In an isosceles triangle the length of the legs is 8 cm and the angle between them is 150°. Find the altitude to the legs and the area of the triangle. Find AH and Aᴬᴮᶜ

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 =

AB = 15.45

In ΔABH,

sin(m∠ABH) =

sin(15°) =

AH = 15.45sin(15°)

AH = 3.998

AH ≈ 4.0 units

HC =

     =

     = 6.93

Area of the triangle ABC =

                                          =

                                          = 16 square units

the answer is: the length is 5/6 yards.

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step-by-step explanation: