Answer with explanation:
Drawing a triangle having angle measurement ,30°,60°,90°.
A.
[tex]tan 60^{\circ}=\frac{k\sqrt{3}}{1k}=\sqrt{3}\\\\ \frac{sin 60^{\circ}}{cos 60^{\circ}}=\frac{\frac{k\sqrt{3}}{2k}}{\frac{k}{2k}}=\sqrt{3}\\\\\text{as},tan A=\frac{sin A}{cos A}[/tex]
B.
[tex]sin^2 30^{\circ} + cos^2 30^{\circ}=[\frac{k}{2k}]^2+[\frac{k\sqrt{3}}{2k}]^2\\\\=\frac{1}{4}+\frac{3}{4}\\\\=\frac{4}{4}=1\\\\ \text{as}, Sin ^2 A +Cos ^2 A=1[/tex]
