Question

Special Triangles

Find the missing side lengths. Leave answers as radicals in the simplest form.

Answer 1

see explanation

Explanation:

Using the cosine and tangent trigonometric ratios and the exact values

cos30° =
`(√(3) )/(2)` and tan30° =
`(1)/(√(3) )` , then

cos30° =
`(adjacent)/(hypotenuse)` =
`(6)/(m)` =
`(√(3) )/(2)` ( cross- multiply )

`√(3)` × m = 12 ( divide both sides by
`√(3)` )

m =
`(12)/(√(3) )` ← rationalise by multiplying numerator/ denominator by
`√(3)` )

m =
`(12)/(√(3) )` ×
`(√(3) )/(√(3) )` =
`(12√(3) )/(3)` = 4
`√(3)`

-------------------------------------------------------------------------------

tan30° =
`(opposite)/(adjacent)` =
`(n)/(6)` =
`(1)/(√(3) )` ( cross- multiply )

`√(3)` × n = 6 ( divide both sides by
`√(3)` )

n =
`(6)/(√(3) )` ← rationalise the denominator

n =
`(6)/(√(3) )` ×
`(√(3) )/(√(3) )` =
`(6√(3) )/(3)` = 2
`√(3)`

-----------------------------------------------------------------------------------