Question

2. A person standing on the bank of a river, observes that the angle of elevation of the
top of a tree, standing on the opposite bank is 45'. When he moves 40 m away from
the bank, he finds the angle of elevation to be 30. Find the height of the tree and the
width of the river. (V3 = 1.732)
4.In right triangle RST, right angled at S, B is a point on ST such that SB =2/3ST.
Prove that 9RB^2 = 9RS^2 + 4ST^2

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Answer 1

Step-by-step explanation:

Let CD=h be the height of the tree and BC=x be the breadth of the river.

From the figure ∠DAC=30

∘

and ∠DBC=60

∘

In right angled triangle △BCD,tan60

∘

=

BC

DC

⇒

3

=

x

h

⇒h=x

3

.....(1)

From the right-angled triangle △ACD

tan30

∘

=

40+x

h

⇒

3

1

=

40+x

h

⇒

3

h=40+x .......(2)

From (1) and (2) we have

3

(x

3

)=40+x

⇒3x=40+x

⇒3x−x=40

⇒2x=40

⇒x=

2

40

=20

From (1) we get h=x

3

=20

3

=20×1.732=34.64m

∴ Height of the tree=34.64 m and width of the river=20m

solution

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