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Featured 3 months ago

In figure

AB = h ft = Height of the building

C is the point on the ground from which angle of elevation of building

D is the point on the ground(50 ft closer to the building from C) from which angle of elevation of building

Let BD = x ft

Now

Again

Featured 3 months ago

From Pythagoras, we have:

#sin^2 theta + cos^2 theta = 1#

If

#theta in [0, pi]#

#sin(theta) >= 0#

Hence:

#sin(arccos(x)) = sin(theta) = sqrt(1 - cos^2 theta) = sqrt(1-x^2)#

Note we can use the non-negative square root since we have already established that

Featured 3 months ago

The inverse sine function

#=cos((pi)/3)/sin((pi)/3)#

#=(1/2)/(sqrt(3)/2)#

#=1/sqrt(3)#

#=sqrt(3)/3#

Featured 3 months ago

The sides measure

We know the length of BD and of AC. However, we don't have an angle that is on the exterior (ex BDC, BCD or DBC), so we don't have enough information to use Cosine's Law. We will approach this problem by a different approach.

We know, by the properties of the parallelogram, that diagonals are cut into two equal parts at the point of intersection. Hence, if

We can now use the Law of Cosines to solve for side

To find the length of

Since

Hence, the two sides of the parallelogram measure

Hopefully this helps!

Featured 2 months ago

Let CB be the cliff . From point A the angle of elevation of the peak C of the cliff is

DF and DE are perpendiculars drawn from D on CB and AB.

Now

Let

For

Now

For

Featured 2 months ago

Given that

Let us consider

so

From (2) and (3) we get

Hence

**(a)**

**(b)**

Now the given equation becomes

Differentiating this equation w,r to t we get the rate of change of the price of stock as

In this equation the value of

So rate of positive change of price of stock will occur in this period i.e. Jan every year this means the stock appreciate s most in this period.

**(c)**

Taking

when t =4

Again

when t =8

So during the period **May -September** growth of price gets diminished i.e the price of the stock is actually **losing in this period.**

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