C-10-Q076: Hence $h = \dfrac{42\sqrt{3}}{\sqrt{3}-1}$ approximately equals—

C-10-Q076: Hence $h = \dfrac{42\sqrt{3}}{\sqrt{3}-1}$ approximately equals— | প্রশ্ন করি

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Hence h=4233−1h = \dfrac{42\sqrt{3}}{\sqrt{3}-1}h=3​−1423​​ approximately equals—