1. ## 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.

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

2. 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.

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

3. 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.

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

4. 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

5. 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.

AND Sin B = cos (90 - B)

6. 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.

AND Sin B = cos (90 - B)
Actually, the more direct one to use would be either
$\displaystyle cos(B) = \pm \sqrt{1 - sin^2(B)}$<-- Choose the "+" since cosine is positive for an acute angle.

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

-Dan

7. 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.

8. 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)

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 $\displaystyle \frac{ \sqrt{3} }{2}$ )

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

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

Take it from here.