Area of isosceles triangle is 63.98 square units.
Solution:
Note: Refer the image attached below
Given that leg of an isosceles triangle is 16; Measured of one of the angle is 150; Need to determine area of triangle.
Consider the figure ABC is required isosceles triangle
From given information,
AC = CB = 16 and [tex]\angle \mathrm{ACB}=150^{\circ}[/tex]
So we have two sides and angle between them. Let us use law of cosine
On applying law of cosine on triangle ABC we get,
[tex]\begin{array}{l}{\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{CB}^{2}-2 \times \mathrm{AC} \times \mathrm{CB} \cos \mathrm{C}} \\\\ {=>\mathrm{AB}^{2}=16^{2}+16^{2}-2 \times 16 \times 16 \cos 150} \\\\ {\Rightarrow \mathrm{AB}^{2}=16^{2}+16^{2}+443.4} \\\\ {\Rightarrow \mathrm{AB}^{2}=955.4} \\\\ {\Rightarrow \mathrm{AB}=30.91}\end{array}[/tex]
Now we are having three sides of triangle,
AB = 30.91; BC = 16 and CA = 16
As three sides are given, we can apply heron’s formula to determine area of triangle ABC. According to herons formula,
[tex]\begin{array}{l}{\text { Area of Triangle }=\sqrt{s(s-a)(s-b)(s-c)}} \\\\ {\text { Where } s=\frac{a+b+c}{2}}\end{array}[/tex]
In our case a = AB = 30.91; b = BC = 16; c = CA = 16
[tex]\begin{array}{l}{\text { So } s=\frac{A B+B C+C A}{2}=\frac{30.91+16+16}{2}=31.455} \\\\ {\text { Area of Triangle } A B C=\sqrt{31.455(31.455-30.91)(31.455-16)(31.455-16)}} \\\\ {\Rightarrow \text { Area of Triangle } A B C=\sqrt{31.455 \times 0.545 \times 15.455 \times 15.455}=\sqrt{4094.72}=63.98}\end{array}[/tex]
