C-10-Q079mediumsingle_mcq

The pole stands $BC=x$ , breaks without separation, and the broken piece $CD$ slopes down to $D$ at distance $BD=10$ m, with $\angle BCD=30^\circ$ . From $\triangle BCD$ , $\tan 30^\circ$ equals—

a $\dfrac{BC}{BD}$
b $\dfrac{BD}{BC}$
c $\dfrac{CD}{BC}$
d $\dfrac{CD}{BD}$

ব্যাখ্যা

In right triangle $BCD$ the right angle is at $B$ (the foot of the pole), so relative to angle $C$ the side $BD$ is opposite and $BC$ is adjacent. Therefore $\tan 30^\circ = \frac{BD}{BC}$ .

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q080Solving for the standing piece, equals—
C-10-Q081Numerically equals—
C-10-Q082From the same triangle, equals—
C-10-Q083So the broken piece equals—
C-10-Q084The total height of the pole is approximately—
C-10-Q085The unbroken (still standing) piece measures approximately—
C-10-Q086The slanted (broken) piece measures exactly—
C-10-Q087A balloon flies above a point between two posts km apart on level ground. From the balloon, the depression of
C-10-Q088The horizontal distance to the foot of the post equals—
C-10-Q089The two horizontal distances together equal the spacing between the posts—

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

The pole stands $BC=x$ , breaks without separation, and the broken piece $CD$ slopes down to $D$ at distance $BD=10$ m, with $\angle BCD=30^\circ$ . From $\triangle BCD$ , $\tan 30^\circ$ equals—

\dfrac{BC}{BD}

\dfrac{BD}{BC}

\dfrac{CD}{BC}

\dfrac{CD}{BD}

ব্যাখ্যা

In right triangle $BCD$ the right angle is at $B$ (the foot of the pole), so relative to angle $C$ the side $BD$ is opposite and $BC$ is adjacent. Therefore $\tan 30^\circ = \frac{BD}{BC}$ .

The pole stands BC=xBC=xBC=x, breaks without separation, and the broken piece CDCDCD slopes down to DDD at distance BD=10BD=10BD=10 m, with ∠BCD=30∘\angle BCD=30^\circ∠BCD=30∘. From △BCD\triangle BCD△BCD, tan⁡30∘\tan 30^\circtan30∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The pole stands BC=xBC=xBC=x, breaks without separation, and the broken piece CDCDCD slopes down to DDD at distance BD=10BD=10BD=10 m, with ∠BCD=30∘\angle BCD=30^\circ∠BCD=30∘. From △BCD\triangle BCD△BCD, tan⁡30∘\tan 30^\circtan30∘ equals—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The pole stands $BC=x$ , breaks without separation, and the broken piece $CD$ slopes down to $D$ at distance $BD=10$ m, with $\angle BCD=30^\circ$ . From $\triangle BCD$ , $\tan 30^\circ$ equals—

The pole stands $BC=x$ , breaks without separation, and the broken piece $CD$ slopes down to $D$ at distance $BD=10$ m, with $\angle BCD=30^\circ$ . From $\triangle BCD$ , $\tan 30^\circ$ equals—