Question

a jogger runs 100m due west ,then changes direction for the second leg of the run .At the end of the run she is 175m away from the starting point at an angle of 15 north of west.What was the direction and length of her second displacement???????

Answer 1

Answer:

The jogger ran for 82.5 m at an angle of 33.28 degrees north of west.

Explanation:

Working:

Draw the points on paper sheet.

Drop a perpendicular on the X-axis, to make a triangle

Base = 175 sin 15 = 169

Height = 175 cos 15 = 45.3

Another triangle will be the final point, point at which he changed the direction and the third point will be the intersection of perpendicular dropped on x-axis.

Base of this triangle = 169 - 100 = 69

Calculate the hypotenuse

sqrt (69^2 + 45.3^2) = 82.3, thats the distance jogger ran after changing the direction after 100 m.

Now to find out the angle, go for any inverse function (sin, cos etc)

acos (69 / 82.3) = 33.28 degrees (inverse of cos)

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Answer 2

Given :

Jogger run 100 m due west from a point

At the end she is 175 m away from the point

To Find : Direction and length of her second displacement

Solution :

Draw the points on paper with x axis and y axis.

Drop a perpendicular on x axis from ending point to make a triangle from   staring point.

Now,

Base of  right angle triangle = 175$\[\cos 15^\circ \]$= 169m

Height of right angle triangle = 175$\[\sin 15^\circ \]$ = 45.3m

Another triangle will be formed as perpendicular from ending point to x-axis will remain same, forming right angle triangle from 1st displacement (100m due west)

Now, in new triangle

base of the right angle triangle = 169 - 100 = 69m

perpendicular of triangle = 45.3m     ( same as height of the 1st triangle)

Calculate the hypotenuse, by Pythagoras theorem

$\[hypotenus{e^2}\]$  = $\[bas{e^2}\]$ + $\[perpendicula{r^2}\]$

$\[hypotenus{e^2}\]$= $\[{69^2} + {45.3^2}\]$

Hypotenuse = 82.3m

Thus , 82.3 m is the direction jogger ran from her first displacement of 100m after changing direction.

Now to find angle from first displacement to end point, use any inverse function

$\[{\sin ^{ - 1}}\]$ =( 45.3/82.3)= $\[33.28^\circ \]$                       ( $\[{\sin ^{ - 1}}\]$is  inverse of sin)

Hence, the length of her second displacement from first displacement are 82.3m at a direction of an angle of 33.28 degree north of west.