C-10-Q119mediumsingle_mcq

Combined equations give—

a $h = \sqrt{3}(h - 60)$
b $h(\sqrt{3} - 1) = 60\sqrt{3}$
c $h = \dfrac{60\sqrt{3}}{\sqrt{3} - 1}$
dall of these

ব্যাখ্যা

At the new point the elevation is $60^\circ$ , so $\tan 60^\circ=\sqrt{3}=\dfrac{h}{h-60}$ , giving $h=\sqrt{3}(h-60)$ . Rearranging gives $h(\sqrt{3}-1)=60\sqrt{3}$ and hence $h=\dfrac{60\sqrt{3}}{\sqrt{3}-1}$ , so every listed form is correct.

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q120The approximate height of the minar is—
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C-10-Q122With initial distance to the tower and elevation , equals—
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

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

Combined equations give—

h = \sqrt{3}(h - 60)

h(\sqrt{3} - 1) = 60\sqrt{3}

h = \dfrac{60\sqrt{3}}{\sqrt{3} - 1}

dall of these