Question

Length of two sides of a triangle are $6$ centimetres and $5$ centimetres. If their included angle is ${50}^o$. What is the area of the triangle?
Also find the length of its third side.
$(\sin {50}^o=0.77,\cos {50}^o=0.64,\tan {50}^o=1.19)$

Answer 1

Consider the following figure of triangle:
Area of triangle $=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times 6\times AD=3\times AD$
From right-angled $△ABD$,
$\sin {50}^o=\frac{AD}{AB}$
$AD=AB\times \sin {50}^o=5\times 0.77=3.85cm$
Thus, area of triangle $3\times 3.85=11.55{cm}^2$
Now, $\cos {50}^o=\frac{BD}{AB}$
$\Rightarrow BD=AB\times \cos {50}^o=5\times 0.64=3.2cm$
Thus, $DC=6-3.2=2.8cm$
$\Rightarrow AC=\sqrt{{\left(2.8\right)}^2+{\left(3.85\right)}^2}=\sqrt{7.84+14.82}=\sqrt{22.66}cm$