find cos D and tan D if sin D =squrae root of 3 over 2?

Question

desperate fof help si there a step by step instructional?

Answer 1

I have that in my memory even without computation:
cos D = 1/2
tan D = sqrt(3)
D = 60 degrees (bonus for you)

Answer 2

sin D=squrae root of 3 over 2... Make a triangle...sin D is the Opposite/Hyponteneuse... so the triangle should look like this >>>>> the side opposite the angle is =squrae root of 3 and the the hypoteneuse of the triangle is 2. Use pythagorean thm. to find other side of right triangle.... Cos D = the adjacent (or side next to angle) / hyponeneuse and Tan D = the opp/adj.

Answer 3

Refer yourself to a unit circle...Go to google and type that in the search bar

Sin D = (sqrt3)/2

then D = 60 degrees

Plug in 60 degrees for D

So then Cos (60) = 1/2
and Tan (60) = sqrt 3

Answer 4

just draw a right angeled triangle then infront of angle 90 put angle d and since the sin rule is opposite over hypotenouse then root 3 is the opposite side infont of d and2 is the side infront of angle 90 so u can get the third side by squaring of root 3 then subtract from it the squaring of 2 and u will get the third side so simply you will have the cosine as (adjacent side to d over the opposite to angle 90) 1 over 2 and tan(opposite to angle d over adjacent to angle d) =root 3

Answer 5

sin D = (sqrt 3)/2 -- the square is 3/4.

Because (sin D)^2 + (cos D)^2 = 1, we have

3/4 + 1/4 = 1, so cos D = +/- sqrt(1/4) = +/- 1/2

next, tan D = sin D / cos D
= +/- sqrt 3

Answer 6

Solve for D, from the equation:

What will be the value of D ( Hint: what is Sine 60)

Then substitute D for Cosine D and Tangent D

That's the answer:

Try for urself: Its fun.

Answer 7

sin(D) = (√3) / 2 = side opposite / hypotenuse

Side adjacent = √(4 - 3) = 1 (from Pythagorean Theorem)

cos(D) = side adjacent / hypotenuse = 1 / 2
tan(D) = side opposite / side adjacent = √3 / 1 = √3

Just use the definitions and Pythagorean theorem and it will get you through about 90% of trig


Doug

Answer 8

since sin^2(D)+cos^2(D)=1
if you replace the sin(D) in the equation, then re-arrange you get:
cos^2(D)=1/4
so: cos(D)=1/(sqrt(2))

for tan(D):
since tan^2(D)+1=1/(cos^2(D)),
if you replace the cos^2(D) in the equation, then re-arrange you get:
tan^2(D)=3
so: cos(D)=sqrt(3)

Answer 9

sinD=opp.side/hypotenuse
=(3)^1/2/2
adjacent side^2=hyp^2-opp.side^2
=4-3=1
cosD=adj.side/hyp=1/2
and tand=sinD/cosD=
(3)^1/2/2)/1/2=(3)^1/2