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Glven: AABC with altitude h shown in the diagram. The altitude is perpendicular to BC.
Prove: Area of AABC = 1/2ab sin(C)

By the definition of the (sine ratio, area of triangle, or perimeter of triangle) , the area of ABC is 1/2 ah, then, sin (C)h/b bye the definition of the sine ratio. Using the (division property, multiplication property, substitution property,) bsin(C) = H. Lastly A= 1/2absin(C) by the (simplification, division property, multiplication property, substitution property)

Glven AABC with altitude h shown in the diagram The altitude is perpendicular to BC Prove Area of AABC 12ab sinC By the definition of the sine ratio area of tri class=

Respuesta :

Answer:

Area of triangle

Multiplication property of equality

Substitution property of equality

Step-by-step explanation:

Hope this helps

So, it is proven that the area of triangle ABC is  [tex]\frac{1}{2} * a*b*SinC[/tex].

What is a triangle?

A triangle is a 2D shape having three sides and the sum of all angles is 180°.

We have,

A ΔABC,

And,

AB = c

BC = a

CA = b

H = Altitude of the triangle,

Now,

Area of triangle [tex]=\frac{1}{2} * Base *Altitude[/tex]

i.e.

Area of triangle ABC [tex]=\frac{1}{2} * a*h[/tex]     .....(i)

Now,

Using trigonometric ratios,

[tex]Sin C = \frac{Perpendicular}{Hypotenuse}[/tex]

i.e.

[tex]Sin C = \frac{h}{b}[/tex]

We get,

h = b * SinC

Now,

putting this value in equation (i),

We get,

Area of triangle ABC [tex]=\frac{1}{2} * a*b*SinC[/tex]  

So,

It is proved that area of triangle ABC [tex]=\frac{1}{2} * a*b*SinC[/tex] ,

Hence, we can say that it is proven that the area of triangle ABC is  [tex]\frac{1}{2} * a*b*SinC[/tex], using trigonometric ratios and the triangle's area formula.

To know more about triangles click here

https://brainly.com/question/2272632

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