The leg of an isosceles triangle is 16 and the measure of one of the angles is 150. Find the area of the triangle.

Area of isosceles triangle is 63.98 square units.Solution:Note: Refer the image attached belowGiven 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 triangleFrom 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 cosineOn 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 = 16As 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]