IBDP Maths SL 3.2 Use of sine, cosine and tangent ratios AA HL Paper 2- Exam Style Questions- New Syllabus

Apply the cosine rule to relate \(x\) and \(\theta\):

\( x^2=15^2+20^2-2(15)(20)\cos\theta \)

\( x=\sqrt{625-600\cos\theta} \)

Differentiate with respect to \(\theta\):

\( \frac{dx}{d\theta} =\frac{1}{2}(625-600\cos\theta)^{-\frac12}(600\sin\theta) =\frac{300\sin\theta}{\sqrt{625-600\cos\theta}} \)

Since \(\theta\) is decreasing,

\( \frac{d\theta}{dt}=-\frac{\pi}{60} \)

Using the chain rule,

\( \frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt} \)

The area of the triangle is

\( \frac12(15)(20)\sin\theta=140 \)

\( 150\sin\theta=140 \)

\( \sin\theta=\frac{14}{15} \)

Since \(0<\theta<\dfrac{\pi}{2}\),

\( \cos\theta = \sqrt{1-\left(\frac{14}{15}\right)^2} = \frac{\sqrt{29}}{15} \)

Hence

\( x = \sqrt{625-600\left(\frac{\sqrt{29}}{15}\right)} = \sqrt{625-40\sqrt{29}} \)

Substitute these values into the related-rates formula:

\( \frac{dx}{dt} = \frac{300\left(\frac{14}{15}\right)} {\sqrt{625-40\sqrt{29}}} \left(-\frac{\pi}{60}\right) \)

\( \frac{dx}{dt} \approx -0.724\text{ cm min}^{-1} \)

The negative sign indicates that the side \(AC\) is becoming shorter as the angle decreases.

✅ Answer: \(\boxed{\dfrac{dx}{dt}\approx -0.724\text{ cm min}^{-1}}\)