Hi everyone, I think that I have got the first part (a) of this question correct but would appreciate someone checking and pointing me in the right direction as I have a feeling that I am missing something as the question seems too simple to be true.
In triangle ABC, angle A=90 and sec B=2
a/ Find cos B - I have got:
sec B=2
1/cos B=2
1=2cos B
cosB=1/2
b/ Find angles B and C
I've got:
cosB=1/2
B=60,300
can't use B=300 as angles in a triangle = 180 so B=60
180-A-B=C so 180-90-60=C=30
I've just looked at the next section of the question and am stuck (I have a feeling its really simple but my brain just doesn't seem to be working today).
c/ Find tan B
I've got tan B = Sin B / Cos B
I've already worked out cos B, but could someone tell me how I find Sin B? Thanks
Hi, I'm still struggling with this one.
I have now got;
sin B = +Squareroot(1-cos^2 B)
but am now confused as I know cos B x cos B = cos^2 B ... so does this mean that cos 60 x cos 60 = 1/2 x 1/2 = 1/4
Then 1-1/4 = 3/4 so sin B = \pm \sqrt{3/4} = 0.866025........ and so B=59.99999 ( I already have worked out that B=60)
I really have no clue what is going on and am going around in circles - would really appreciate someone showing me how its done - Thanks.