Question

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

ɴᴏᴛᴇ :- ᴘʟᴇᴀsᴇ ɢɪᴠᴇ ғᴜʟʟ sᴏʟᴜᴛɪᴏɴ​

Answer 1

Answer:

From fig, PB = 4m, QB = 9m. Let angle of elevation from P be α and angle of elevation from Q be β. Therefore, AB = 6. Hence, height of tower is 6m.

Answer 2

Answer:

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⇒ Let the height of the tower (CD) = h m

⇒ AC be a horizontal line on a ground.

⇒ A and B be the two points on a line at a distance of 9m and 4m from base of tower

Let angle CBD = θ ; then angle CAD = 90 - θ

(The complementary means the sum of 2 angles is 90°)

From right angled ΔACD,

${ \sf{Tan(90-θ) = \frac{ CD }{AC} }}$

${ \sf{Cotθ = \frac{h}{9} .....(1)}}$

From right angled Δ BCD,

${ \sf{Tanθ = \frac{CD }{BC} }}$

${ \sf{Tanθ = \frac{h}{4} .....(2)}}$

From (1) & (2) we get,

${ \to{ \sf{Tanθ \times \: Cotθ = \frac{h}{9} \times \frac{h}{4} }}}$

${ \to{ \sf{1 = \frac{ {h}^{2} }{36} }}}$

${ \to{ \sf{ {h}^{2} = 36}}}$

${ \to{ \sf{h = \sqrt{36} = 6}}}$

Therefore,

• Height of tower = 6m