Question

Calculate
4sin30-6cos60+5tg45
2Sin60-4cos30+3tg60​

Answer 1

 

$\displaystyle\bf\\a)\\sin30^o=\frac{1}{2}\\\\cos60^o=\frac{1}{2}\\\\tg45^o=1\\\\4sin30^o-6cos60^o+5tg45^o=\\\\=4\times\frac{1}{2}-6\times\frac{1}{2}+5\times1=\\\\=\frac{4}{2}-\frac{6}{2}+5=\\\\=2-3+5=\boxed{\bf4}$

.

$\displaystyle\bf\\b)\\sin60^o=\frac{\sqrt{3}}{2}\\\\cos30^o=\frac{\sqrt{3}}{2}\\\\tg60^o=\sqrt{3}\\\\2Sin60-4cos30+3tg60=\\\\2\times\frac{\sqrt{3}}{2}-4\times\frac{\sqrt{3}}{2}+3\times\sqrt{3}=\\\\=\frac{2\sqrt{3}}{2}-\frac{4\sqrt{3}}{2}+3\sqrt{3}=\\\\=-\frac{2\sqrt{3}}{2}+3\sqrt{3}=\\\\=-\sqrt{3}+3\sqrt{3}=\boxed{\bf2\sqrt{3}}$