C-10-Q049mediumsingle_mcq

Numerically $75 \cdot \tan 30^\circ$ equals—

a $\dfrac{75}{\sqrt{3}}$ m
b $75\sqrt{3}$ m
c $25$ m
d $75$ m

ব্যাখ্যা

Substituting $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$ gives $h = 75 \cdot \dfrac{1}{\sqrt{3}} = \dfrac{75}{\sqrt{3}}$ m. This is the value before rationalizing the denominator.

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q050After rationalizing, becomes—
C-10-Q051The approximate height of the tower in Example 1 is—
C-10-Q052To rationalize , the numerator and denominator are multiplied by—
C-10-Q053In Example 2, with tree m, foot distance , and , equals—
C-10-Q054The distance satisfies—
C-10-Q055simplifies to—
C-10-Q056The approximate ground distance of the point in Example 2 is—
C-10-Q057The relation tells us that the side opposite the angle is times the—
C-10-Q058With the same height m, if the elevation were instead of , the foot distance would be—
C-10-Q059A ladder of length m leans against a wall making with the ground; with m and wall height , equals—

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

Numerically $75 \cdot \tan 30^\circ$ equals—

\dfrac{75}{\sqrt{3}}

75\sqrt{3}

25

75

ব্যাখ্যা

Substituting $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$ gives $h = 75 \cdot \dfrac{1}{\sqrt{3}} = \dfrac{75}{\sqrt{3}}$ m. This is the value before rationalizing the denominator.

Numerically 75⋅tan⁡30∘75 \cdot \tan 30^\circ75⋅tan30∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Numerically 75⋅tan⁡30∘75 \cdot \tan 30^\circ75⋅tan30∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Numerically $75 \cdot \tan 30^\circ$ equals—

Numerically $75 \cdot \tan 30^\circ$ equals—