Q) A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7m. From a point on the plane, the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the flagstaff is 45°. Find the height of the tower.    

Ans:

Let’s start from the diagram for the question:

A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7m. From a point on the plane, the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the flagstaff is 45°.

Let ‘s take CD as the tower of height H and AD be the flagstaff of 7 m.

Point B is Q distance away from C.

Step 1: Let’s start from In Δ BCD, tan ∠ DBC = \frac{CD}{BC}

∴ tan 300 = \frac{1}{\sqrt 3}  = \frac{H}{Q}

∴ Q = H √3 …………. (i)

Step 2: Next, in Δ ABC, tan ∠ ABC = \frac{AC}BC}

∴ tan 450 = \frac{H + 7}{Q}

∴ 1 = \frac{H + 7}{Q}

∴ Q = H + 7  …………. (ii)

(Note: Here we calculate Q in terms of H. When we will get all H terms together and value of H will be calculated.)

Step 3: From equation (i) and equation (ii), we get:

H √3 = H + 7

∴ H √3 – H = 7

∴ H (√3 – 1) = 7

∴ H = \frac {7}{\sqrt 3 - 1}

∴ H = \frac {7}{\sqrt 3 - 1} \times \frac{ \sqrt 3 + 1}{\sqrt 3 + 1}

∴ H = \frac {7 (\sqrt 3 + 1)}{(\sqrt 3 - 1)(\sqrt 3 + 1)}

∴ H = \frac {7(\sqrt 3 + 1)}{(\sqrt 3)^2 - (1)^2}      (by (a  + b) ( a – b) = a2 – b2 )

∴ H = \frac {7(\sqrt 3 + 1)}{2}

∴ H = \frac {7(1.73  + 1)}{2}

∴ H = \frac{19.11}{2}

∴ H = 9.56 m

Therefore, height of the tower is 9.56 m

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