The angles of elevation of a tower from two points at a distance of $4 m$ and $9 m$ from the base of the tower and in the same straight line with it are complimentary. Prove that the height of the tower is $6 m .$

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Solution

Given $α + β = 90$

h = Height of tower

Now $t a n α = \frac{h}{P B} - - - - - - (1)$

And $t a n β = \frac{h}{B Q}$

$t a n (90 - α) = \frac{h}{B Q}$

$c o t α = \frac{h}{B Q}$

$t a n α = \frac{B Q}{h} - - - (2)$

From (1) and (2)

$\frac{h}{P B} = \frac{B Q}{h}$

$⟹ h^{2} = P B \times B Q$

$= 4 \times 9$

$= 36$

$h = 6 m$

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The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. The height of the tower is ___ m.

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The angles of elevation of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complimentary. Prove that the height of the tower is 6m.

Given α+β=90 h = Height of tower Now tanα=hPB−−−−−−(1) And tanβ=hBQ tan(90−α)=hBQ cotα=hBQ tanα=BQh−−−(2) From (1) and (2) hPB=BQh ⟹h2=PB×BQ =4×9 =36 h=6m

The angles of elevation of a tower from two points at a distance of $4 m$ and $9 m$ from the base of the tower and in the same straight line with it are complimentary. Prove that the height of the tower is $6 m .$