If cos theta = 5/13, theta in quadrant II, find sin theta and tan theta?
You are right!!!! In Q2, cosΘ is negative. Assuming the mistake was in omitting the negative sign, this is a 5 - 12 - 13 right triangle in Q2, with cosΘ = -5/13 sinΘ = 12/13 tanΘ = -12/5
If tan theta = 2/5 and theta is in quadrant 3, find cos theta?
tan ∅ = 2/5 r² = 2² + 5² = 29 r = √29 cos ∅ = - 5 / √29
If tan(theta) = -1/2 for theta in quadrant II, find sin(theta) + cos(theta). Please explain process.?
theta = arctan(-.5) + 180degrees = 153.4349488 degrees sin(153.4349488 degrees) + cos(153.4349488 degrees) = -.4472135955 Answer: -.4472135955
If tan of theta = -7/24 and theta is in the second quadrant find sin theta?
If theta is in the 2nd quadrant, it means that the 'x' is negative and y is positive. This means that it is supposed to be (7)/(-24) The length of the diagonal of the triangle formed is always positive, ie: sqrt( 7^2 + (-24)^2 ) = 25 Hence sin of theta = opposite/hypoteneus = 7/25 [positive]
If sin theta=4/5 and theta is in quadrant 2, what is the value of tan theta?
1) tan theta is definetely negative ! In 2 squadrant all are negative except sin .. So u can find cos by the fundimental rule ==> cos^2=1-sin^2 that gives cos=-3/5 Tan theta is sin/cos = -4/3 2)as i remember that third quadrand ony cot and tan positive so sintheta is negative and sin^2 =1/(1+cot^2) And u find sine :D
If tan theta = 1.5 and theta is in quadrant 3, is the value of sin theta -3?
No, it is notIf θ is in third quadrant, thentan(θ) = - 15/-10= - 3/-2Hypotenuse = √((-3)^2 + (-2)^2)= √(13)sin(θ) = -3/√(13)
If sec theta+tan theta=2/3, then how would one find the quadrant in which theta lies and as well as sin theta?
Answer should be -5/13 as theta lies in 4th quadrant
Find sin 2theta. tan theta=8/15, theta lies in quadrant 3?
We have that tan(θ) = 8/15 We know that tan(θ) = opposite / adjacent (from SOHCAHTOA) so we draw ourself a nice right angled triangle where the opposite side = 8 and the adjacent side = 15. Now we use Pythagoras to find the length of the hypotenuse: c² = 8² + 15² c² = 64 + 225 c² = 289 c = √289 = 17 So we have all lengths for the triangle. sin(2θ) = 2cos(θ)sin(θ) Using the triangle we found, we have that sin(θ) = opposite / hypotenuse = 8/17 and cos(θ) = adjacent / hypotenuse = 15/17 hence sin(2θ) = 2(8/17)(15/17) = 240 / 289 Hope that helps :-) EDIT: As θ lies in quadrant 3, sin(2θ) must be negative. So sin(2θ) = -240 / 249 Sorry about that!
Find sin theta if tan theta = -√3/3 and falls in Quadrant 4?
Tan theta = Opposite / Adjacent Tan theta = -√3/3 Opposite = -√3 Adjacent = 3 Hypotenuse = sqrt((-√3)^2 + 3^2) Hypotenuse = sqrt(3 + 9) Hypotenuse = sqrt(12) Sin = Opposite / Hypotenuse Sin = -√3 / √12 Sin = -√3 / (√3 * √4) Sin = -√3 / (√3 * 2) Sin = - 1 / (2) Sin = - 1/2 Answer is C cos = Adjacent / Hypotenuse cos = 3/√12 cos = 3*√12/12 cos = √12/4 cos = (√3*√4)/4 cos = (2 * √3)/4 cos = √3/2 Answer is A