Answer:
Perimeter of ΔABC = 16.46 units
Step-by-step explanation:
By applying cosine rule (CAH) in the given triangle ADB,
cos(∠BAC) = [tex]\frac{\text{Adjacent side}}{\text{Hypotenuse}}[/tex]
cos(55°) = [tex]\frac{3}{AB}[/tex]
AB = 5.23 units
Since, ΔABC is an isosceles triangle
AB = BC = 5.23 units
AD = DC = 3 units [Perpendicular drawn from a vertex of an isosceles triangle is a perpendicular bisector of the opposite side]
Therefore, AC = AD + DC = 6 units
Now perimeter of ΔABC = AB + BC + AC
= 5.23 + 5.23 + 6
= 16.46 units
