WAEC 2008 · Paper 2 · Q9

Trigonometric ratios

In the diagram, a ladder $LN$ , $10\text{ m}$ long, rests on a wall $4.5\text{ m}$ high such that $2.5\text{ m}$ of it projects beyond the wall.

Not to scale.

(a)
Calculate, correct to one decimal place, the angle which the ladder makes with the ground.
(b)
How high above the ground is the upper end of the ladder?
(c)
If the foot of the ladder is moved $2\text{ m}$ further away from the wall, calculate, correct to the nearest degree, the angle which the ladder makes with the ground.

Worked solution (try it first)

(a)

The part of the ladder from the foot $L$ to the top of the wall $T$ is $10 - 2.5 = 7.5\text{ m}$ .
In the right-angled triangle $LPT$ , the wall $PT = 4.5$ is opposite $\theta$ and $LT = 7.5$ is the hypotenuse: $\sin\theta = \frac{4.5}{7.5} = 0.6$ .
So $\theta = 36.87^\circ$ , which is $36.9^\circ$ to one decimal place.

(b)

The top end $N$ is $10\text{ m}$ along the ladder: its height is $10\sin\theta = 10 \times 0.6 = 6\text{ m}$ .

(c)

First find $LP$ by Pythagoras: $LP = \sqrt{7.5^2 - 4.5^2}$
$= \sqrt{36}$
$= 6\text{ m}$ .
Moving the foot $2\text{ m}$ further makes $LP = 8\text{ m}$ , with the ladder still resting on top of the wall.
Now $\tan\theta = \frac{4.5}{8} = 0.5625$ .
So $\theta = 29.36^\circ$ , which is $29^\circ$ to the nearest degree.

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