1. Problem statement: In the parallelogram $ABCD$, $AB=10$ cm, $AD=7$ cm and $\angle DAB=60^\circ$.\n 2. Diagram suggestion: Place $A$ at the origin, let $AB$ lie on the positive $x$-axis so $B=(10,0)$, and place $D$ so that $AD=7$ and $\angle DAB=60^\circ$, for example $D=(7\cos60^\circ,7\sin60^\circ)$.\n 3. Formula used: For a parallelogram with adjacent sides $a,b$ and included angle $\theta$ the diagonals are given by the vector sums and differences, which give the formulas below.\n $$AC=\sqrt{a^2+b^2+2ab\cos\theta}$$\n $$BD=\sqrt{a^2+b^2-2ab\cos\theta}$$\n 4. Important rules to remember: Opposite sides of a parallelogram are equal and parallel.\n Opposite angles are equal and adjacent angles are supplementary.\n The diagonals of a parallelogram bisect each other.\n 5. Apply the formula to compute $AC$.\n $$AC^2=10^2+7^2+2\cdot10\cdot7\cdot\cos60^\circ$$\n $$=100+49+140\cdot\frac{1}{2}$$\n $$=100+49+\frac{140}{2}$$\n $$=100+49+\frac{\cancel{140}}{\cancel{2}}=100+49+70$$\n $$=219$$\n $$AC=\sqrt{219}\approx14.81\,\text{cm}$$\n 6. Apply the formula to compute $BD$.\n $$BD^2=10^2+7^2-2\cdot10\cdot7\cdot\cos60^\circ$$\n $$=100+49-140\cdot\frac{1}{2}$$\n $$=100+49-\frac{140}{2}$$\n $$=100+49-\frac{\cancel{140}}{\cancel{2}}=100+49-70$$\n $$=79$$\n $$BD=\sqrt{79}\approx8.89\,\text{cm}$$\n 7. Final answers: The lengths of the diagonals are $AC=\sqrt{219}\approx14.81$ cm and $BD=\sqrt{79}\approx8.89$ cm.\n 8. Figure note for a complete white background: Draw points $A(0,0)$, $B(10,0)$, $D(7\cos60^\circ,7\sin60^\circ)$ and then draw $C=B+D-A$ to complete the parallelogram; the background should be plain white and labels clear for class VIII ICSE level.\n