Let a, β, y be the three interior angles of any triangle, and try to prove that cos(a+β+y) = -1. prove \: that \: \cos( \alpha + \beta + \gamma ) = - 1 ​

Question

Let a, β, y be the three interior angles of any triangle, and try to prove that cos(a+β+y) = -1.


`prove \: that \: \cos( \alpha + \beta + \gamma ) = - 1`
​

Answer 1

Given:

`\alpha, \beta, \gamma` are three interior angles of any triangle.

To prove:

`\cos (\alpha+\beta+\gamma)=-1`

Solution:

According to the angle sum property of a triangle, the sum of all three interior angle of a triangle is always 180 degrees.

It is given that,
`\alpha, \beta, \gamma` are three interior angles of any triangle.

Using the angle sum property of triangles, we get

`\alpha+\beta+\gamma =180^\circ` ...(i)

We need to prove,

`\cos (\alpha+\beta+\gamma)=-1`

Taking LHS, we get

`LHS=\cos (\alpha+\beta+\gamma)`

`LHS=\cos (180^\circ)` [Using (i)]

`LHS=-1`

`LHS=RHS`

Hence proved.