A school authority constructed a slide for its children below the age of 12 years.

The constructed slide has a height of 4 m above the ground and is inclined at an angle of 30° to the ground.

Use the given information to answer each of the following :

(i) the length of slide AB is :

(a) 8 m
(b) 6 m
(c) 5 m
(d) 10 m

(ii) the value of sin² 30° + cos² 60° is :

(a) $\dfrac{1}{4}$

(b) $\dfrac{1}{2}$

(c) $\dfrac{3}{4}$

(d) $\dfrac{3}{2}$

(iii) if cos A = $\dfrac{1}{2}$, then the value of 12 cot² A - 2 is :

(a) 5
(b) 4
(c) 3
(d) 2

Given,

Height = 4 m

Angle with the ground = 30°

(i) Calculating,

⇒ sin 30° = $\dfrac{AC}{AB}$

⇒ $\dfrac{1}{2} = \dfrac{4}{AB}$

⇒ AB = 4 × 2 = 8 m.

Length of slide = 8 m.

Hence, option (a) is the correct option.

(ii) We know that:

sin 30° = $\dfrac{1}{2}$ and cos 60° = $\dfrac{1}{2}$

sin² 30° + cos² 60°

$= \Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{1}{4} + \dfrac{1}{4}\\[1em] = \dfrac{1}{2}.$

Hence, option (b) is the correct option.

(iii) Given,

cos A = $\dfrac{1}{2}$

So, A = 60°

sin A = $\dfrac{\sqrt{3}}{2}$

$\text{cot A} = \dfrac{cos A}{sin A} \\[1em] = \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} \\[1em] = \dfrac{1}{\sqrt{3}}$

According to the question,

12 cot² A - 2

= 12 $\Big(\dfrac{1}{\sqrt{3}}\Big)^2$ - 2

= 12 × $\dfrac{1}{3}$ - 2

= 4 - 2 = 2.

Hence, option (d) is the correct option.