The upper part of a coconut tree is broken part touches the ground without being detached.The top of the broken part touches the ground at an angle 45° at a point 10 m .From the foot of tree.Find the height of the tree.
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In the traingle formwd between base of tree, the fallen point and the area where it is bent,
tan45°=AB/10
1=AB/10
AB=10m.
Also sin45°= AB/AC
1/√2 = 10/AC
AC=10√2m
Therefore height of tree=AB+AC=10+10√2=10(1+√2)=10(1+1.41)=10(2.41)=24.1m(Ans)
tan45°=AB/10
1=AB/10
AB=10m.
Also sin45°= AB/AC
1/√2 = 10/AC
AC=10√2m
Therefore height of tree=AB+AC=10+10√2=10(1+√2)=10(1+1.41)=10(2.41)=24.1m(Ans)
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