sir this is the question from the book -ML aggarwal( avichal publication) class 10th , chapter20, heights and distances ……..
we have given that 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 38degree 30’ at a point 6 m from the foot of the tree.
we have to Calculate (i) the height at which the tree is broken. (ii) the original height of the tree correct to two decimal places.
ML Aggarwal ICSE, Heights and distances
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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° 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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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 380 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 380 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 380 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