Question

One person goes 10 m east and changes direction 5 m to the north-east
The distance is gone. What is the mate movement of that person?​

Answer 1

Explanation:

2 votes

THANKS 2

:)

Thanks for helping! ☺

Thank you selecting as brainliest

Abhi178★ Brainly Teacher ★

Let car A moves in East and car B moves in North ,

Let car A covered distance x km in East direction

and Car B covered distance (2x + 16) km in North direction .

because north and east are perpendicular

so, at that time right angle trainlge form and hypotenuse is distance between car at that time

distance between the cars = √(x)² + (2x +16)²

again ,

√{ x ² + (2x + 16)² } = 3x -16

take square both sides

x² + 4x² + 64x +256 = 9x² - 96x +256

4x² -96x -64x = 0

x = 40 km , 0 km

but x ≠ 0 km

so, x = 40 Km

hence distance travelled by car A = 40 km

and distance travelled by car B = 2{40} +16 = 96 km

so, distance between car = √(40² +96)² or 3(40) -16 = 120 -16 = 104 km

hence distance between car at that time = 104 km

Answer 2

One person goes 10 m east and changes direction 5 m to the north-east

The distance is gone. What is the mate movement of that person?