Question

A boat sails directly toward a 250-m skyscraper that stands on the
edge of a harbor. The angular size of the building is the angle
formed by lines from the top and bottom of the building to the
observer (see figure). Complete parts (a) and (b) below.

Answer 1

a) The value of  $\theta$ as a function of "x" is $\theta = tan^{-1}\frac{250}{x}$

b) The rate of change of the angle with respect to x is -0.000308rad/m

Find the complete question and diagram attached

Form the right-angled triangle shown:

Adjacent side = x

Opposite side = 250m

angle of elevation  = $\theta$

Using the SOH CAH TOA identity

$tan\theta = \frac{opposite}{hypotenuse} \\tan \theta = \frac{250}{x} \\\theta = tan^{-1}\frac{250}{x}$

Hence the value of  $\theta$ as a function of "x" is $\theta = tan^{-1}\frac{250}{x}$

b) If x = 450m

$\theta = tan^{-1}\frac{250}{x}\\tan \theta = \frac{250}{x}$

Differentiate both sides with respect to "x"

$sec^2 \theta \frac{d\theta}{dx} = \frac{-250}{x^2} \\ \frac{d\theta}{dx} = \frac{1}{sec^2\theta} \times \frac{-250}{x^2}$

Substitute x = 450m and $\theta = 60^0$ into the resulting expression to have:

$\frac{d\theta}{dx} = \frac{1}{sec^260^0} \times \frac{-250}{450^2}\\\frac{d\theta}{dx} = (cos60)^2\times \frac{-250}{202,500}\\\frac{d\theta}{dx} = 0.5^2\times \frac{-250}{202,500}\\\frac{d\theta}{dx} =-0.000308$

Hence the rate of change of the angle with respect to x is -0.000308rad/m

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