C-10-Q132mediumsingle_mcq

With standing piece $x$ and slanting piece $CD$ touching ground at $D$ where $BD=12$ m and $\angle BCD=30^\circ$ , $\tan 30^\circ = \dfrac{12}{x}$ gives $x = ?$

a $12 \cdot \tan 30^\circ$
b $\dfrac{12}{\tan 30^\circ}$
c $12 \cdot \sin 30^\circ$
d $\dfrac{12}{\sin 30^\circ}$

ব্যাখ্যা

Since $\tan 30^\circ=\dfrac{12}{x}$ relates the opposite side $12$ to the adjacent side $x$ , solving for $x$ moves $\tan 30^\circ$ to the denominator, giving $x=\dfrac{12}{\tan 30^\circ}$ .

এই অধ্যায়ের আরও প্রশ্ন

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

With standing piece $x$ and slanting piece $CD$ touching ground at $D$ where $BD=12$ m and $\angle BCD=30^\circ$ , $\tan 30^\circ = \dfrac{12}{x}$ gives $x = ?$

12 \cdot \tan 30^\circ

\dfrac{12}{\tan 30^\circ}

12 \cdot \sin 30^\circ

\dfrac{12}{\sin 30^\circ}

ব্যাখ্যা

Since $\tan 30^\circ=\dfrac{12}{x}$ relates the opposite side $12$ to the adjacent side $x$ , solving for $x$ moves $\tan 30^\circ$ to the denominator, giving $x=\dfrac{12}{\tan 30^\circ}$ .

With standing piece xxx and slanting piece CDCDCD touching ground at DDD where BD=12BD=12BD=12 m and ∠BCD=30∘\angle BCD=30^\circ∠BCD=30∘, tan⁡30∘=12x\tan 30^\circ = \dfrac{12}{x}tan30∘=x12​ gives x=?x = ?x=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

With standing piece xxx and slanting piece CDCDCD touching ground at DDD where BD=12BD=12BD=12 m and ∠BCD=30∘\angle BCD=30^\circ∠BCD=30∘, tan⁡30∘=12x\tan 30^\circ = \dfrac{12}{x}tan30∘=x12​ gives x=?x = ?x=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

With standing piece $x$ and slanting piece $CD$ touching ground at $D$ where $BD=12$ m and $\angle BCD=30^\circ$ , $\tan 30^\circ = \dfrac{12}{x}$ gives $x = ?$

With standing piece $x$ and slanting piece $CD$ touching ground at $D$ where $BD=12$ m and $\angle BCD=30^\circ$ , $\tan 30^\circ = \dfrac{12}{x}$ gives $x = ?$