Trig Questions? How to I do this crap!? :)?

1.) cos is Adj./Hyp. so cosA= 12/13 rearranging gives us A=cos^-1(12/13)=22.62 Degrees

2.)sin= opp./hyp. so sin(70)=x/14 rearranging gives us 14sin(70)=x=13.16

3.)sinx=10/4 so sin^-1(10/14)=45.58

4.)tan=opp/adj so tan(58)=4/x rearranging gives 4/tan(58)=x=2.5

5.) tan(x)=5/8 rearranging gives tan^-1(5/8)=x=32.0

6.)this can be done two different ways you can add the two sides together and use 30 degrees or you can use the smaller triangle and 20 degrees. i will use the smaller triangle.

tan(40)=x/20 so 20tan(40)=x=16.8

1) To find the cos of A remeber that Cos=adj/hyp

so Cos of A=12/13 (note that 13 is the hypotenuse also known as the longest side of a triangle)

2) To find X look at the triangle from the perspective of the angle given, what sides are provided?? The sides provided is "X" which is opposite the angle and "14" which is on the longest side of the triangle(hyp). You ask yourself what notion is opp/hyp?? The answer is Sin

So it would be Sin70=X/14

note that sin70= .9397

so, .9397=X/14 ==> .9397*14=X

Thus, X= 13.156

Using the same notions as before apply it to the other problems:

3) What is given is SinX=10/14 (now solve for X)

SinX= .7143 (now do the inverse of .7143)

so X=45.6

4)What is given is Tan58=4.0/X (Tan58= 1.6)

1.6=4.0/X =>1.6*X=4.0==> X=4.0/1.6=2.5

5) What is given is tanX=5/8

5/8= .625 [so do the inverse tan of .625]

X=32

6) Using the diagram, the bottom side which is 20

Use the 40 degree as the angle:

so we want to find BC:

Cos40=20/X

cos40= .766

so, .766=20/X

.766*X=20

X=20/ .866===> X=23= the length of BC

I hope this helped.

First of all:

SohCahToa.

Sine=Opposite/Hypotenuse

Cosine=Adjacent/Hypotenuse

Tangent=Opposite/Adjacent.

1) Cos(A)=12/13.

2) sin(70)=x/14. 14sin(70)=x.

3) sin(x)=10/14. Input sin-1(10/14) into your calculator, and you will get a value of x.

4) tan(58)=4/x. xtan(58)=4. x=4/tan(58).

5) tan(x)=5/8. Same idea with inverse tangent of 5/8.

6) The 30˚ triangle, ADB, is absolutely irrelevant to the problem. sin(40)=(BC)/20. 20sin(40)=(BC).

What grade are you in? You really should begin thinking this through for yourself, as it will haunt you for the rest of your schooling career if you don't.

cos (A) = adjacent / hypotenuse = 12 / 13

sin (70) = x / 14 so,

sin (70) * 14 = x

sin (x) = 10 / 14 so,

1 / sin (10 / 14) = x

cot (58) = x / 4.0 so,

cot (58) * 4.0 = x

tan (x) = opposite / adjacent = 5 / 8 so,

1 / tan (5 / 8) = x

tan (30) = BC / 30 and tan (40) = BC / 20 so,

30 * tan (30) = 20 * tan (40) = BC

I miss Trigonometry. Fun stuff!

1st part

cos A=base/hypotneuse

=12/13

see the side opposite the angle is the perpendicular

the side opposite to the right angle is hypotneuse and the third side is the base.

2nd part

i am taking the 70deg angle

the side opposite the angle is unknown which is the perpendicular

and the hypotneuse is known which is 14

so sin 70=x/14

so x=14sin70

the formula for sinX=perpendicular/hypotneuse

tanX=perpendicular/base

x is the angle

hope this helps try the others on ur own u wll get them

Alright, Add up the two angles...

Like that one was 70 and 14...

So add 70+14 which equals 84

Then a triangle always equals 180 degrees/.

SO then 180 - 84 which equals 96 degrees.

So then X=96

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