IB Maths AA HL 3.2 Geometry and Trigonometry Ahl Content Topic Practice

$\tan \theta=\frac{50}{y-x}$ OR $\cot \theta=\frac{y-x}{50}$$y=x+50 \cot \theta$
AG
Note: y-x may be identified as a length on a diagram, and not written explicitly.

(b) EITHER
choosing two appropriate angles, for example $60^{\circ}$ and $45^{\circ}$$\sin 105^{\circ}=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}$ and
$\cos 105^{\circ}=\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ}$$\sin 105^{\circ}+\cos 105^{\circ}=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}$$=\frac{1}{\sqrt{2}}$
OR
attempt to square the expression
$\left(\sin 105^{\circ}+\cos 105^{\circ}\right)^{2}=\sin ^{2} 105^{\circ}+2 \sin 105^{\circ} \cos 105^{\circ}+\cos ^{2} 105^{\circ}$$\left(\sin 105^{\circ}+\cos 105^{\circ}\right)^{2}=1+\sin 210^{\circ}$$=\frac{1}{2}$$\sin 105^{\circ}+\cos 105^{\circ}=\frac{1}{\sqrt{2}}$

EITHER
$\sin x+\sin 3 x=\cos x \Rightarrow \frac{1-\cos 4 x}{2 \sin x}=\cos x$$\Rightarrow 1-\cos 4 x=2 \sin x \cos x,(\sin x \neq 0)$$\Rightarrow 1-\left(1-2 \sin ^{2} 2 x\right)=\sin 2 x$$\Rightarrow \sin 2 x(2 \sin 2 x-1)=0$$\Rightarrow \sin 2 x=0$ or $\sin 2 x=\frac{1}{2}$$2 x=\pi, 2 x=\frac{\pi}{6}$ and $2 x=\frac{5 \pi}{6}$

OR
$\sin x+\sin 3 x=\cos x \Rightarrow 2 \sin 2 x \cos x=\cos x$$\Rightarrow(2 \sin 2 x-1) \cos x=0,(\sin x \neq 0)$$\Rightarrow \sin 2 x=\frac{1}{2}$ or $\cos x=0$$2 x=\frac{\pi}{6}, 2 x=\frac{5 \pi}{6}$ and $x=\frac{\pi}{2}$

THEN
$\therefore x=\frac{\pi}{2}, x=\frac{\pi}{12}$ and $x=\frac{5 \pi}{12}$