CAIE IGCSE Mathematics Extended 6 Trigonometry Topic Practice

$\frac{AB}{18}=\cos60^\circ$ or $AB=18\cos60^\circ$$\cos60^\circ=\frac{1}{2}$ and AB=9
Correct use of Pythagoras' theorem, i.e. $CD^2=6^2+(17-\text{their }AB)^2$ oe
Correct evaluation for their AB: $CD^2=36+64$ or $CD=\sqrt{36+64}$100=10

$39+9\sqrt{3}$
B3 for $BE=9\sqrt{3}$ or for answer equivalent to $39+9\sqrt{3}$ but not in required form
OR M2 for $\frac{BE}{18}=\frac{\sqrt3}{2}$ oe or better
or M1 for $\frac{BE}{18}=\sin60^\circ$ oe or better
M1 for 18+17+10+ their BE-6 oe
M2 for $\sqrt{18^2-\text{their }9^2}$ oe
or M1 for $BE^2+(\text{their }9)^2=18^2$ oe

$\cos ACB=\frac{15^2+8^2-20^2}{2\times15\times8}$
117.54 to 117.55
A1 for $-\frac{37}{80}$, $-\frac{111}{240}$, or -0.4625

53.2 or 53.19 to 53.23
M1 for $0.5\times8\times15\times\sin117.5^\circ$ oe

15.5 or 15.52 to 15.53
M1 for $15^2+4^2$ oe

7.1 or 7.13 or 7.125 to 7.126
M2 for $\tan P=\frac{4-3}{8}$ oe or M1 for vertical line =4-3

11.5 nfww or 11.48 to 11.49...
B1 for height 3.5
M2 for $15^2+4^2-2\times15\times4\cos117.5^\circ$
M1 for $\tan=\frac{3.5}{\text{their }17.216...}$ oe

8.09 or $8.094 \ldots$

M1 for $\tan 34=\frac{A B}{12}$ oe

98.3 or 98.28 to 98.31

M1 for $\frac{1}{2} \times 10 \times 12 \sin 56$ oe

M1 for $\frac{1}{2} \times 12 \times$ their (a) oe

43[.0] to 43.1

M2 for $[B C=] \sqrt{10^{2}+12^{2}-2 \times 10 \times 12 \cos 56}$ or $\mathbf{M 1}$ for $\left[B C^{[2]}=\right] 10^{2}+12^{2}-2 \times 10 \times 12 \cos 56$

M2 for $[A D=] \frac{12}{\cos 34}$ oe or $\mathbf{M 1}$ for $\cos 34=\frac{12}{A D}$ oe

6.71 or 6.706 to $6.710 \ldots$

M2 for $\sin 34=\frac{\text { dist }}{12}$ oe or
$\frac{1}{2} \times 12 \times$ their $(a)=\frac{1}{2} \times$ theirAD × dist oe
or M1 for recognition of perpendicular distance