Mathematics College

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

Answers

Answer 1

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

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What is probability?

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

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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

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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