Further Pure Mathematics 4PM1 / Trigonometry / 1.591.59: Sine Rule and Triangle Area4PM1/1R/June/2019 — Question 2 · 6 marksMark as done · Save for laterFigure 2 shows triangle ABC\displaystyle ABCABC in whichAB=2x cm,AC=3x cm,BC=4x cm.AB=2x\text{ cm}, \qquad AC=3x\text{ cm}, \qquad BC=4x\text{ cm}.AB=2x cm,AC=3x cm,BC=4x cm.(a) Show thatsinABC=31516.\sin ABC=\frac{3\sqrt{15}}{16}.sinABC=16315.(4)Given that the area of triangle ABC\displaystyle ABCABC is751564 cm2,\frac{75\sqrt{15}}{64}\text{ cm}^2,647515 cm2,(b) find the value of x\displaystyle xx.(2)▸ Mark scheme← 1.58: 4PM1/1R/June/2019 — Question 11.60: 4PM1/2/June/2019 — Question 4 →