C-10-Q076mediumsingle_mcq

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

a $42$ m
b $99.37$ m
c $60.62$ m
d $84$ m

ব্যাখ্যা

Substituting $x = \frac{42}{\sqrt{3}-1}$ into $h = \sqrt{3}\,x$ gives $h = \frac{42\sqrt{3}}{\sqrt{3}-1}$ . Evaluating with $\sqrt{3}\approx 1.732$ gives $\frac{72.74}{0.732}\approx 99.37$ m.

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—
C-10-Q086The slanted (broken) piece measures exactly—

Chapter 10: Distance and Elevation — সব প্রশ্ন দেখুন (160টি) →

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

42

99.37

60.62

84

ব্যাখ্যা

Substituting $x = \frac{42}{\sqrt{3}-1}$ into $h = \sqrt{3}\,x$ gives $h = \frac{42\sqrt{3}}{\sqrt{3}-1}$ . Evaluating with $\sqrt{3}\approx 1.732$ gives $\frac{72.74}{0.732}\approx 99.37$ m.

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

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

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

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

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

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