AC = the truck bed;
AB = the length of the ramp;
CB = length of the rear of the truck from the point where the ramp touches the ground.
a) tan35∘=ACCBtan 35^\circ={\cfrac{AC}{CB}}tan35∘=CBAC ;
CB=ACtan35∘=1.50.7≈2.14mCB ={\cfrac{AC}{tan35^\circ}}={\cfrac {1.5}{0.7}}\approx 2.14 mCB=tan35∘AC=0.71.5≈2.14m ;
b)
sin35∘=ACABsin 35^\circ={\cfrac{AC}{AB}}sin35∘=ABAC ;
AB=ACsin35∘=1.50.5736≈2.62mAB ={\cfrac{AC}{sin 35^\circ}}={\cfrac {1.5}{0.5736}}\approx 2.62 mAB=sin35∘AC=0.57361.5≈2.62m .
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments