Question

If in a triangle ABC, a = 2, b = 3 and c = 4 then find the value of cos2A

Answer 1

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$\bold{\underline{Question-}}$

If in a triangle ABC, a = 2, b = 3 and c = 4 then find the value of cos2A


$\bold{\underline{Answer-}}$


Solution : there a = 2, b = 3, c = 4


Now, using the formula $cos \: A = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc}$



the, we put the above values


$= \frac{ {3}^{2} + {4}^{2} - {2}^{2} }{2 \times 3 \times 4} \\ \\ \\ = \frac{9 + 16 - 4}{2 \times 3 \times 4} \\ \\ \\ = \frac{25 - 4}{2 \times 3 \times 4} \\ \\ \\ = \frac{21}{24} = \frac{7}{8}$


then, we know that formula cos2A = 2 cos²A - 1

now,


$= 2 \times \frac{49}{64} - 1 \\ \\ \\ = \frac{49}{32} - 1$




$cos \: 2A \: = \frac{17}{32}$



Hence, solved

Answer 2

Thanks for question !


Here give values are
A=2
B=3
C=4

Using formula of
$cos$
A
$= {b}^{2} + {c}^{2} - {a}^{2}$
Now we putting value
$\frac{9 + 16 - 4}{2 \times 3 \times 4}$
After solving this we get value
$\bf{ \frac{7}{8} }$
then, we know that formula cos2A = 2 cos²A - 1

Putting value we get answer
$2 \times \frac{49}{64} - 1$
$= \frac{49}{32} - 1$
Then we get
$cos$

A
$= \frac{17}{32}$
Here u get required value !

hopes that helps u !