C-10-Q109mediumsingle_mcq

The equation $\tan 30^\circ = \dfrac{h}{20}$ gives $h = ?$

a $\dfrac{20}{\sqrt{3}}$ m
b $20\sqrt{3}$ m
c $20 \cdot \tan 30^\circ$ m
dboth (a) and (c)

ব্যাখ্যা

From $\tan 30^\circ = \frac{h}{20}$ , we get $h = 20\tan 30^\circ$ , and since $\tan 30^\circ = \frac{1}{\sqrt{3}}$ this also equals $\frac{20}{\sqrt{3}}$ . Both forms (a) and (c) are therefore correct.

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q110is approximately—
C-10-Q111After rationalization, becomes—
C-10-Q112Initial elevation of tower top is at with . Then equals—
C-10-Q113After moving m back to point where elevation is , equals—
C-10-Q114Combining the two equations gives
C-10-Q115Then equals—
C-10-Q116The approximate height of the tower is—
C-10-Q117From the initial point , elevation of the minar top is . The distance to the foot equals—
C-10-Q118After moving m towards the minar, the new distance to the foot equals—
C-10-Q119Combined equations give—

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

The equation $\tan 30^\circ = \dfrac{h}{20}$ gives $h = ?$

\dfrac{20}{\sqrt{3}}

20\sqrt{3}

20 \cdot \tan 30^\circ

dboth (a) and (c)

The equation tan⁡30∘=h20\tan 30^\circ = \dfrac{h}{20}tan30∘=20h​ gives h=?h = ?h=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The equation tan⁡30∘=h20\tan 30^\circ = \dfrac{h}{20}tan30∘=20h​ gives h=?h = ?h=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The equation $\tan 30^\circ = \dfrac{h}{20}$ gives $h = ?$

The equation $\tan 30^\circ = \dfrac{h}{20}$ gives $h = ?$