Question

Please help

Simple geometry / trig problem

Answer 1

`\bf{b.}\ \ \it cos30^o=(\sqrt3)/(2)\ \ (True)\\ \\ \bf{d.}\ \ \it sin60^o=cos(90^o-60^o)=cos30^o\ \ (True) \\ \\ \bf{c.}\ \it cos60^o\cdot tan60^o=cos60^o\cdot(sin60^o)/(cos60^o)=(cos60^o)/(cos60^o)\cdot sin60^o=1\cdot sin60^o=cos30^o\ (T)\\ \\ \\ \bf{a.}\ \ \it (tan30^o)/(sin30^o)=tan30^o\cdot(1)/(sin30^o)=(sin30^o)/(cos30^o)\cdot(1)/(sin30^o)=(sin30^o)/(sin30^o)\cdot(1)/(cos30^o)=\\ \\ \\ =1\cdot(1)/(cos30^o)=(1)/(cos30^o)\ \\e\ cos30^o`

Answer 2

• b, c, d

Explanation:

a. tan 30/ sin 30 = (sin 30/cos 30) / sin 30 = 1 / cos 30, no

b. cos 30 = √3/2, yes

c. cos 60 tan 60 = cos 60 (sin 60/cos 60) = sin 60 = cos 30, yes

d. sin 60 = cos (90 - 60) = cos 30, yes