Answer:
AH = 4 units
Area of ΔABC = 16 square units
Step-by-step explanation:
Since ΔABC is an isosceles triangle,
AC = BC
∠CAB ≅ ∠ABC [Opposite angles of the equal sides in a triangle are equal]
In ΔABC,
m∠CAB + m∠ABC + m∠ACB = 180°
m∠ABC + m∠ABC + 150° = 180°
2m∠ABC = 30°
m∠ABC = 15°
Now we apply cosine rule in ΔABC,
AB² = AC² + BC² - 2(AB)(BC)cos(m∠ACB)
AB² = 8² + 8² - 2(8)(8)cos(150°)
AB² = 128 + 110.85
AB = [tex]\sqrt{238.85}[/tex]
AB = 15.45
In ΔABH,
sin(m∠ABH) = [tex]\frac{AH}{AB}[/tex]
sin(15°) = [tex]\frac{AH}{15.45}[/tex]
AH = 15.45sin(15°)
AH = 3.998
AH ≈ 4.0 units
HC = [tex]\sqrt{8^{2}-4^2}[/tex]
= [tex]\sqrt{48}[/tex]
= 6.93
Area of the triangle ABC = [tex]\frac{1}{2}(BC)(AH)[/tex]
= [tex]\frac{1}{2}(8)(4)[/tex]
= 16 square units
