Question

Question #2c9a8

Answer 1

`13/85`

Explanation:

If `sin(A)=3/5`,which is positive, and `A` is obtuse, we know `A` is in `Q`II.

The ratio `3/5` uses the Pythagorean triple `3, 4, 5`. We know that `cos(A) = -4/5`.

Since `B` is acute we know `B` is in `Q`I. The ratio `15/17` comes from the Pythagorean triple `8, 15, 17`, so we know that `sin(B) = 8/17`.

The sum formula for sine is:

`sin(A+B) = sin(A)cos(B)+cos(A)sin(B)`

`sin(A+B) = sin(A)cos(B)+cos(A)sin(B)`

`=(3/5)(15/17) + (-4/5)(8/17)`

`=13/85`

Answer 2

`sin(A+B)=13/85`

Explanation:

`sinA=3/5 , CosB=15/17`

Acute angles are smaller than `90^0`. Obtuse angles are larger

than 90°, but less than `180^0`. Since `/_A` is obtuse angle,

it is on second quadrant. Base `B^2= H^2-P^2` or

`B^2=5^2-3^2=5^2-3^2=4^2or B=4; H=5 , P=3`

`sin A=P/H=3/5 :. cos A =B/H= -4/5`

Since `/_B` is acute angle,it is on first quadrant.

Perpendicular `P^2= H^2-B^2` or

`P^2=17^2-15^2=64or P=8; H=17 , B=15`

`cos B =B/H= 15/17 :.sin B=P/H=8/17`

`sin(A+B)=sinA cosB+cosA sinB` or

`sin(A+B)=3/5*15/17+(-4/5)* 8/17` or

`sin(A+B)=9/17-32/85= (45-32)/85=13/85` [Ans]

Answer 3

`13/85`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)sin^2x+cos^2x=1`

`rArrsinx=+-sqrt(1-cos^2x)`

`rArrcosx=+-sqrt(1-sin^2x)`

`•color(white)(x)sin(A+B)=sinAcosB+cosAsinB`

`"to obtain the expansion we require the values of"`

`cosA" and "sinB`

`"since A is obtuse then "cosA<0`

`rArrcosA=-sqrt(1-(3/5)^2)=-sqrt(16/25)=-4/5`

`"since B is acute then "sinB>0`

`rArrsinB=sqrt(1-(15/17)^2)=sqrt(64/289)=8/17`

`rArrsin(A+B)`

`=(3/5xx15/17)+(-4/5xx8/17)`

`=45/85-32/85`

`=13/85`