Given triangle ABC with sides a,b,c, show that the area of the triangle is given by A=(a^2 sinB sinC)/(2sin(B+C))

Question

Need help! 

Answer 1

Hey Ed, here is how I got it:

 

To get it, you have to use right triangle but the key is that there is no angle A which means a is hypotenuse.

 

If a is hypotenuse, b or c are either base or height.

 

a² = b² + c²

 

From trigonometry, definition of sine is opp / hyp.  So, you will have sin B & sin C as either b/a or c/a (depends how you draw and label them but it will be the same at the end).

 

a² = (b/a)² + (c/a)²

 

From trig identities, sin (x + y) = sinx cosy + siny cosx

We can rewrite this as sin (B + C) = sinB cosC +sinC cosB

 

As we put this in order, (sin B) * (sin C) becomes (bc / a²).

Which means that (sin B) * (sin C) = (bc / a²) --->  a² sin B sin C = bc

Remember, a must be hypotenuse which means b or c are either base or height!!!  we got it (not just yet)

 

For bottom part,

From trig identities, sin (x + y) = sinx cosy + siny cosx
We can rewrite this as sin (B + C) = sinB cosC +sinC cosB

 

Let's do it.

 

sin B = (b/a) sin C= (c/a)  cos B = (c/a) cos C = (b/a)

sin (B + C) = sin B cos C +sin C cos B

sin (B + C) = (b/a) * (b/a) + (c/a) * (c/a) = b²/a² + c²/a² = (b² + c²) / a² = a² / a² = 1

 

Put everything we have and see if it is true:

 

Before we do that, revisit the formula for an area of triangle: base * height / 2.

 

(sin B) * (sin C) = (bc / a²) ---> a² sin B sin C = bc  (remember b or c are either base or height)

Which means bc is base * height!!!

 

sin (B + C) = (b/a) * (b/a) + (c/a) * (c/a) = b²/a² + c²/a² = (b² + c²) / a² = a² / a² = 1

sin (B + C) = 1

 

base * height / 2 = a² sin B sin C / 2 sin (B + C)

 

Rewrite this again: a² sin B sin C / 2 (1) = a² sin B sin C / 2

 

So, we finally have something that is very familiar: base * height / 2 = Area of triangle

 

Revisit the original problem and analyze:

 

If a² sin B sin C is base * height and sin (B + C) = 1, then

 

Area of triangle can be written as:

 

Area (triangle) = a² sin B sin C / 2 sin (B + C)

 

Yes, it is.

 

Good luck to you.  I hope you got it and hope you understand how I did it.

 

Answer 2

The area of the triangle is given by each of (1/2)absinC=(1/2)bcsinA=(1/2)acsinB.

B+C=π-A, since A+B+C=π

We now want to show that Area=a²sinB sinC)/(2sin(π-A)) but sin(π-A)=sinA

We need to show Area=a²sinB sinC)/(2sinA) Multiplying numerator and denominator by (b/2)(c/2)

rewrite as Area=(1/2)acsinB×(1/2)absinC/2(b/2)(c/2)sinA=Area²/Area=Area