The upper part of a tree broken by wind, falls to the
ground without being detached. The top of the broken
part touches the ground at an angle of 30 degree at a
point 8m from the foot of the tree. Calculate:
1) the height at which the tree is broken
ii) the original height of the tree correct to the two
decimal places
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10th
Maths
Some Applications of Trigonometry
Heights and Distances
The upper part of a tree br...
MATHS
The upper part of a tree broken by wind falls to the ground without being detached. The top of the broken part touches the ground at an angle of 38
o
30
′
at a point 6 m from the foot of the tree. Calculate
(i) the height at which the tree is broken.
(ii) the original height of the tree correct to two decimal places.
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ANSWER
Consider
TR as the height of the tree,
TP as the broken part which touches the ground at a distance of 6m from the foot of the tree which makes an angle of 38
o
30
′
with the ground.
Take PR=x and PQ=PT=y
Thus, TR=x+y
In right triangle PQR,
tanθ=
QR
PR
Substituting the values,
⇒tan38
o
30
′
=
6
x
⇒0.7954=
6
x
⇒x=0.7954×6=4.7724
sinθ=
PQ
PR
Substituting the values,
⇒sin38
o
30
′
=
y
x
⇒0.6225=
y
4.7724
⇒y=
0.6225
4.7724
=7.6665
Hence,
Height of the tree =4.7724+7.6665=12.4389=12.44 m.
Height of the tree at which it is broken =4.77 m.
solution
Answer:
Consider TR as the total height of the tree
TP as the broken part which touches the ground at a distance of 6 m from the foot of the tree which makes an angle of 38° 30’ with the ground
Take PR = x and TR = x + y
PQ = PT = y
In right triangle PQR
tan θ = PR/QR
Substituting the values
tan 38° 30’ = x/6
x/6 = 0.7954
By cross multiplication
x = 0.7954 × 6 = 4.7724
We know that
sin θ = PR/PQ
Substituting the values
sin 38° 30’ = x/y
So we get
0.6225 = 4.7724/y
y = 4.7724/0.6225 = 7.6665
Here
Height of the tree = 4.7724 + 7.6665 = 12.4389 = 12.44m
Height of the tree at which it is broken = 4.77 m
