wrong solution (Question:837674.0)

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ankit sharma

· started a discussion

· 1 Months ago

wrong solution (Question:837674.0)

sin30 = y/x
so ans will be 34+24rt3

Kapil Motiyani

· started a discussion

· 1 Months ago

wrong solution (Question:837674.0)

sin30 must be y/x not x/y

Question:

A tall tree AB and a tower CD are standing opposite to each other. A portion of the tree breaks off and falls on top of the tower making an angle of 30º. After some tine it falls again to the ground in front of the building, 6 m away from foot of the tree, making on angle of 45º, the height of the building is 10 m. Find the total height of the tree in meters before it broke.

Options:

$(22 + 12 \sqrt{3}) m$

$15 \sqrt{3} + 21$

$22 + 15 \sqrt{3}$

can’t determined

Solution:

Let the height of the broken portion of tree FB = x

Hence, BF = BD = BE = x

From the figure, total height of the tree = x+y+10.

In $Δ B E A$ ,

tan 45º = $\frac{A B}{A E}$

AE = y + 10 m

Now, In $Δ$ BDG,

$\frac{B G}{G D}$ = tan 30º, $\frac{y}{A E + 6} = \frac{1}{\sqrt{3}}$

$\sqrt{3}$ y = y + 16

$y = \frac{16}{\sqrt{3} - 1}$ $\frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$ $\Rightarrow$ y = 8 $(\sqrt{3} + 1)$

Now, In $Δ$ BDG,

Sin 30º = $\frac{x}{y}$

y = 2x

x = $4 (\sqrt{3} + 1)$

Total height of the tree = x + y + 10

= $4 (\sqrt{3} + 1) + 8 (\sqrt{3} + 1) + 10$

= $8 \sqrt{3} + 4 \sqrt{3} + 4 + 8 + 10$

= $22 + 12 \sqrt{3}$

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