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

1 Answer
Sep 28, 2017

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#