# Question 2c9a8

Feb 4, 2018

$\frac{13}{85}$

#### Explanation:

If $\sin \left(A\right) = \frac{3}{5}$,which is positive, and $A$ is obtuse, we know $A$ is in $Q$II.

The ratio $\frac{3}{5}$ uses the Pythagorean triple $3 , 4 , 5$. We know that $\cos \left(A\right) = - \frac{4}{5}$.

Since $B$ is acute we know $B$ is in $Q$I. The ratio $\frac{15}{17}$ comes from the Pythagorean triple $8 , 15 , 17$, so we know that $\sin \left(B\right) = \frac{8}{17}$.

The sum formula for sine is:

$\sin \left(A + B\right) = \sin \left(A\right) \cos \left(B\right) + \cos \left(A\right) \sin \left(B\right)$

$\sin \left(A + B\right) = \sin \left(A\right) \cos \left(B\right) + \cos \left(A\right) \sin \left(B\right)$

$= \left(\frac{3}{5}\right) \left(\frac{15}{17}\right) + \left(- \frac{4}{5}\right) \left(\frac{8}{17}\right)$

$= \frac{13}{85}$

Feb 4, 2018

$\sin \left(A + B\right) = \frac{13}{85}$

#### Explanation:

$\sin A = \frac{3}{5} , C o s B = \frac{15}{17}$

Acute angles are smaller than ${90}^{0}$. Obtuse angles are larger

than 90°, but less than ${180}^{0}$. Since $\angle A$ 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

$\sin A = \frac{P}{H} = \frac{3}{5} \therefore \cos A = \frac{B}{H} = - \frac{4}{5}$

Since $\angle B$ 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

$\cos B = \frac{B}{H} = \frac{15}{17} \therefore \sin B = \frac{P}{H} = \frac{8}{17}$

$\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$ or

$\sin \left(A + B\right) = \frac{3}{5} \cdot \frac{15}{17} + \left(- \frac{4}{5}\right) \cdot \frac{8}{17}$ or

$\sin \left(A + B\right) = \frac{9}{17} - \frac{32}{85} = \frac{45 - 32}{85} = \frac{13}{85}$ [Ans]

Feb 4, 2018

$\frac{13}{85}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)sin^2x+cos^2x=1

$\Rightarrow \sin x = \pm \sqrt{1 - {\cos}^{2} x}$

$\Rightarrow \cos x = \pm \sqrt{1 - {\sin}^{2} x}$

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

$\text{to obtain the expansion we require the values of}$

$\cos A \text{ and } \sin B$

$\text{since A is obtuse then } \cos A < 0$

$\Rightarrow \cos A = - \sqrt{1 - {\left(\frac{3}{5}\right)}^{2}} = - \sqrt{\frac{16}{25}} = - \frac{4}{5}$

$\text{since B is acute then } \sin B > 0$

$\Rightarrow \sin B = \sqrt{1 - {\left(\frac{15}{17}\right)}^{2}} = \sqrt{\frac{64}{289}} = \frac{8}{17}$

$\Rightarrow \sin \left(A + B\right)$

$= \left(\frac{3}{5} \times \frac{15}{17}\right) + \left(- \frac{4}{5} \times \frac{8}{17}\right)$

$= \frac{45}{85} - \frac{32}{85}$

$= \frac{13}{85}$