C-10-Q071mediumsingle_mcq

Building height $AB=h$ , $BC=x$ , with $\angle ACB=60^\circ$ . Then $\tan 60^\circ$ equals—

a $\dfrac{x}{h}$
b $\dfrac{h}{x}$
c $\dfrac{h}{x+42}$
d $\dfrac{x+42}{h}$

ব্যাখ্যা

In right triangle $ABC$ , the angle $C$ is at the base with the building $AB=h$ opposite to it and $BC=x$ adjacent. Since tangent equals the opposite side over the adjacent side, $\tan 60^\circ = \frac{AB}{BC} = \frac{h}{x}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q072From the equation , we get
C-10-Q073Moving back m to point gives elevation , so equals—
C-10-Q074Therefore also equals—
C-10-Q075Combining gives
C-10-Q076Hence approximately equals—
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—

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

Building height $AB=h$ , $BC=x$ , with $\angle ACB=60^\circ$ . Then $\tan 60^\circ$ equals—

\dfrac{x}{h}

\dfrac{h}{x}

\dfrac{h}{x+42}

\dfrac{x+42}{h}

Building height AB=hAB=hAB=h, BC=xBC=xBC=x, with ∠ACB=60∘\angle ACB=60^\circ∠ACB=60∘. Then tan⁡60∘\tan 60^\circtan60∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Building height AB=hAB=hAB=h, BC=xBC=xBC=x, with ∠ACB=60∘\angle ACB=60^\circ∠ACB=60∘. Then tan⁡60∘\tan 60^\circtan60∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Building height $AB=h$ , $BC=x$ , with $\angle ACB=60^\circ$ . Then $\tan 60^\circ$ equals—

Building height $AB=h$ , $BC=x$ , with $\angle ACB=60^\circ$ . Then $\tan 60^\circ$ equals—