Question

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Answer 1

Answer:

Height of the tree is 27.39m

Step-by-step explanation:

REFER THE FIGURE ATTACHED

CD=20m

AB=h(height of tree)

CB=x

DB=20+x

w.r.t  Δ ABC

        tanФ=AB/CB

         tan45=h/x

         1= h/x

          x=h...............(1)

w.r.t ΔABD

         tanФ=AB/DB

         tan30=h/20+x

          1/√3=h/20+x

          20+h-√3h=0

          20+h(1-1.73) =0     [taking h common and putting the value of √3 i.e.1.73]

          20 - 0.73h = 0

          h=20/0.73

          h=27.39m

         

Answer 2

Answer:

27.3 m

Step-by-step explanation:

Let height of the tree = h and river width = x.

(i) In ΔABC,

tan 45° = h/x

1 = h/x

h = x

(ii) In ΔABD,

tan 30° = h/20 + x

(1/√3) = h/20 + x

√3h = 20 + x

√3x = 20 + x

x = 10(√3 + 1).

Substitute x in (i), we get

⇒ h = 10(√3 + 1)

      = 27.3 m.

Hope it helps!