Question

Sin (A+2B)=√3/2 and cos(A+4B)=0 Find A and B

Answer 1

The given information states that $sin(A+2b)= \frac{ \sqrt{3} }{2} }$ and $cos(A+4B)=0$ .

We can easily calculate that $sin(60)= \frac{ \sqrt{3} }{2}$ and $cos(90)=0$ .

Using this information, we can set up a system of equations.
A+2B=60
A+4B=90

Now, we can solve this system using the substitution method.

First, we will isolate the variable A on one side of the first equation.
A=60-2B

Now, we will substitute 60-2B in for A in the second equation.
60-2B+4B=90

Combine like terms
60+2B=90

Subtract 60 from both sides of the equation
2B=30

Divide both sides of the equation by 2.
B=15

Now, we can substitute 15 in for B in the first equation and solve for A.

A+2B=60
A+2(15)=60
A+30=60
A=30

A=30 and B=15