Question

Given that sin A=12/3 and cos B=4/5 where A is an acute angle and B is a reflex angle. Finda) tan Bb) sin (A+B)

Answer 1

Let us draw a sketch to understand the question

Since A is an acute angle, then it lies on the 1st quadrant

Since B is a reflex angle, then it lies on the 3rd quadrant

In the first quadrant, all ratios are +ve

In the 3rd quadrant sin and cos - ve but the tan is +ve

We will use the identity

`\sin ^2B+\cos ^2B=1`

To find the value of sin B because tan B = sin B/cos B

Since cos B = -4/5, then

`\begin{gathered} \sin ^2B+(-(4)/(5))^2=1 \\ \sin ^2B+(16)/(25)=1 \end{gathered}`

Subtract 16/25 from both sides

`\begin{gathered} \sin ^2B+(16)/(25)-(16)/(25)=1-(16)/(25) \\ \sin ^2B=(9)/(25) \end{gathered}`

Take a square root for both sides

`\begin{gathered} \sqrt[]{\sin^2B}=\pm\sqrt[]{(9)/(25)} \\ \sin B=-(3)/(5) \end{gathered}`

We take the negative value because sin in the 3rd quadrant is -ve

Divide sin B by cos B to find tan B

`\begin{gathered} \tan B=(\sin B)/(\cos B) \\ \tan B=(-(3)/(5))/(-(4)/(5)) \\ \tan B=(3)/(4) \end{gathered}`

a) tan B = 3/4

Since sin(A + b) = sinA cosB + SinB cosA --------(1)

Then we have to find the value of cos A, we will use the identity above

`\begin{gathered} \sin ^2A+\cos ^2A=1 \\ ((12)/(13))^2+\cos ^2A=1 \\ (144)/(169)+\cos ^2A=1 \end{gathered}`

Subtract 144/169 from both sides to find the value of cos^2A

`\begin{gathered} (144)/(169)-(144)/(169)+\cos ^2A=1-(144)/(169) \\ \cos ^2A=(25)/(169) \end{gathered}`

Take a square root for both sides, then

`\begin{gathered} \sqrt[]{\cos^2A}=\pm\sqrt[]{(25)/(169)} \\ \cos A=(5)/(13) \end{gathered}`

Substitute the values of sin A, cos A, and sin B, cos B in expression (1) above to find sin (A + B)

`\begin{gathered} \sin (A+B)=\sin A\cos B+\sin B\cos A \\ \sin (A+B)=(12)/(13)*(-(4)/(5))+(-(3)/(5))*((5)/(13)) \end{gathered}`

Simplify it

`\begin{gathered} \sin (A+B)=(-48)/(65)+((-15)/(65)) \\ \sin (A+B)=-(63)/(65) \end{gathered}`

b) sin(A + b) = -63/65