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suppose a triangle has 2 sides of lengths 32 and 35 and that the angle between these 2 sides is 120 what is the length of the 3rd side of the triangle

Sagot :

mitak5575

Let the Vertices of the  Δ  be  A , B ,  and  C

We will follow the Usual Notation for  Δ A B C ,  e.g., the side

opposite to the Vertex  A  will be denoted by  a , m ∠ A = A ,  etc.

In this notation, let us assume that,

a = 32 , b = 35 , & , C = 120 ° & we have to find  c

Using Cosine-Rule for  Δ A B C , we have,  

c²= a²+b² - 2 ab cos C = 32 ²+35² - 2 x 32 x 35 x cos 120° =

1024 + 1225 − 2240  cos ( 180 °− 60 °) = 2249 - 2240(-cos 60°)

2249+ 2240 (1/2)= 2249 + 1120=  3369

Answer: C= √3369 is about 58.04

Answer:  58.043087 (approximate)

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Explanation:

Refer to the diagram below. We can use the law of cosines to solve for c

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

c^2 = 32^2 + 35^2 - 2*32*35*cos(120)

c^2 = 3369

c = sqrt(3369)

c = 58.043087 which is approximate

Round this value however you need to.

View image jimthompson5910
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