IB DP Math MAI SL : IB Style Mock Exams – Set 3 Paper 1

(a)
Use the cosine rule in \(\triangle NPV\):
\(NV^2 = NP^2 + PV^2 – 2(NP)(PV)\cos(\widehat{VPN})\)
\(NV^2 = 20^2 + 25^2 – 2(20)(25)\cos 55.5^\circ\)
\(NV^2 = 400 + 625 – 1000\cos 55.5^\circ\)
\(\cos 55.5^\circ \approx 0.565772\)
\(NV^2 \approx 1025 – 565.772 = 458.228\)
\(NV = \sqrt{458.228} \approx 21.4 \text{ m} \quad (21.4148\ldots)\)
\(\boxed{21.4\ \text{m}}\)

(b)
Apply the sine rule in \(\triangle NPV\):
\(\frac{\sin(\widehat{PNV})}{PV} = \frac{\sin(\widehat{VPN})}{NV}\)
\(\frac{\sin(\widehat{PNV})}{25} = \frac{\sin 55.5^\circ}{21.4148\ldots}\)
\(\sin(\widehat{PNV}) = \frac{25 \times \sin 55.5^\circ}{21.4148\ldots} \approx \frac{25 \times 0.824126}{21.4148} \approx 0.9619\)
\(\widehat{PNV} \approx \sin^{-1}(0.9619) \approx 74.2^\circ \quad (74.1749\ldots^\circ)\)
\(\boxed{74.2^\circ}\)

(c)
The shortest distance from \(P\) to line \(NV\) is the perpendicular distance.

In \(\triangle PNV\), using side \(NP = 20\) and \(\widehat{PNV} \approx 74.2^\circ\):
\(d = NP \times \sin(\widehat{PNV})\)
\(d \approx 20 \times \sin 74.1749^\circ\)
\(\sin 74.1749^\circ \approx 0.9619\)
\(d \approx 20 \times 0.9619 \approx 19.2 \text{ m}\)
(Alternatively, use area: \( \text{Area} = \frac12 (NP)(PV)\sin 55.5^\circ\), then \(d = \frac{2\times\text{Area}}{NV}\).)
\(\boxed{19.2\ \text{m}}\)