C-10-Q122mediumsingle_mcq

With initial distance $x$ to the tower and elevation $60^\circ$ , $\tan 60^\circ$ equals—

a $\dfrac{h}{x}$
b $\dfrac{x}{h}$
c $\dfrac{h}{32}$
d $\dfrac{32}{x}$

ব্যাখ্যা

The height $h$ is opposite the angle and the horizontal distance $x$ is adjacent, so $\tan 60^\circ=\dfrac{h}{x}$ . This is the standard opposite-over-adjacent ratio.

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q123After moving m back, equals—
C-10-Q124Solving the two equations gives
C-10-Q125The tower height m is approximately—
C-10-Q126The width of the river (from observer to opposite bank under tower) equals—
C-10-Q127With standing piece and broken (slanting) piece falling from the break to the ground, the original pole length
C-10-Q128The vertical drop of the break equals the standing piece, giving
C-10-Q129Substituting and solving gives
C-10-Q130The approximate length of the broken piece is—
C-10-Q131The rationalized form equals—
C-10-Q132With standing piece and slanting piece touching ground at where m and , gives

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

With initial distance $x$ to the tower and elevation $60^\circ$ , $\tan 60^\circ$ equals—

\dfrac{h}{x}

\dfrac{x}{h}

\dfrac{h}{32}

\dfrac{32}{x}

ব্যাখ্যা

The height $h$ is opposite the angle and the horizontal distance $x$ is adjacent, so $\tan 60^\circ=\dfrac{h}{x}$ . This is the standard opposite-over-adjacent ratio.

With initial distance xxx to the tower and elevation 60∘60^\circ60∘, tan⁡60∘\tan 60^\circtan60∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

With initial distance xxx to the tower and elevation 60∘60^\circ60∘, tan⁡60∘\tan 60^\circtan60∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

With initial distance $x$ to the tower and elevation $60^\circ$ , $\tan 60^\circ$ equals—

With initial distance $x$ to the tower and elevation $60^\circ$ , $\tan 60^\circ$ equals—