Given that sin x 2/3 and is an acute angle find tan x
We can start by using the Pythagorean identity:
sin^2 x + cos^2 x = 1
Since sin x is given as 2/3, we can solve for cos x:
(2/3)^2 + cos^2 x = 1
4/9 + cos^2 x = 1
cos^2 x = 5/9
Taking the positive square root since x is an acute angle:
cos x = √(5/9)
Now we can use the tangent identity:
tan x = sin x/cos x
Substituting the values we know:
tan x = (2/3)/√(5/9)
Rationalizing the denominator:
tan x = (2/3)√(9/5)
tan x = (2/3)(3/√5)
tan x = 2√5/5
To find tan(x), we can use the relationship between sin(x) and tan(x) in a right triangle.
We are given that sin(x) = 2/3 and we know that sin(x) = opposite / hypotenuse in a right triangle.
Let's label the sides of the triangle as follows:
Opposite side = 2
Hypotenuse = 3
Using the Pythagorean theorem, we can find the third side (adjacent side) of the triangle:
a^2 + b^2 = c^2
a^2 + 2^2 = 3^2
a^2 + 4 = 9
a^2 = 5
a = √5
Now that we have the opposite and adjacent sides of the triangle, we can find tan(x) using the formula tan(x) = opposite / adjacent:
tan(x) = 2/√5
To simplify this expression, we can multiply the numerator and denominator by √5:
tan(x) = (2/√5) * (√5/√5)
= 2√5 / 5
Therefore, tan(x) = 2√5 / 5.