Question

In triangle DEF with right angle at F, where EF = 1 and DF = 2, the sine of angle E is:

a) 1/2
b) 2/√5
c) 1/√5
d) √5/2

Answer 1

(b)

Explanation:

The first step is to calculate the hypotenuse DE of the right triangle

using Pythagoras identity

DE² = EF² + DF² ( substitute values )

DE² = 1² + 2² = 1 + 4 = 5 ( take square root of both sides )

`√(DE^2)` =
`√(5)` , then

DE =
`√(5)`

Then

sin E =
`(opposite)/(hypotenuse)` =
`(DF)/(DE)` =
`(2)/(√(5) )`