12(7x)^1/5
This is simplification into index form, the index itself is the 5th root of 7x if you just wanted to state what it is
I think it’s the first one , sorry if i’m wrong
We are given: cos theta = 1/3. Thus, adj side = 1 and hyp = 3. Using the Pythagorean Theorem to find the length of the opposite side, represented by x.
1^2 + opp^2 = 3^2, or 1 + x^2 = 9, or x^2 = 8. Here, x could be either sqrt(8) or -sqrt(8): sqrt(8) if angle theta is in QI, and -sqrt(8) if in QIV.
Thus, the sine of angle theta could be either sqrt(8)/3 or -sqrt(8)/3.
Answer:
VYW
Step-by-step explanation:
Vertical angles form an x shape when two lines cross. The vertical angle is across from its partner angle.
Answer:
Step-by-step explanation:
I honestly have no idea what you mean by answer by formula, but I'm going to give it my best. I began by squaring both sides to get:
(a² - b²) tan²θ = b² and then distributed to get:
a² tan²θ - b² tan²θ = b² and then got the b terms on the side to get:
a² tan²θ = b² + b² tan²θ and then changed the tans to sin/cos to get:
and isolated the sin-squared on the left to get:
and distributed to get:
***
*** and factored the right side to get:
and utilized a trig Pythagorean identity to get:
and then solved for sinθ in the following way:
so
This, along with the *** expression above will be important. I'm picking up at the *** to solve for cosθ:
and get the cos²θ alone on the right by subtracting to get:
and divide by b² to get:
and factor on the left to get:
and take the square root of both sides to get:
and simplify to get:
and go back to the identity we found for sinθ and sub it in to get:
and simplifying a bit gives us:
That's my spin on things....not sure if it's what you were looking for. If not.....YIKES
