If 5cosA+3=0 and 180(degrees)<A<360(degrees)
Determine
1. Tan^2A
And
2. SinA over cosA
see your previous post, in related questions below.
Also, recall that tanA = sinA/cosA
To find the values of Tan^2A and SinA over CosA, let's solve the equation 5cosA + 3 = 0 first.
Step 1: Rewriting the equation
5cosA = -3
Step 2: Solving for cosA
cosA = -3/5
Now, let's find the values of Tan^2A and SinA over CosA based on the given information.
1. Tan^2A:
To find Tan^2A, we need to use the trigonometric identity:
Tan^2A = (Sin^2A) / (Cos^2A)
Step 1: Finding SinA:
To find SinA, we can use the Pythagorean identity in a right triangle:
Sin^2A = 1 - Cos^2A
Sin^2A = 1 - (-3/5)^2
Sin^2A = 1 - 9/25
Sin^2A = 16/25
Step 2: Calculating Tan^2A:
Tan^2A = (Sin^2A) / (Cos^2A)
Substituting the values of Sin^2A and CosA, we have:
Tan^2A = (16/25) / (-3/5)^2
Tan^2A = (16/25) / (9/25)
Tan^2A = 16/25 * 25/9
Tan^2A = 16/9
So, the value of Tan^2A is 16/9.
2. SinA over CosA:
To find SinA over CosA, we can simply use the definition of TanA:
SinA / CosA = TanA
Substituting the value of CosA, we have:
SinA / (-3/5) = TanA
So, SinA over CosA is equal to TanA, or (-SinA)/3/5.
To summarize:
1. Tan^2A = 16/9
2. SinA over CosA = (-SinA)/(3/5)
To determine the values of Tan^2A and SinA over cosA, we first need to find the value of cosA.
Given the equation 5cosA + 3 = 0, we can solve for cosA by isolating it. Subtracting 3 from both sides of the equation gives us:
5cosA = -3
Dividing both sides by 5 gives us:
cosA = -3/5
Now that we have the value of cosA, we can find the values of Tan^2A and SinA over cosA.
1. To find Tan^2A, we can use the identity: Tan^2A = (SinA)^2 / (CosA)^2.
Using the value of cosA that we found earlier, cosA = -3/5, we can substitute it into the equation:
Tan^2A = (SinA)^2 / (-3/5)^2
Since we only have the value of cosA, we need to determine the value of sinA. To do this, we can use the Pythagorean identity:
(SinA)^2 + (CosA)^2 = 1
Substituting the value of cosA = -3/5, we have:
(SinA)^2 + (-3/5)^2 = 1
Simplifying:
(SinA)^2 + 9/25 = 1
(SinA)^2 = 1 - 9/25
(SinA)^2 = 25/25 - 9/25
(SinA)^2 = 16/25
(SinA)^2 = 4/5
Now we can substitute this value back into the equation for Tan^2A:
Tan^2A = (4/5) / (-3/5)^2
Tan^2A = 4/5 / 9/25
Tan^2A = 4/5 * 25/9
Tan^2A = 100/45
Tan^2A simplifies to 20/9.
Therefore, Tan^2A is equal to 20/9.
2. To find SinA over cosA, we can use the identity: SinA / cosA = TanA.
Substituting the value of cosA = -3/5, we have:
SinA / (-3/5) = TanA
To determine the value of SinA, we can use the Pythagorean identity once again:
(SinA)^2 + (CosA)^2 = 1
Substituting the value of cosA = -3/5, we have:
(SinA)^2 + (-3/5)^2 = 1
Simplifying:
(SinA)^2 + 9/25 = 1
(SinA)^2 = 1 - 9/25
(SinA)^2 = 25/25 - 9/25
(SinA)^2 = 16/25
(SinA)^2 = 4/5
Taking the square root of both sides gives us:
SinA = ±√(4/5)
Now we can substitute these values into the equation SinA / cosA = TanA:
(±√(4/5)) / (-3/5) = TanA
Simplifying:
±√(4/5) * (-5/3) = TanA
±√20/√5 * (-5/3) = TanA
±√(20/5) * (-5/3) = TanA
±√4 * (-5/3) = TanA
±2 * (-5/3) = TanA
-10/3 = TanA
Therefore, SinA / cosA is equal to -10/3.