WAEC 2013 · Paper 2 · Q13

(a)
When one end of a ladder $LM$ is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of $37^\circ$ with the horizontal ground. (i) Represent this information in a diagram. (ii) Calculate, correct to 3 significant figures, the length of the ladder. (iii) If the foot of the ladder is pushed towards the wall by 2 metres, calculate, correct to the nearest degree, the angle which the ladder now makes with the ground.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw the wall vertical and the ground horizontal.
The ladder $LM$ runs from $L$ , 5 m up the wall, to $M$ on the ground, making $37^\circ$ with the ground at $M$ .

The height, 5 m, is opposite the $37^\circ$ angle, and the ladder is the hypotenuse: $\sin 37^\circ = \frac{5}{|LM|}$ .
So $|LM| = \frac{5}{\sin 37^\circ}$
$= \frac{5}{0.6018}$
$\approx 8.31\text{ m}$ .

First find how far the foot was from the wall: $\tan 37^\circ = \frac{5}{d}$ , so $d = \frac{5}{0.7536} \approx 6.635$ m.
Pushed 2 m closer, the foot is $6.635 - 2 = 4.635$ m from the wall.
The ladder is still 8.308 m long, and it is now the hypotenuse with 4.635 m adjacent to the new angle: $\cos\theta = \frac{4.635}{8.308}$
$\approx 0.5579$ .
So $\theta \approx 56^\circ$ .