Question

Help me to find AC please

Answer 1

see explanation

Explanation:

Since the triangle is right use cosine or sine to solve for AC

note cos30° =
`(√(3) )/(2)`, thus

cos30° =
`(adjacent)/(hypotenuse)` =
`(AC)/(BC)` =
`(AC)/((10)/(3) )` and

`(AC)/((10)/(3) )` =
`(√(3) )/(2)`, that is

`(3x)/(10)` =
`(√(3) )/(2)` ( cross- multiply )

6x = 10
`√(3)` ( divide both sides by 6 )

x = AC =
`(10)/(6)`
`√(3)` =
`(5)/(3)`
`√(3)`