Given that sinθ = 2/3 and is an acute angle.
Find:
Tanθ giving your answer in surd form
θ = tan^(-1)〖(2/√5)= 41.81°〗
Sec^22 θ
θ= cos^(-1)〖(3/√5〗)=
sinØ= 2/3 , Ø is in quad I
in my triangle, x = ? , y = 2, r = 3
x^2 + 2^2 = 3^2
x^2 = 5
x = √5
tanØ = y/x = 2/√5 or, if rationalized, 2√5 /5
To find Tanθ, we can use the formula Tanθ = sinθ / cosθ. Since we already know that sinθ = 2/3, we need to find cosθ.
To find cosθ, we can use the Pythagorean identity: sin^2θ + cos^2θ = 1. Since we know that sinθ = 2/3, we can substitute it into the equation:
(2/3)^2 + cos^2θ = 1
4/9 + cos^2θ = 1
cos^2θ = 1 - 4/9
cos^2θ = 5/9
Taking the square root of both sides, we can find cosθ:
cosθ = √(5/9)
cosθ = √5/3
Now we can substitute the values of sinθ and cosθ into the formula for Tanθ:
Tanθ = sinθ / cosθ
Tanθ = (2/3) / (√5/3)
To simplify, we can multiply the numerator and denominator by the conjugate of the denominator:
Tanθ = (2/3) * (3/√5)
Tanθ = 2/√5
Therefore, Tanθ in surd form is 2/√5.
For the second part of the question, to find sec^2θ, we can use the identity sec^2θ = 1/cos^2θ. Since we already know that cosθ = √5/3, we can substitute it into the equation:
sec^2θ = 1 / ( √5/3)^2
sec^2θ = 1 / (5/9)
sec^2θ = 9/5
Therefore, sec^2θ = 9/5.
To find the value of tanθ, we can use the identity tanθ = sinθ / cosθ. Since we know that sinθ = 2/3, we need to find cosθ.
To do that, we can use the Pythagorean identity sin^2θ + cos^2θ = 1. Plugging in sinθ = 2/3, we have (2/3)^2 + cos^2θ = 1.
Solving for cos^2θ, we get cos^2θ = 1 - (2/3)^2 = 1 - 4/9 = 5/9.
Taking the square root of both sides, we find that cosθ = √(5/9) = √5/3.
Now we can calculate tanθ using the formula tanθ = sinθ / cosθ. Plugging in sinθ = 2/3 and cosθ = √5/3:
tanθ = (2/3) / (√5/3) = (2/3) * (3/√5) = 2 / √5 = √5/5.
So the value of tanθ in surd form is √5/5.
Next, to find sec^2θ, we use the identity sec^2θ = 1 / cos^2θ. Since we found cosθ = √5/3, we can substitute it into the identity:
sec^2θ = 1 / (cosθ)^2 = 1 / (√5/3)^2 = 1 / (5/9) = 9/5.
Therefore, sec^2θ = 9/5.