C-10-Q113mediumsingle_mcq

After moving $25$ m back to point $D$ where elevation is $30^\circ$ , $\tan 30^\circ$ equals—

Moving $25$ m back makes the new horizontal distance $x+25$ , while the height $h$ stays the same. Therefore $\tan 30^\circ=\dfrac{h}{x+25}$ .

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এই অধ্যায়ের আরও প্রশ্ন

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—

\dfrac{x+25}{h}

\dfrac{25}{h}

∘

x+25h