the trunk of a tree cracks at a point 10 m above the ground the tree Falls and its top touches the ground at a distance of 24 m from the base of the tree find the actual height of the tree
tnwramit1:
can u tell me the angle
Angle of elevation and angle of depression????
ans is 26
use pgp
24^2+10^2=height^2
then height =26
aww..Pythagorean theorem??
10²=24²+b²
10m above the ground
24²+10²=height ²
Answers
Answered by
53
Let a triangle be ABC
AB=10 M
AC =24M
BC^2=AC^2+AB^2
=100+576=
676
BC=26M
LENGTH OF THE BROKE TREE IS 26 M
TOTAL Height OF THE TREE IS (26+10)M
=36M
AB=10 M
AC =24M
BC^2=AC^2+AB^2
=100+576=
676
BC=26M
LENGTH OF THE BROKE TREE IS 26 M
TOTAL Height OF THE TREE IS (26+10)M
=36M
right
thanks
Answered by
10
the actual height of the tree = 36 m
Step-by-step explanation:
Let say the actual height of the tree = H m
the trunk of a tree cracks at a point 10 m above the ground the tree Falls and its top touches the ground at a distance of 24 m from the base of the tree
Base = 24 m
perpendicular = 10 m
Hypotenuse = H - 10 m
applying Pythagorean theorem
(H - 10)^2 = 24^2 + 10^2
=> (H-10)^2 = 576 + 100
=> (H -10)^2 = 676
=> (H-10)^2 = 26^2
=> H -10 = 26
=> H = 36
the actual height of the tree = 36 m
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