities:

[sin (B/2)]^2 = (1 - cos B)/2

[cos (B/2)]^2 = (1 + cos B)/2

[tan (B/2)]^2 = (1 - cos B)/2/(1 + cos B)/2

We'll simplify and we'll get:

[tan (B/2)]^2 = (1 - cos B)/(1 + cos B)

But, from enunciatin, we know the followings:

[tan (B/2)]^2 = (1-sinC)/(1+sinC)

Comparing, we'll get:

(1 - sinC)/(1 + sinC) = (1 - cos B)/(1 + cos B)

1 - sin C = 1 - cos B

sin C = cos B

If sin C = cos B, then the sum of angles B+C = 90, therefore the measure of the angle A is of 90 degrees.

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