A shadow of a tower standing on a level ground is found to be 40√3 meters longer when the Sun's altitude is 30° than when it is 60°. The height of the tower is:A50 mB60 mC70 mD40 mAnswer: B. 60 m Read Explanation: tan60=3=hxtan60=\sqrt3=\frac hxtan60=3=xhtan30=13=h403+xtan 30= \frac{1}{\sqrt3}=\frac{h}{40\sqrt3 +x}tan30=31=403+xhx=h3x=\frac{h}{\sqrt3}x=3h403+h3=3h40\sqrt3+\frac{h}{\sqrt3}=\sqrt3h403+3h=3h120+h=3h120+h=3h120+h=3h120=2h120=2h120=2hh=60h=60h=60 Read more in App