Question Number 43205 by Tawa1 last updated on 08/Sep/18
Answered by $@ty@m last updated on 08/Sep/18
$$\because\angle{SRQ}=\angle{RTS}\:\left({alternate}\:{segment}\:{theorem}\right) \\ $$$$\therefore\angle{SRQ}=\mathrm{28}^{\mathrm{o}} \\ $$$${We}\:{have} \\ $$$$\angle{VRM}+\angle{VRS}+\angle{SRQ}=\mathrm{180}^{\mathrm{o}} \\ $$$$\Rightarrow\mathrm{46}^{\mathrm{o}} +\angle{VRS}+\mathrm{28}^{\mathrm{o}} =\mathrm{180}^{\mathrm{o}} \\ $$$$\Rightarrow\angle{VRS}+\mathrm{106}^{\mathrm{o}} =\mathrm{180}^{\mathrm{o}} ...\left(\mathrm{1}\right) \\ $$$${We}\:{know}\:{that}\:{opposite}\:{angles}\:{of} \\ $$$${a}\:{cyclic}\:{quadrilateral}\:{are}\:{supplementary}. \\ $$$$\therefore\:\angle{VRS}+\angle{VUS}=\mathrm{180}^{\mathrm{o}} ....\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right), \\ $$$${we}\:{get} \\ $$$$\angle{VUS}=\mathrm{106}^{\mathrm{o}} \:{Ans}. \\ $$
Commented by Tawa1 last updated on 08/Sep/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$