Asked by Anonymous
                The angles from Quadrant 1 satisfying the following equations:
2sin𝑥+3tan𝑦=4√3 and 6sin𝑥−tan𝑦=2√3 are:
            
        2sin𝑥+3tan𝑦=4√3 and 6sin𝑥−tan𝑦=2√3 are:
Answers
                    Answered by
            mathhelper
            
    let sinx = a , just for easier typing
let tany = b
2a + 3b = 4√3
6a - b = 2√3 ----- times 3 ----> 18a - 3b = 6√3
add this to the first:
20a = 10√3
a = √3/2 , so sinx = √3/2
x = π/3
sub into first:
2a + 3b = 4√3
√3 + 3b = 4√3
3b = 3√3
b = √3
so tany = √3
y = π/3
    
let tany = b
2a + 3b = 4√3
6a - b = 2√3 ----- times 3 ----> 18a - 3b = 6√3
add this to the first:
20a = 10√3
a = √3/2 , so sinx = √3/2
x = π/3
sub into first:
2a + 3b = 4√3
√3 + 3b = 4√3
3b = 3√3
b = √3
so tany = √3
y = π/3
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