Step-by-step explanation:
Required Answer:-
Let the shortest leg=x
other leg=2x
According to Pythagoras thereon
[tex]{\boxed{\sf p^2+b^2=h^2}}[/tex]
[tex]{:}\dashrightarrow [/tex][tex]\sf (x)^2+(2x)^2=(5)^2 [/tex]
[tex]{:}\dashrightarrow [/tex][tex]\sf x^2+2x^2=25 [/tex]
[tex]{:}\dashrightarrow [/tex][tex]\sf 3x^2=25 [/tex]
[tex]{:}\dashrightarrow [/tex][tex]\sf x^2={\dfrac{25}{3}}[/tex]
[tex]{:}\dashrightarrow [/tex][tex]\sf x^2=8.3 [/tex]
[tex]{:}\dashrightarrow [/tex][tex]\sf x=\sqrt {8.3}[/tex]
[tex]{:}\dashrightarrow [/tex][tex]\sf x=2.8cm [/tex]
[tex]{:}\dashrightarrow [/tex][tex]\sf x=3cm (Approx)[/tex]
[tex]\therefore[/tex]The shorter leg is 3cm long
Learn more:-
Trigonometric Values table:-
[tex]\Large {\begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} $ \sin $ & 0 & $\dfrac{1}{2 }$ & $\dfrac{1}{ \sqrt{2} }$ & $\dfrac{ \sqrt{3}}{2}$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ \dfrac{ \sqrt{ 3 }}{2} } $ & $ \dfrac{1}{ \sqrt{2} } $ & $ \dfrac{ 1 }{ 2 } $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ \dfrac{1}{ \sqrt{3} } $ & 1 & $ \sqrt{3} $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ \sqrt{3} $ & 1 & $ \dfrac{1}{ \sqrt{3} } $ &0 \\ \cline{1 - 6} \sec & 1 & $ \dfrac{2}{ \sqrt{3}} $ & $ \sqrt{2} $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ \sqrt{2 } $ & $ \dfrac{ 2 }{ \sqrt{ 3 } } $ & 1 \\ \cline{1 - 6}\end{tabular}}[/tex]
