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

k = 1

P(t) = (t + 20)²

Step-by-step explanation:

P' = k√P = k P⁰•⁵

To solve for k,

P' = 10 rabbits/month

P = 100 rabbits

10 = k √100

10 = 10k

k = 1

To solve for P(t)

P' = dP/dt

(dP/dt) = kP⁰•⁵

dP/P⁰•⁵ = k dt

P⁻⁰•⁵ dP = k dt

∫ P⁻⁰•⁵ dP = ∫ k dt

2P⁰•⁵ = kt + c

At t = 0, P = 100

2(100)⁰•⁵ = 0 + c

2 × 10 = c

c = 20

2P⁰•⁵ = kt + c

2P⁰•⁵ = kt + 20

Recall, k = 1

2P⁰•⁵ = t + 20

P⁰•⁵ = (t + 20)

P(t) = (t + 20)²