C-10-Q083: So the broken piece $CD = \dfrac{BD}{\sin 30^\circ} = \dfrac{10}{1/2}$… | প্রশ্ন করি

C-10-Q083mediumsingle_mcq

So the broken piece $CD = \dfrac{BD}{\sin 30^\circ} = \dfrac{10}{1/2}$ equals—

From $\sin 30^\circ = \frac{BD}{CD}$ with , the broken slanted piece is m.

83/160

এই অধ্যায়ের আরও প্রশ্ন

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—
C-10-Q090Solving gives
C-10-Q091Approximate height of the balloon above the ground in metres—
C-10-Q092Given , the ratio stick : shadow equals—

25

25

\sin 30^\circ = \frac{1}{2}

sin 3 0^{\circ} = \frac{1}{2}

CD = \frac{10}{1/2} = 20