Question

Suppose that sec(B) = - 8⁄3 and that 180o < B < 270o. Find sin(B).

Answer 1

Suppose that sec(B) = - 12⁄5 and that 180o < B < 270o. Find sin(B)

Answer 2

since sec B=1/cos B
so 1st find cos B.
Then use \(sin^2 B+cos ^2 B=1\)
to find sin B

Answer 3

thanks hartnn

Answer 4

welcome. :) tell me what answer u got,though.

Answer 5

k hang on

Answer 6

@ hartnn this is what i get sin b =13/12

Answer 7

nopes,how did u get that?
whats cos B u got?

Answer 8

cos b = 5/-12

Answer 9

can u work it and i c where i went wrong? please

Answer 10

u might have got sin^2 B = 169/144 right ??

Answer 11

ya

Answer 12

then i found the root...is that not what I was supposed to do?

Answer 13

nopes, even thats not correct.....let me work it out..
cos B =-5/12
sin^2 B= 1- 25/144
sin^2 B = 119/144
clear till here ???

Answer 14

how do u get 119?

Answer 15

144-25
\(1-\frac{25}{144}=\frac{144-25}{144}=\frac{119}{144}\)

Answer 16

okay gat you..

Answer 17

continue please

Answer 18

now as 180o < B < 270o.
B is in 3rd Quadrant, in this quadrant, sine is negative.
so
sin B=-\(\sqrt{\frac{119}{144}}=-\frac{\sqrt{119}}{12}\)
ok ?

Answer 19

wow!! you're a genius thanks alot

Answer 20

I have some more please don go away..

Answer 21

welcome :)
ask if any more doubts.

Answer 22

can you make up on so we can try it again and c if i get the same answer as yours?

Answer 23

ofcourse,try the one u have asked in the question.....

Answer 24

cool

Answer 25

1-9/64=55/64
so \(sin B=-\sqrt{\frac{55}{64}}=-\frac{\sqrt{55}}{8}\)