Question

Plz guys solve this question

Answer 1

∆ABC is Right triangle than,
AC^2=AB^2+BC^2
Suppose that AC = x
${x}^{2} = {(20)}^{2} + {(20 \sqrt{3)} }^{2} \\ {x}^{2} = 400 + 400 \times 3 \\ {x}^{2} = 400 + 1200 \\ {x}^{2} = 1600 \\x = \sqrt{1600} \\ x = 40$
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Answer 2

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