Question Number 120946 by mnjuly1970 last updated on 04/Nov/20
Answered by Jamshidbek2311 last updated on 04/Nov/20
$$\frac{{sin}\beta}{{sin}\mathrm{20}°}=\frac{{sin}\left(\mathrm{110}°−\beta\right)}{{sin}\mathrm{10}°}\:\Rightarrow\:\beta=\mathrm{80}° \\ $$$$ \\ $$