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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Answer:

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Step-by-step explanation:

?
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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

xddd

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What is probability?

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The question is incomplete.

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

It is given that:

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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