crisisyt
Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

Consider this equation.


sin(θ)= 3sqrt10/10


If θ is an angle in quadrant II, what is the value of tan(θ) ?

Consider This Equationsinθ 3sqrt1010If Θ Is An Angle In Quadrant II What Is The Value Of Tanθ class=

Sagot :

altavistard

Answer:

tan x = -3 (Answer A)

Step-by-step explanation:

We want to find the tangent of this angle "theta," and recall the trig identity

(sin x)^2 + (cos x)^2 = 1.

               3√10

If sin x = -----------

                   10

                             90

then (sin x)^2 = ----------- = 9/10

                             100

and (cos x)^2 = 1 - 9/10 = 1/10

 

                      sin x       3√10/10

Then tan x = ---------- = -------------- = -3 (Answer A)

                       cos x      1√10/10

The tangent function is negative in Quadrant II.  In Quadrant I tan x = +3

Answer:

A

Step-by-step explanation:

Using the trig identity

sin²x + cos²x = 1 , then cos x = ± [tex]\sqrt{1-(\frac{3\sqrt{10} }{10})^2 }[/tex]

Given

sinθ = [tex]\frac{3\sqrt{10} }{10}[/tex] , then

cosθ = ± [tex]\sqrt{1-(\frac{3\sqrt{10} }{10})^2 }[/tex]

        = ± [tex]\sqrt{1-\frac{9}{10} }[/tex]

        = ± [tex]\sqrt{\frac{1}{10} }[/tex]

Since θ is in second quadrant where cosθ < 0 , then

cosθ = - [tex]\frac{1}{\sqrt{10} }[/tex]

Then

tanθ = [tex]\frac{sin0}{cos0}[/tex] = [tex]\frac{\frac{3\sqrt{10} }{10} }{\frac{-1}{\sqrt{10} } }[/tex] = - 3 → A

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.