C-10-Q112mediumsingle_mcq

Initial elevation of tower top is $60^\circ$ at $C$ with $BC=x$ . Then $\tan 60^\circ$ equals—

a $\dfrac{h}{x}$
b $\dfrac{x}{h}$
c $\dfrac{h}{x+25}$
d $\dfrac{25}{h}$

ব্যাখ্যা

In the right triangle, the tower height $h$ is the side opposite the angle and $BC=x$ is the adjacent side. Since tangent equals opposite over adjacent, $\tan 60^\circ=\dfrac{h}{x}$ .

এই অধ্যায়ের আরও প্রশ্ন

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—
C-10-Q120The approximate height of the minar is—
C-10-Q121The exact (rationalized) form equals—
C-10-Q122With initial distance to the tower and elevation , equals—

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

Initial elevation of tower top is $60^\circ$ at $C$ with $BC=x$ . Then $\tan 60^\circ$ equals—

\dfrac{h}{x}

\dfrac{x}{h}

\dfrac{h}{x+25}

\dfrac{25}{h}

ব্যাখ্যা

In the right triangle, the tower height $h$ is the side opposite the angle and $BC=x$ is the adjacent side. Since tangent equals opposite over adjacent, $\tan 60^\circ=\dfrac{h}{x}$ .

Initial elevation of tower top is 60∘60^\circ60∘ at CCC with BC=xBC=xBC=x. Then tan⁡60∘\tan 60^\circtan60∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Initial elevation of tower top is 60∘60^\circ60∘ at CCC with BC=xBC=xBC=x. Then tan⁡60∘\tan 60^\circtan60∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Initial elevation of tower top is $60^\circ$ at $C$ with $BC=x$ . Then $\tan 60^\circ$ equals—

Initial elevation of tower top is $60^\circ$ at $C$ with $BC=x$ . Then $\tan 60^\circ$ equals—