From the given figure, find :

(i) cos x°

(ii) x°

(iii) $\dfrac{1}{\text{tan}^2 \text{ x°}} - \dfrac{1}{\text{sin}^2 \text{ x°}}$

(iv) Use tan x°, to find the value of y.

(i) cos x°

cos x° = $\dfrac{Base}{Hypotenuse}$

$\dfrac{Base}{Hypotenuse} = \dfrac{BC}{AC} = \dfrac{10}{20} = \dfrac{1}{2}$

Hence, cos x° = $\dfrac{1}{2}$

(ii) x°

cos x° = $\dfrac{1}{2}$

cos x° = cos 60°

Hence, x° = 60°.

(iii)

$\dfrac{1}{\text{tan}^2 \text{ x°}} - \dfrac{1}{\text{sin}^2 \text{ x°}}\\[1em] = \dfrac{1}{\text{tan}^2 \text{ 60°}} - \dfrac{1}{\text{sin}^2 \text{ 60°}}\\[1em] = \dfrac{1}{(\sqrt3)^2} - \dfrac{1}{\Big(\dfrac{\sqrt3}{2}\Big)^2}\\[1em] = \dfrac{1}{3} - \dfrac{1}{\dfrac{3}{4}}\\[1em] = \dfrac{1}{3} - \dfrac{4}{3}\\[1em] = \dfrac{1 - 4}{3}\\[1em] = \dfrac{- 3}{3}\\[1em] = - 1$

Hence, $\dfrac{1}{\text{tan}^2 \text{ x°}} - \dfrac{1}{\text{sin}^2 \text{ x°}}$ = -1.

(iv) tan x° = $\dfrac{Perpendicular}{Base}$

⇒ tan 60° = $\dfrac{AB}{AC}$

⇒ $\sqrt3 = \dfrac{y}{10}$

⇒ y = 10 $\sqrt3$ = 17.32

Hence, y = 17.32 cm.