Question

(10.05 LC)

Answer 1

Answer:

Cos x = $\frac{base}{hypotenuse}=\frac{\sqrt{3} }{2}$

Tan x = $\frac{perpendicular}{base} = \frac{1}{\sqrt{3} }$

Step-by-step explanation:

Sin x = $\frac{perpendicular}{hypotenuse} = \frac{1}{2}$

So, Perp = 1 and hyp = 2

Putting this in Pythagorean theorem to get the base

$c^2 = a^2+b^2$

$a^2 = c^2-b^2$

base ² = 4-1

base² = 3

Taking square root on both sides

Base = $\sqrt{3}$

Now

Cos x = $\frac{base}{hypotenuse}=\frac{\sqrt{3} }{2}$

Tan x = $\frac{perpendicular}{base} = \frac{1}{\sqrt{3} }$