Question

1) complete the special right triangle
what is length of BD
what is length of BC

Answer 1

`BD = 4`

`BC =4\sqrt 3`

Explanation:

Given

The attached triangle

Required

Find BD and BC

Solving BD

Considering angle at D, we have:

`\cos(D) = (Adjacent)/(Hypotenuse)`

`\cos(60) = (BD)/(8)`

Solve for BD

`BD = 8 * \cos(60)`

`\cos(60) = 0.5` So:

`BD = 8 * 0.5`

`BD = 4`

To solve for BC, we make use of Pythagoras theorem

`CD^2 = BC^2 + BD^2`

This gives

`8^2 = BC^2 + 4^2`

`64 = BC^2 + 16`

Collect like terms

`BC^2 =64-16`

`BC^2 =48`

Take square roots

`BC =\sqrt{48`

Expand

`BC =\sqrt{16*3`

Split

`BC =√(16)*\sqrt 3`

`BC =4\sqrt 3`