# Area of the right triangle

In the right triangle $ABC$ the hypotenuse $AB = 4\sqrt{2}$ and $\angle BAC = 75°$ are given. What is the area?

I solved this using $\sin(30°/2)$ and finding the side length, but that took 3-4 minutes, there is supposedly an easier solution, but I am unable to find it.

• Find $cos(75)$ and $sin(75)$ from the Triangle and use $sin2x=2sinxcosx$ – Ekaveera Kumar Sharma Jan 7 '16 at 14:58
• Let $t=\tan 15^{\circ}$. Then $\sin 30^{\circ}=\frac{1}{2}=\frac{2t}{1+t^2}$. Solve the quadratic equation to find $t$. Then solve the system $t=\frac{a}{b}$ and $a^2+b^2=16$ to find the side lengths. – user236182 Jan 7 '16 at 15:05
• I meant $a^2+b^2=32$. – user236182 Jan 7 '16 at 15:27

In the right triangle $\triangle ABC$, find $CA$ and $BC$ to use in $A=\frac12 bh$. So, $A=16 \cos 75 \sin 75$.
$$\sin 30^{\circ}=2\sin 15^{\circ}\cos 15^{\circ}=2\cdot \frac{AC}{4\sqrt{2}}\cdot \frac{BC}{4\sqrt{2}}=\frac{1}{2}$$
$$\implies \frac{AC\cdot BC}{2}=4$$
Draw median $CM$ and altitude $CH$, we have $\angle CMH =30^\circ$ and we know $CM = \frac{AB}{2}$ So $CH=\frac{CM}{2}=\frac{AB}{4}$
Use cosine rule or sine rule or use $\cos(A+B)=\cos A\cos B-\sin A\sin B$ or when one side, two angles are given $\text{area}=\dfrac{c^2\sin(A-B)\sin(A+B)}{4\sin C}$. So here $C=90$, $A=75$, $B=15$, $c=4\sqrt2$. Thats all.