Question

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

Answer 1

Answer:

A = 61.22 degrees

B =

C =

Explanation:

Parameters:

a = 16 ft

b = 18 ft

c = 6 ft

A = ?

B = ?

C = ?

Using cosine rule, we have:

1.

$\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ 16^2=18^2+6^2-2(18\times6)\cos A \\ 256=360-216\cos A \\ 256-360=-216\cos A \\ -104=-216\cos A \\ \cos A=\frac{-104}{-216}=\frac{13}{27} \\ \\ A=\cos ^{-1}(\frac{13}{27})=61.22^o \end{gathered}$

2.

$\begin{gathered} b^2=a^2+c^2-2ac\cos B \\ 18^2=16^2+6^2-2(16\times6)\cos B \\ 324=292-192\cos B \\ 324-292=-192\cos B \\ 32=-192\cos B \\ \cos B=\frac{32}{(-192)}=-\frac{1}{6} \\ \\ B=\cos ^{-1}(-\frac{1}{6})=99.59^o \end{gathered}$

3.

$\begin{gathered} c^2=a^2+b^2-2ab\cos C \\ 6^2=16^2+18^2-2(16\times18)\cos C \\ 36=580-576\cos C \\ 36-580=-576\cos C \\ -544=-574\cos C \\ \cos C=\frac{544}{574}=\frac{272}{287} \\ \\ C=\cos ^{-1}(\frac{272}{287})=18.61^o \end{gathered}$