Triangle Angle Problem

• Sep 22nd 2007, 08:17 AM
Tom G
Triangle Angle Problem
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.:D

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
• Sep 22nd 2007, 08:18 AM
topsquark
Quote:

Originally Posted by Tom G
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.:D

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

Too simple, perhaps, but also correct. :)

-Dan
• Sep 22nd 2007, 08:20 AM
janvdl
Quote:

Originally Posted by Tom G
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.:D

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

Yes, seems fine (Nod)
• Sep 22nd 2007, 08:29 AM
Tom G
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
• Sep 22nd 2007, 08:32 AM
janvdl
Quote:

Originally Posted by Tom G
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

You've just worked B out to be 30 degrees didn't you? :)

Now use your special triangles to aid you. (Nod)

AND Sin B = cos (90 - B)
• Sep 22nd 2007, 09:09 AM
topsquark
Quote:

Originally Posted by janvdl
You've just worked B out to be 30 degrees didn't you? :)

Now use your special triangles to aid you. (Nod)

AND Sin B = cos (90 - B)

Actually, the more direct one to use would be either
$cos(B) = \pm \sqrt{1 - sin^2(B)}$<-- Choose the "+" since cosine is positive for an acute angle.

or
$cos(B) = sin(90 - B)$

-Dan
• Sep 23rd 2007, 03:46 AM
Tom G
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):confused:

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.
• Sep 23rd 2007, 05:17 AM
janvdl
Quote:

Originally Posted by Tom G
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):confused:

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.

Sin B = 0.866025 (which is actually $\frac{ \sqrt{3} }{2}$ )

And $Cos B = \frac{1}{2}$

And $Tan B = \frac{Sin B}{Cos B}$

Take it from here.