# What will be the Height of Tower?If

The shadow of a tower standing on a level plane is found to be 40m longer when the sun’s altitude is 45^{o} than when it is 60^{o}. The height of the tower is:

- 30(3+sqrt. 3)m
- 40 (3+sqrt. 3)m
- 10(3+sqrt. 3)
- 20 (3+sqrt. 3) m

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**Answer:(4) (4) 20(3 + Ö3).**

**Explanation:-**In the figure-

Let PQ be the height of the tower is h m and QR be its shadow of length x m when the sun’s altitude is 60

^{o}.

And QS be the length of the shadow, when the angle of elevation is 30

^{o}.

According to the question,

QS is 40 m longer than QR.

Therefore, QS = (40 + x) m

Now, in the right triangle ABC,

tan60^{o} = PQ/QR

Therefore, Ö3 = h/x { tan60^{o} = Ö3}

Then, h = xÖ3 …………………………….. (1)

In right triangle PQS,

tan45^{o} = PQ/QS

Therefore, 1 = h/(x + 40) {tan45^{o} = 1}

h = (x + 40) ………………….. (2)

From equation (1) & (2),

xÖ3 = (x + 40)

xÖ3 = x + 40

xÖ3 = x + 40

xÖ3 – x = 40

x(Ö3 – 1) = 40

x = 40/(Ö3 – 1)

h = 40 x Ö3/(Ö3 – 1) { From equation (1) }

h = 40Ö3(Ö3 + 1)/( Ö3 – 1)( Ö3 + 1)

h = {40Ö3 x Ö3 + 40Ö3}/(3 – 1) {a^{2} – b^{2} = (a + b)(a – b)}

h = {40 x 3 + 40Ö3}/2

h = 20(3 + Ö3)

Hence, the answer is (4) 20(3 + Ö3).