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Q.4.
A flagstaff of height 7 metres stands on the top of a tower. The angles subtended by the tower and the flagstaff to a point on the ground are 45° and 15° respectively. Find the height of the tower.with figure

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2.56m⠀

StepbyStepexplanation:

Let AC be the 7 meters high Flagstaff

And BC be the tower

Let the height of BC = x

The height of AB will be (x + 7)m

∠ADB = 45° and ∠CDB = 15°

(as given in the question)

Byusingtrigonometryformula

$$\sf \tan(A) = \frac{perpendicular}{base}$$

So tan 45 will be

$$\sf \implies \tan45{ \degree} = \frac{AB}{BD} \\ \\ \sf \implies \tan45{ \degree} = \frac{x + 7}{BD} \\ \\ \sf \implies 1 = \frac{x + 7}{BD} \\ \\ \sf \implies BD = (x + 7)m$$

And tan 15 will be

$$\sf \implies \tan15{ \degree} = \frac{BC}{BD} \\ \\ \sf \implies \tan15{ \degree} = \frac{x}{BD} \\ \\ \sf \implies 0.2679 = \frac{x}{BD} \\ \\ \sf \implies BD = \frac{x}{0.2679} \\ \\ \sf \implies BD = 3.7327x$$

Using both Values of BD to find x

$$\sf x + 7 = 3.7327x \\ \\ \sf 7 = 2.7327x \\ \\ \sf \frac{7}{2.7327} = x \\ \\ \sf \boxed{2.56 = x}$$

Theheightofthetoweris2.56m
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