If $3 tan A - 5 cos B = s q r t 3$ and $B = 90^{0}$ , find the value of angle $A$ in degrees

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Solution

Given: $B = 90^{\circ}$
$3 t a n A - 5 c o s B = \sqrt{3}$
$3 t a n A - 5 c o s 90 = \sqrt{3}$
$3 t a n A - 0 = \sqrt{3}$
$t a n A = \frac{1}{\sqrt{3}}$
$t a n A = t a n 30$
$A = 30^{\circ}$

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