How to draw this triangle?

Question

The statement is : In a triangle $ABC$, angle $A$ is an obtuse angle such that $\sin A= \dfrac{3}{5}$ and $\sin B = \dfrac{5}{13}$.

I wonder how to draw it.

Answer 1

If you want $\sin(A)=3/5$ draw a circle with radius $5$, see where it intersects with a point of height $3$ -- that will give you the angle. Same for angle $B$. This is what it looks like :

Answer 2

As $A$ is obtuse, $A>90^o$

So $\sin(A)$ is positive and $\cos(A)$ is negative ( $2^{nd}$ quadrant).
Find $\cos(A) < 0$

$B<90^o$ and $C<90^o$ and $B+C <90^o$ (as $A+B+C = 180^o$)

Calculate

$\bullet \cos(B)>0$ (in $1^{st}$ quadrant)
$\bullet \sin(A+B) = \sin A \cos B+ \cos A\sin B = \sin(180^o - C) = \sin(C)$

Then find $\angle A,\angle B, \angle C$ from the sines.

Answer 3

Since $\angle A$ is obtuse, we obtain: $$\cos\measuredangle A=-\sqrt{1-\sin^2\measuredangle A}=-\frac{4}{5}.$$ Also, $$\cos\measuredangle B=\frac{12}{13}$$ and use $$\sin\measuredangle C=\sin(180^{\circ}-\measuredangle A-\measuredangle B)=\sin(\measuredangle A+\measuredangle B)=...$$