Question

if the vertices of a triangle A,B,C are A(0,0)B(2,1)C(9,-2) then cos B=

Answer 1

It is very simple just go through it

Answer 2

So, The value of $cosB$ is $\frac{-11}{\sqrt{290} }$

Step-by-step explanation:

Given,

Vertices of a $\triangle ABC$ are $A(0,0)$, $B(2,1)$ and $C(9,-2)$ then $cosB=?$

From figure,

$AB=\sqrt{(2-0)^{2}+(1-0)^{2} }$

⇒$AB=\sqrt{2^{2}+1^{2} }$

⇒$AB=\sqrt{4+1}$

⇒$AB=\sqrt{5}$

$BC=\sqrt{(9-2)^{2}+(-2-1)^{2} }$

⇒$BC=\sqrt{(7)^{2}+(-3)^{2} }$

⇒$BC=\sqrt{49+9}$

⇒$BC=\sqrt{58}$

And $AC=\sqrt{(9-0)^{2}+(-2-0)^{2} }$

⇒$AC=\sqrt{81+4 }$

⇒$AC=\sqrt{85}$

By Applying Cosine Rules,

$cosB=\frac{BC^{2}+AB^{2}-AC^{2}}{2\times BC\times AB}$

⇒$cosB=\frac{(\sqrt{58} )^{2}+(\sqrt{5}) ^{2}-(\sqrt{85}) ^{2}}{2\times \sqrt{58} \times \sqrt{5} }$

⇒$cosB=\frac{58+5-85}{2\sqrt{290} }$

⇒$cosB=\frac{-22}{2\sqrt{290} }$

∴$cosB=\frac{-11}{\sqrt{290} }$

∴ The value of $cosB$ is $\frac{-11}{\sqrt{290} }$