Question

A car is driven 110 km west and then 40 km southwest, how far is the car from the point of origin?

Answer 1

Answer:

141.147 km

Explanation:

$sin45=\frac{BD}{BC}\\\Rightarrow BD=BCsin45\\\Rightarrow BD=40\times sin45\\\Rightarrow BD=28.28\ km$

Now, AE=BD also CD=BD as the angle B and C are equal

So, from Pythagoras theorem

AC² = AE²+(CD+DE)²

$AC=\sqrt{AE^2+(CD+DE)^2}\\\Rightarrow AC=\sqrt{(40\times sin45)^2+(110+40\times sin45)^2}\\\Rightarrow AC=141.147\ km$

Displacement of the car is 141.147 km