Thursday, October 8, 2015

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

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.

No comments:

Post a Comment

What accomplishments did Bill Clinton have as president?

Of course, Bill Clinton's presidency will be most clearly remembered for the fact that he was only the second president ever...