Question

PLEASE HELP!!!!!

List the sides of triangle BCD in order from least to greatest if angle B=(10x-28)°, angle C=(6x-1)°, and angle D=(8x-7)°​

Answer 1

BD < CD < BC

Explanation:

In a triangle, the sum of the interior angles is always 180 degrees. So, for triangle BCD:

`\sf \angle B + \angle C + \angle D = 180^\circ`

Given that:

`\sf \angle B = (10x-28)^\circ`

`\sf \angle C = (6x-1)^\circ`

`\sf \angle D = (8x-7)^\circ`

Now, set up the equation:

`\sf (10x-28) + (6x-1) + (8x-7) = 180`

Combine like terms:

`\sf 10x + 6x + 8x - 28 - 1 - 7 = 180`

`\sf 24x - 36 = 180`

Now, solve for
`\sf x`:

`\sf 24x = 216`

`\sf x = (216)/(24)`

`\sf x = 9`

Now that we have the value of
`\sf x`, we can substitute it back into the expressions for the angles to find their values:

`\sf \angle B = (10x-28) = (10 * 9 - 28) = 62^\circ`

`\sf \angle C = (6x-1) = (6 * 9 - 1) = 53^\circ`

`\sf \angle D = (8x-7) = (8 * 9 - 7) = 65^\circ`

Now, list the angles in order from least to greatest:

`\sf \angle C, \angle B, \angle D`

Since the side is directly proportional to the opposite angle to them.

So,

The opposite side of
`\sf \angle C` is BD

The opposite side of
`\sf \angle B` is CD

The opposite side of
`\sf \angle D` is BC

So, the list of the sides of triangle BCD in order from least to greatest is:

BD < CD < BC