Question

A man goes 15 m due west and then 8 m due north. How far is he from the starting point? (i) A ladder 17 m long reaches the window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.​

Answer 1

first the man goes to the west 15 m from point A to point B and then it turns north from point B and walks 8 m to point C.

This forms a right angled triangle ABC

So,by pythagoras theorem

AC^{2} = AB^{2}+ BC^{2}AC

2

=AB

2

+BC

2

SO,

AC^{2} = 15^{2} + 8^{2}AC

2

=15

2

+8

2

AC^{2} = 289AC

2

=289

AC^{2}=225+64AC

2

=225+64

AC= \sqrt{289}AC=

289

AC = 17 metres.

So the man is now 17metres away from his starting point.