Question

A triangle has sides of length $2cm,8cm$ and $9cm$ .Calculate the value of the largest angle in this triangle.

Answer 1

Given sides of triangle, $a=2cm,b=8cm,c=9cm$

cosine law

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

$\cos B=\frac{a^2+c^2-b^2}{ac}$

$\cos C=\frac{b^2+a^2-c^2}{2ab}$

$cosA=\frac{8^2+9^2-2^2}{\left(2\right)\left(8\right)\left(9\right)}=\frac{141}{144}$

$cosB=\frac{2^2+9^2-8^2}{\left(2\right)\left(2\right)\left(8\right)}=\frac{21}{36}$

$cosC=\frac{8^2+2^2-9^2}{\left(2\right)\left(2\right)\left(8\right)}=\frac{-13}{32}$

$A=co{s}^{-1}\left(\frac{141}{144}\right),B=co{s}^{-1}\left(\frac{21}{36}\right),C=co{s}^{-1}\left(\frac{-13}{32}\right)$

$\therefore$ Largest angle $\to C=co{s}^{-1}\left(\frac{-13}{32}\right)$