The angle of elevation of the of a tower from point a distance of 100 metres from its foot on a horizontal plane is found to be displaystyle 60-circ- Find the height of the tower173-2 m196 m144 m123-2 m

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Toppr Admin · Accepted Answer

Let the height of the tower AC be h metres Given distance BC - 100 m-xA0-x2220-ABC-60-x2218-x2234-tan60-x2218-ACBC-h100-x21D2-h-100-xD7-tan60-x2218-100-xD7-x221A-3-metre-100-xD7-1-732-173-2m

The angle of elevation of the top of a tower from point at a distance of $100$ metres from its foot on a horizontal plane is found to be $60^{\circ}$ . Find the height of the tower
$173.2 m$
$196 m$
$144 m$
$123.2 m$

Let the height of the tower AC be h metres Given distance BC = 100 m, $∠ A B C = 60^{\circ}$
$∴ tan 60^{\circ} = \frac{A C}{B C} = \frac{h}{100} \Rightarrow h = 100 \times tan 60^{\circ} = (100 \times \sqrt{3}) m e t r e$
$= 100 \times 1.732 = 173.2 m$

The angle of elevation of the top of a tower from point at a distance of 100 metres from its foot on a horizontal plane is found to be 60∘. Find the height of the tower173.2m196m144m123.2m

Let the height of the tower AC be h metres Given distance BC = 100 m, ∠ABC=60∘∴tan60∘=ACBC=h100⇒h=100×tan60∘=(100×√3)metre=100×1.732=173.2m

The angle of elevation of the top of a tower from point at a distance of $100$ metres from its foot on a horizontal plane is found to be $60^{\circ}$ . Find the height of the tower
$173.2 m$
$196 m$
$144 m$
$123.2 m$

Let the height of the tower AC be h metres Given distance BC = 100 m, $∠ A B C = 60^{\circ}$
$∴ tan 60^{\circ} = \frac{A C}{B C} = \frac{h}{100} \Rightarrow h = 100 \times tan 60^{\circ} = (100 \times \sqrt{3}) m e t r e$
$= 100 \times 1.732 = 173.2 m$