Question

10. 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.​

Answer 1

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 38o30′ with the ground.

Take PR=x and PQ=PT=y

Thus, TR=x+y

In right triangle PQR,

tanθ=QRPR

Substituting the values,

⇒tan38o30′=6x

⇒0.7954=6x

⇒x=0.7954×6=4.7724

sinθ=PQPR

Substituting the values,

⇒sin38o30′=yx

⇒

Answer 2

Step-by-step explanation:

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

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