Thread: Trigonometry to seld complex shapes

1. Trigonometry to seld complex shapes

how do i work out the angles for A, B and C

all in mm

cheers

also how do i find the angles A & B in this triangle ?

is it 20cos60?

2. Originally Posted by turbod15b
how do i work out the angles for A, B and C

all in mm

cheers

also how do i find the angles A & B in this triangle ?

is it 20cos60?
For the first triangle use either the Law of Sines or the Law of Cosines:
Law of Sines:
sin(angle A)/a = sin(angle B)/b = sin(angle C)/c
where a, b, and c are the sides across the triangle from angles A, B, and C respectively.

Law of Cosines:
a^2 = b^2 + c^2 - 2bc*cos(angle A)
b^2 = a^2 + c^2 - 2ac*cos(angle B)
c^2 = a^2 + b^2 - 2bc*cos(angle C)

For the second triangle, is what "20cos60?"

By definition of the sine and cosine of an angle we have that
sin(angle B) = 20/60 ==> angle B = asn(1/3) = 19.4712 degrees
and
cos(angle A) = 20/60 ==> angle A = acs(1/3) = 70.5288 degrees

As a quick check, note that the sum of these two angles is 90 degrees, as it should be for a right triangle.

-Dan

3. Hello, turbod15b!

The first problem has: .a = 31, b = 27, c = 37.

Given no angles, we must use variations of the Law of Cosines.

. . . . . . . b² + c² - a² . . .27² + 37² - 31²
cos A .= .-------------- .= .------------------- .= .0.569069069
. . . . . . . . . .2bc . . . . . . . . 2(27)(37)

Hence: .A .= .55.31466541 . . . . A . .55.3°

. . . . . . . a² + c² - b² . . .31² + 37² - 27²
cos B .= .-------------- .= .------------------- .= .0.697907585
. . . . . . . . . .2ac . . . . . . . . 2(31)(37)

Hence: .B .= .45.74063061 . . . . B . .45.7°

Then: .C .= .180° - 55.3° - 45.7° . . . . C .= .79.0°

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you have time, check angle C by using the Law of Cosines . . .

. . . . . . . a² + b² - c² . . . 31² + 27² - 37²
cos C .= .--------------- .= .------------------ .= .0.191756272
. . . . . . . . . 2ab . . . . . . . . .2(31)(27)

Hence: .C .= .78.94470398 . . . . C . .78.9° .close enough!

4. i dont understand how u get

Cos A = 0.569069069

Cos B = 0.697907585

5. Originally Posted by turbod15b
i dont understand how u get

Cos A = 0.569069069

cos(A):
26^2 + 37^2 - 31^2 = 1137
2*27*37 = 1998

cos(a) = 1137/1998 = 0.569069

Now you try cos(B).

-Dan

6. 31^2+37^2-27^2=1601
2*31*37=2294

1601/2294=0.697907585

i get that bit, but i dont get how u get it to

Hence: .A .= .55.31466541 . . → . . A .≈ .55.3°

Hence: .B .= .45.74063061 . . → . . B .≈ .45.7°

7. Originally Posted by turbod15b
31^2+37^2-27^2=1601
2*31*37=2294

1601/2294=0.697907585

i get that bit, but i dont get how u get it to

Hence: .A .= .55.31466541 . . → . . A .≈ .55.3°

Hence: .B .= .45.74063061 . . → . . B .≈ .45.7°
Oh! Sorry.

Your calculator probably has a "2nd" key or something like it. You need to plug in the 0.569069069, and hit "2nd cos" (to get to the function above the "cos" button.) This is the inverse cosine function and should give you the correct answer. If it does not, then you are likely in "radian" mode and need to learn how to change your calculator into "degree" mode.

-Dan

8. thanks

my second buttons are shift sin-1, cos-1,tan-1