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Mathematics

Angle of Depression

A tower is ob...

Question

A tower is observed from two stations $A$ and $B$ where $B$ is East of $A$ at a distance $100$ metres. The tower is due North of $A$ and due North-West of $B$ . The angles of elevation of the tower from $A$ and $B$ are complementary. Find the height of the tower.

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Solution

$h = A P tan θ, h = B P tan (90 - θ)$ .
$∴ h^{2} = A P . B P$ ... $(1)$
Clearly $A P = A B = 100$ .
$∴ B P \sqrt{100^{2} + 100^{2}} = 100 \sqrt{2}$ .
Putting the values of $A P$ and $B P$ in $(1)$ , we get
$h^{2} = 100 \times 100 \sqrt{2}$
$∴ h = 100 {(2)}^{1 / 4}$ .

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A tower is observed from two stations A and B where B is East of A at a distance 100 metres. The tower is due North of A and due North-West of B. The angles of elevation of the tower from A and B are complementary. Find the height of the tower.

h=APtanθ,h=BPtan(90−θ).∴h2=AP.BP ...(1)Clearly AP=AB=100.∴BP√1002+1002=100√2.Putting the values of AP and BP in (1), we geth2=100×100√2∴h=100(2)1/4.

A tower is observed from two stations $A$ and $B$ where $B$ is East of $A$ at a distance $100$ metres. The tower is due North of $A$ and due North-West of $B$ . The angles of elevation of the tower from $A$ and $B$ are complementary. Find the height of the tower.

$h = A P tan θ, h = B P tan (90 - θ)$ .
$∴ h^{2} = A P . B P$ ... $(1)$
Clearly $A P = A B = 100$ .
$∴ B P \sqrt{100^{2} + 100^{2}} = 100 \sqrt{2}$ .
Putting the values of $A P$ and $B P$ in $(1)$ , we get
$h^{2} = 100 \times 100 \sqrt{2}$
$∴ h = 100 {(2)}^{1 / 4}$ .