Question

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:

1. the length of slide AB is:
   1. 8 m
   2. 6 m
   3. 5 m
   4. 10 m
2. the value of sin² 30° + cos² 60° is:
   1. `1/4`
   2. `1/2`
   3. `3/4`
   4. `3/2`
3. if cos A = `1/2`, then the value of 12 cot² A − 2 is:
   1. 5
   2. 4
   3. 3
   4. 2

Answer 1

In right-angled triangle ABC formed by the slide, the height AC = 4 m and the angle of inclination at B is ∠B = 30°.

(i)

Using the trigonometric ratio:

\[\sin(30^\circ) = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AC}{AB}\]

\[\frac{1}{2} = \frac{4}{AB}\]

\[AB = 4 \times 2 = 8\text{ m}\]

Therefore, the length of slide AB is 8 m.

(ii)

Substitute the known trigonometric values:

\[\sin 30^\circ = \frac{1}{2}\]

\[\cos 60^\circ = \frac{1}{2}\]

\[\sin^2 30^\circ + \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\]

Therefore, the value of sin² 30° + cos² 60° is `1/2`.

(iii)

Given \[\cos A = \frac{1}{2}\], we know that angle A = 60°.

Now, find cot A:

\[\cot(60^\circ) = \frac{\cos(60^\circ)}{\sin(60^\circ)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\]

Substitute cot A into the expression:

\[12\cot^2 A - 2 = 12\left(\frac{1}{\sqrt{3}}\right)^2 - 2\]

\[= 12\left(\frac{1}{3}\right) - 2\]

= 4 − 2 = 2

Therefore, if cos A = `1/2`, then the value of 12 cot² A − 2 is 2.