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

C-10-Q076mediumsingle_mcq

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

Substituting $x = \frac{42}{\sqrt{3}-1}$ into gives . Evaluating with gives m.

76/160

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q077Rationalizing, is equivalent to—
C-10-Q078As the observer moves backwards on level ground, the angle of elevation of a fixed building top—
C-10-Q079The pole stands , breaks without separation, and the broken piece slopes down to at distance m, with . From ,
C-10-Q080Solving for the standing piece, equals—
C-10-Q081Numerically equals—
C-10-Q082From the same triangle, equals—
C-10-Q083So the broken piece equals—
C-10-Q084The total height of the pole is approximately—
C-10-Q085The unbroken (still standing) piece measures approximately—

84

84

84

h = \sqrt{3}\,x

h = \frac{42\sqrt{3}}{\sqrt{3}-1}

\sqrt{3}\approx 1.732

\frac{72.74}{0.732}\approx 99.37