Question

A boat sails directly toward a 250-m skyscraper that stands on the

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

Learn more here: brainly.com/question/15580615