Question

A triangle with sides AB = 6, BC = 5, CA= 6. Find Cos A ​

Answer 1

Answer:

The answer is $\dfrac{47}{72}$

Step-by-step explanation:

Given side lengths are

$\text{AB}=6=c,\text{BC}=5=a,\text{CA}=6=b$

Therefore using cosine rule :

     $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

              $=\dfrac{6^2+6^2-5^2}{2(6)(6)}\\=\dfrac{36+36-25}{72}\\=\dfrac{47}{72}$