Question

In triangle ABC , angleA =60 .prove that BC2= AB2 + AC2-AB.AC
Please answer

Answer 1

angle A=60 degree
so it may be be equilateral triangle
so all sides are equal.
let the side be X
bc2= ab2 + ac2 - ab
X2= X2 + X2 - x
0=0

Answer 2

Answer:

Given data:

In ∆ABC,

Angle A = 60°

To prove: BC²= AB2 + AC² - AB.AC

Proof:

Let’s draw a line from C to AB meeting at D such that CD is perpendicular to AB.

In ∆CBD, using the Pythagoras theorem, we have

BC² = CD² + BD²

⇒ CD² = BC² - BD²…… (i)

Also,

In ∆CAD, using the Pythagoras theorem, we have

AC² = CD² + AD²

⇒ CD² = AC² - AD² …… (ii)

Equating (i) & (ii), we get

BC² - BD² = AC² - AD²

⇒ BC² = AC² - AD² + BD²

⇒ BC² = AC² - AD² + (AB - AD)²

⇒ BC² = AC²- AD² + AB² – 2*AB*AD + AD²

⇒ BC² = AC² + AB² – 2*AB*AD …… (iii)

Now, in ∆CAD, using the trigonometry property of triangles, we get

cos 60° = AD / AC

⇒ ½ = AD / AC

⇒ AD = ½ * AC ….. (iv)

Thus,

On substituting (ii) in (i), we get

BC² = AC² + AB² – (2 * AB * ½ * AC)

⇒ BC² = AB² + AC² – AB * AC

Hence proved