Answer:
Let the angle be x and its compliment be\[90{}^\circ -x\] According to question, \[x=\frac{2}{3}({{90}^{\circ }}-x)\] \[x=\frac{2}{3}{{90}^{\circ }}-\frac{2}{3}x\] \[x+\frac{2}{3}x={{60}^{\circ }}\] \[\frac{3x+2x}{3}={{60}^{\circ }}\] \[5x={{180}^{\circ }}\] \[x=\frac{{{180}^{\circ }}}{5}={{36}^{\circ }}\]
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