Question

Can you help me finding the three angles of the triangle in number 21?

Answer 1

Answer:

A = 61.22 degrees

B =

C =

Step-by-step explanation:

Parameters:

a = 16 ft

b = 18 ft

c = 6 ft

A = ?

B = ?

C = ?

Using cosine rule, we have:

1.

`\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ 16^2=18^2+6^2-2(18*6)\cos A \\ 256=360-216\cos A \\ 256-360=-216\cos A \\ -104=-216\cos A \\ \cos A=(-104)/(-216)=(13)/(27) \\ \\ A=\cos ^(-1)((13)/(27))=61.22^o \end{gathered}`

2.

`\begin{gathered} b^2=a^2+c^2-2ac\cos B \\ 18^2=16^2+6^2-2(16*6)\cos B \\ 324=292-192\cos B \\ 324-292=-192\cos B \\ 32=-192\cos B \\ \cos B=(32)/((-192))=-(1)/(6) \\ \\ B=\cos ^(-1)(-(1)/(6))=99.59^o \end{gathered}`

3.

`\begin{gathered} c^2=a^2+b^2-2ab\cos C \\ 6^2=16^2+18^2-2(16*18)\cos C \\ 36=580-576\cos C \\ 36-580=-576\cos C \\ -544=-574\cos C \\ \cos C=(544)/(574)=(272)/(287) \\ \\ C=\cos ^(-1)((272)/(287))=18.61^o \end{gathered}`