Question Number 202911 by Ari last updated on 05/Jan/24
Answered by AST last updated on 05/Jan/24
Commented by AST last updated on 05/Jan/24
$${WLOG},{let}\:{the}\:{side}\:{of}\:{the}\:{square}\:{be}\:\mathrm{1} \\ $$$$\frac{{sin}\:\mathrm{70}}{\mathrm{1}}=\frac{{sin}\left(\mathrm{90}\right)}{{b}}\Rightarrow{b}=\frac{\mathrm{1}}{{cos}\left(\mathrm{20}\right)} \\ $$$$\frac{{sin}\left(\mathrm{65}\right)}{\mathrm{1}}=\frac{{sin}\left(\mathrm{90}\right)}{{a}}\Rightarrow{a}=\frac{\mathrm{1}}{{cos}\left(\mathrm{25}\right)} \\ $$$$\Rightarrow{cos}\left(\mathrm{25}\right){sin}\left({x}\right)={sin}\left(\mathrm{135}−{x}\right){cos}\left(\mathrm{20}\right)\Rightarrow{x}=\mathrm{70}° \\ $$
Answered by mr W last updated on 05/Jan/24
Commented by mr W last updated on 05/Jan/24
$$\Delta{ABD}\equiv\Delta{ABC} \\ $$$${x}=\mathrm{90}°−\mathrm{20}°=\mathrm{70}° \\ $$
Commented by AST last updated on 05/Jan/24
$${How}\:{did}\:{you}\:{get}\:\angle{ABD}={x}°? \\ $$
Commented by mr W last updated on 05/Jan/24
$$\angle{DAB}=\angle{CAB}=\mathrm{45}° \\ $$$${AD}={AC} \\ $$$$\Rightarrow\Delta{ABD}\equiv\Delta{ABC} \\ $$$$\Rightarrow\angle{ABD}=\angle{ABC}={x} \\ $$