In the given figure, $\mathrm{AOB}$  is a straight line. If  $\angle \mathrm{AOC}\text{ =}{\left(3\mathrm{x}+10\right)}^{°}$ and  $\angle \mathrm{BOC}={\left(4\mathrm{x}–26\right)}^{°}$ , then the value of $\angle \mathrm{BOC}$  is

It is given that $\mathrm{AOB}$  is a straight line. 

And,  $\angle \mathrm{AOC}\text{ =}\left(3\mathrm{x}+10\right)°$ and  $\angle \mathrm{BOC}=\left(4\mathrm{x}–26\right)°$ .

The angles $\angle \mathrm{AOC},\angle \mathrm{BOC}$  form a linear pair. So,

$\angle \mathrm{AOC}+\angle \mathrm{BOC}={180}^{°}$  ----[Linear pair]

Calculate the above equation to get the value of angle x ,

$\angle \mathrm{AOC}+\angle \mathrm{BOC}={180}^{°}$ 

$(3\mathrm{x}+{10}^{°})+(4\mathrm{x}-{26}^{°})={180}^{°}$ 

$7\mathrm{x}-{16}^{°}={180}^{°}$ 

$7\mathrm{x}={196}^{°}$ 

$\mathrm{x}={28}^{°}$ 

Calculate the value of $\angle \mathrm{BOC}=4\mathrm{x}-{26}^{°}$ 

$\angle \mathrm{BOC}=4\mathrm{x}-{26}^{°}$ 

$\angle \mathrm{BOC}=(4\times {28}^{°})-{26}^{°}$ 

$\angle \mathrm{BOC}={112}^{°}-{26}^{°}$ 

$\angle \mathrm{BOC}={86}^{°}$ 

Therefore, the value of $\angle \mathrm{BOC}$  is $86°$ .