Question

A tree grows at a uniform rate vertically. Angle of elevation of its top is 30° from a point 30 m away from its foot on ground. If the angle of elevation changes to 60° in 100 days then the rate at which tree is growing is equal to
pls tell fast

Answer 1

Step-by-step explanation:

tan60

∘

=

x

3000

3

x=

3

3000

3

=3000m

tan30

∘

=

1/

3

3000

3

=9000m

distance covered = 9000- 3000

= 6000 m.

speed =

30

6000

=200m/s

hope it will help u

Answer 2

Answer: The rate at which tree is growing is 0.25 m/day.

Step-by-step explanation:

Given: Angle of elevation = 30°

Distance between foot and ground: 30 m

Let's find the initial length of the tree($X^{1}$)

We will use trigonometry to find the initial length of the tree, refer to the attached image.

We have, the measure of the angle, and it's adjacent side. We need to find the hypotenuse($X^{1}$).

The trigonometric formula which relates the adjacent and hypotenuse is cosine. It is given by,

cosine = $\frac{adjacent}{hypotenuse}$

∴ cos 30° =  $\frac{30}{X^{1} }$

we have, cos 30° =  $\frac{\sqrt{3} }{2}$

hence, $\frac{\sqrt{3} }{2}$ =  $\frac{30}{X^{1} }$

solving, we get,  $X^{1}$ = $\frac{60}{\sqrt{3} }$

multiplying numerator and denominator by $\sqrt{3}$

i.e, $X^{1}$ = $\frac{60\sqrt{3}}{ 3 }$ = 20$\sqrt{3}$

∴ $X^{1}$= 20$\sqrt{3}$ m

Now let's find the final length of the tree($X^{2}$)  after 100 days

We will use the same trigonometric formula, with angle 60°

(to find the final length of the tree, refer to the second figure in the attached image.)

∴ cos 60° =  $\frac{30}{X^{2} }$

we have, cos 60° =  $\frac{1 }{2}$

hence, $\frac{1 }{2}$ =  $\frac{30}{X^{2} }$

solving, we get,  $X^{2}$ = 60 m

Now, rate of growth = (final length - initial length) / number of days

∴ Rate of growth = ($X^{2}$ -  $X^{1}$ ) / 100

i.e, (60 - 20$\sqrt{3}$ ) / 100

i.e, (60 - 20 x 1.732 ) / 100

= 25.36 /100

= 0.25 m/day

Hence the rate at which tree is growing is 0.25 m/day.