When Bermuda \({\text{(B)}}\), Puerto Rico \({\text{(P)}}\), and Miami \({\text{(M)}}\) are joined on a map using straight lines, a triangle is formed. This triangle is known as the Bermuda triangle.

According to the map, the distance \({\text{MB}}\) is \(1650\,{\text{km}}\), the distance \({\text{MP}}\) is \(1500\,{\text{km}}\) and angle \({\text{BMP}}\) is \(57^\circ \).

Calculate the distance from Bermuda to Puerto Rico, \({\text{BP}}\).

Calculate the area of the Bermuda triangle.

\({\text{B}}{{\text{P}}^2} = {1650^2} + {1500^2} - 2 \times 1650 \times 1500\,\cos \,(57^\circ )\)                  (M1)(A1)

\(1510\,({\text{km}})\,\,\,\left( {1508.81...\,({\text{km}})} \right)\)       (A1)     (C3)

Notes: Award (M1) for substitution in the cosine rule formula, (A1) for correct substitution.

\(\frac{1}{2} \times 1650 \times 1500 \times \sin \,57^\circ \)        (M1)(A1)

\( = 1\,040\,000\,({\text{k}}{{\text{m}}^2})\,\,\,\left( {1\,037\,854.82...\,({\text{k}}{{\text{m}}^2})} \right)\)       (A1)    (C3)

Note: Award (M1) for substitution in the area of triangle formula, (A1) for correct substitution.

Question 6: Non-right angle trigonometry.
Instead of using the law of cosines weaker candidates substituted into Pythagoras’ theorem and likewise used \(A = \frac{1}{2}bh\) instead of \(A = \frac{1}{2}ab\sin C\). Those that did select the correct formula almost always made correct substitutions but were not always able to calculate the correct answer.

Question 6: Non-right angle trigonometry. Instead of using the law of cosines weaker candidates substituted into Pythagoras’ theorem and likewise used \(A = \frac{1}{2}bh\) instead of \(A = \frac{1}{2}ab\sin C\). Those that did select the correct formula almost always made correct substitutions but were not always able to calculate the correct answer.