Question

The obtuse angel B is such that tanB = -(5/12). How do you find the following values?

1. sin B
2. cos B
3. sin 2B
4. cos 2B

Answer 1

Given that the angle is obtuse , `B in "quadrant II"`

Again `tanB=-5/12=>B~~157.3^@`

`sinB->+ve`

`cosB->-ve`

`sin2B->-ve" as " 2B in" quadrant IV"`

`cos2B->+ve" as " 2B in" quadrant IV"`

Now `cosB=1/secB=-1/sqrt(1+tan^2B)`

`=-1/(1+5^2/12^2)=-12/13`

`sinB=tanBxxcosB=(-5/12)xx(-12/13)=5/13`

`sin(2B)=(2tanB)/(1+tan^2B)==(-2xx5/12)/(1+(-5/12)^2)=-120/169`

`cos(2B)=(1-tan^2B)/(1+tan^2B)==(1-(-5/12)^2)/(1+(-5/12)^2)=119/169`