Find the exact value of the remaining trigonometric functions if cosA= -3/5 and sinA < 0

if tan(theta) = -3 and cos(theta) < 0, then tan(theta/2) = ?

so

so

 (we take the negative case because cosine is negative for this example)
now

Therefore...


which simplifies to

Solve the triangle: side a=3, side b=2, side c= 2.

Solve each right triangle A=35(degrees) and side a=5

Find exact value cos(2x) = 0 on the interval [0, 2pi]

Find area of triangle: a=2,b=1,theta=100(degrees) <--the angle between them

let cos x = 1/sqrt(10) with x in quadrant IV. find cos 2x.

Using a double angle identity:

so we can write:

but cosine is positive in Quad IV, so...

find reference angle for 215 degrees

so our angle is in Quad III
and so

How do i find the measure of each angle for tan B = 0.6000

use inverse tangent function:

 

Triangle ABC is inscribed in a circle. AB is 12, AC is 6 and BC is 6 square root 3. Find arc BC.

These measures tell us that the triangle is a 30-60-90 triangle with a hypotenuse opposite the side AB and since we can see that arc BC is the arc of least measure, it must be 60 degrees

In ∆ABC, if the length of sides a, b, and c are 3.5 centimeters, 5.5 centimeters, and 6 centimeters respectively, what is m C to two decimal places?

law of cosines:

C=arccos(13/77)=80.28 degrees

assuming that cos(theta)=2/7 for an acute angle theta, find sin(theta)

Given:

the fastest way is to use the pythagorean identity:


which simplifies to...

Simplify the trigonometric expression. sin^2 theta /(1+ cos theta)

it is equal to...

and since the numerator is a difference of squares

If cos(feta)=0.7 find exact values for tan(feta) and sin(feta)

In a right triangle with g the right angle, a = 69.2° and c = 28.3. What is b?

so


so...
b=10.05

In a ΔABC, angle B = 90 degrees, AB = 6 units and AC = 10 units then the length of angle bisector AD from angle A is

 cosA=6/10=3/5 

A=arccos(3/5) - about 53.13 deg
so half is about A2=26.57 deg
therefore
cosA2=6/AD
cos26.57=6/AD
AD=6/cos26.57=6.71

cos x°=sin20° also tan 20°­ = cot(x+30)°

x=arccos(sin(20))=70 

for the second:
cot(x)=1/tan(x)
so we can write
tan(x+30)=cot(20)
x+30=arctan(tan(x+30)=arctan(cot(20) 
x+30=70
x=40

solve triangle XYZ where angle X=40, x=12 and y=6

Law of Sines: sinY=(6/12)sin40
sinY=0.3214
Y=arcsine(0.3214)
Y=18.747

Since all angles must add to 180 degrees:
Z=180-40-18.747=121.25
now we find z...
z=12(sin121.25/sin40)=15.96 

If tan x = .593, then sin x = ?

 

 

find the complement and supplement of an angle with the given measure. a.)10° 10' 30" , b.)1° 40' 19"

a)
90°=89° 59' 60"
so...
complement:
89° 59' 60"-10° 10' 30"=
79° 49' 30"
supplement:
169° 49' 30"
b)
complement
89° 59' 60"-1° 40' 19"=
88° 19' 41"
supplement:
178° 19' 41"