Determine if there are zero, one, or two triangles for the following:

Answer with explanation:
1.
In Δ ABC
m ∠A=48°
a=10 m
b=12 m
Using Sine Law
[tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\\frac{10}{\sin 48^{\circ}}=\frac{12}{\sin B}\\\\ \ sinB^{\circ}=\frac{12 \times \sin 48^{\circ}}{10}\\\\\ sinB^{\circ}=\frac{12 \times 0.7431}{10}\\\\\ sinB^{\circ}=0.9\\\\B=65^{\circ}[/tex]
So, Angle C can be obtained by using angle sum property of triangle.
So,This is one kind of triangle, having measure of three Angles are , 48°, 65°,67°.
2.
≡≡Now, we will use cosine law to find if there is another triangle having measure of one angle 48°.
[tex]\cos 48^{\circ}=\frac{12^2+c^2-10^2}{2 \times 12 \times c}\\\\0.67=\frac{144+c^2-100}{24 c}\\\\16.08 c=44+c^2\\\\100 c^2 -1608 c +4400=0\\\\25 c^2-402 c+1100=0\\\\c=12.58\\\\c=3.50[/tex]
Sum of two sides of triangle should always be greater than third side.
So, c=3.50
and, c= 12.58
Triangle of two type is possible having dimension, a=10 m, b=12 m, c=12.58 m and a=10 m, b=12 m, c=3.50 m.
But,when side c=12.58 m, then ∠C=67°.
So, There are two triangles possible in this case,out of which one is Congruent with triangle obtained in case 1.
So,There are two distinct Triangles.
⇒⇒⇒Two Triangles