Respuesta :
Answer:
There are no triangles possible here!
Step-by-step explanation:
Answer with explanation:
The Description of triangle is :
A=51°, b=11 and c=7
We will use cosine law to find length of third side
[tex]\cos A=\frac{b^2+c^2-a^2}{2\times b \times c}\\\\ \ cos 51^{\circ}=\frac{11^2+7^2-a^2}{2\times 11 \times 7}\\\\0.63=\frac{121+49-a^2}{154}\\\\ 154 \times 0.63=170-a^2\\\\97.02=170 -a^2\\\\a^2=170 -97.02\\\\a^2=72.98\\\\a=8.55[/tex]
a=8.55(approx)
Now, we will use Sine law to find other angle.
[tex]\frac{a}{\sin A^{\circ}}=\frac{b}{\sin B^{\circ}}\\\\ \frac{8.55}{\sin 51^{\circ}}=\frac{11}{\ sinB^{\circ}}\\\\ \ sinB^{\circ}=\frac{11 \times0.78}{8.55}\\\\\sin B^{\circ}=\frac{8.55}{8.55} \\\\ sinB^{\circ}=1\\\\B=90^{\circ}[/tex]
We will use angle sum property of triangle to find measure of angle C .
→∠A + ∠B+∠C=180°
→ 51°+90°+∠C=180°
→∠C=180° - 141°
→∠C=39°
Side Opposite to Angle A= 8.55 unit
∠ B=90°
∠ C=39°
