Question

A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?​

Answer 1

he is 17 meters far from starting point

how I did it

well B = 15

P = 8

H = ?

H²= P²+B²

H²=225+64

H=17

H = 17 meters

Answer 2

Given:-

• A man goes 15m due west .
• He then goes 8nm due north.

To Find:-

• The distance of man from starting point.

Formula Used:-

We will use Pythagoras Theorem stated as ,

• In a right angled triangles sum of squares of perpendicular and base is equal to the square to hypontenuse.

$\large\blue{\boxed{\bf{\red{\dag} (Hypontenuse)^2=(perpendicular)^2+(base)^2\red{\dag}}}}$

Answer:-

[ For figure refer to attachment : ]

His distance from the starting point is equal to the hypotenuse of triangle ABC.

Now , in the ∆ ABC ,

$\tt:\implies BC^2=AB^2+AC^2$

$\tt:\implies BC^2=15m^2+8m^2$

$\tt:\implies BC^2=225m^2+64m^2$

$\tt:\implies BC^2 = 289m^2$

$\tt:\implies BC=\sqrt{289m^2}$

$\tt:\implies BC = \sqrt{17m\times17m}$

$\tt:\implies BC=\pm17m$

$\underline\blue{\boxed{\red{\tt{\longmapsto \:\:BC\:\:=\:\:17m\:\:}}}}$

$\pink{\boxed{\purple{\tt{\dag Hence\: Distance\:from\: starting\:point\:is\:17m.}}}}$