From the figure given, the triangle is a right triangle.
Since it is a right triangle, ∠C = 90 degrees
Assuming the figure is drawn to scale and ∠A is greater than ∠B, then we have the following:
m∠A = 90° - m∠B
Using the triangle angle sum theorem, ∠A + ∠B + ∠C = 180
since m∠C = 90°,
m∠B + m∠A = 90°
Thus, m∠A = 90 - m∠B,
Also,
[tex]\tan B=\frac{\sin B}{\cos B}[/tex]
Since ∠A is greater than ∠B, then ∠A = 60 and ∠B = 30
Thus,
[tex]\begin{gathered} \tan B=\frac{\sin B}{\cos B} \\ \\ \tan 30=\frac{\sin 30}{\cos 30} \end{gathered}[/tex]
ANSWER:
[tex]\begin{gathered} \tan B=\frac{\sin B}{\cos B} \\ \\ \\ m\angle A=90-m\angle B \end{gathered}[/tex]
