Given C= 61 degrees, a=55, and b= 29, find the area of triangle, ABC?

Answer 1

Your 'triangle' doesn't add up to 180 degrees. Plus, you need the distance of at least one side in order to find the area.

Doh! Maybe I should read closer.

You use the law of cosines to find side c:

c^2 = a^2 + b^2 - 2ab cos(C)

or

4689.34 = 55^2 + 29^2 - 2*55*29*cos(61)

so c = edit: (48.2)

Use Heron's formula to find the area.

s = 1/2 (48 + 55 + 29) = 66

A = sqrt( 66* (66 - 55) * (66 - 29) * (66 - 48)
= 700

Answer 2

Law of Cosines

c^2 = a^2 + b^2 - 2bc(cosC)

c^2 = 55^2 + 29^2 - 2(55 * 29)cos(61)
c^2 = 3025 + 841 - 2(1595)cos(61)
c^2 = 3866 - 3190cos(61)
c = about 48

Now that we know this is a Scalene Triangle, we use the formula

Area = (ab * sin(C))/2
A = (55 * 29 * sin(61))/2
A = about 697.5

Answer 3

if a is 10 and b is 8, and c is 30 tiers, then i think of you're working with a 30/60/ninety triangle, with sides of length ratio 3/4/5. this makes factor c 6, so which you have a 6/8/10 triangle. 6*8 = 40 8, 40 8/2 = 24. the portion of the triangle is 24.

Answer 4

abc=145. I know this because 55+61+29=145(a+b+c=145).

Answer 5

c^2=55^2+29^2-2*55*29*cos 61=3025+841-3190*cos 61
c^2=2319.4573114141848772495390044937
c=48
s=(55+29+48)/2=66
A=√66(66-55)(66-29)(66-48)=√66*11*37*18=695

Answer 6

The area of the triangle is 725.165 un.^2

Answer 7

Use Heron's Forumula. This is sometimes called Hero's forumula. Just search it on google and use it to find the area with side lengths a, b and c, and with semiperimeter s.

Answer 8

If I'm not mistaken, you find the area by multiplying A, B, and C. Then you'll have your answer - :)

Answer 9

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