# The chainage of the intersection point of two straights is $1585.60 \mathrm{~m}$ and the angle of intersection is $140^{\circ}$. If the radius of a circular curve is $600.00 \mathrm{~m}$, the tangent distance (in $\mathrm{m}$ ) and length of the curve (in m), respectively are (A) 418.88 and 1466.08 (B) 218.38 and 1648.49 (C) 218.38 and 418.88 (D) 418.88 and 218.38

## Question ID - 156477 :- The chainage of the intersection point of two straights is $1585.60 \mathrm{~m}$ and the angle of intersection is $140^{\circ}$. If the radius of a circular curve is $600.00 \mathrm{~m}$, the tangent distance (in $\mathrm{m}$ ) and length of the curve (in m), respectively are (A) 418.88 and 1466.08 (B) 218.38 and 1648.49 (C) 218.38 and 418.88 (D) 418.88 and 218.38

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Answer Key : (C) -

218.38 and 418.88

Next Question :

Integration of $\int_{0}^{D / 2} \frac{2 R^{2} x d x}{\left(R^{2}+x^{2}\right)^{2}}$ is
(A) $\frac{R^{2}}{R^{2}+D^{2}}$
(B) $\frac{4 R^{2}}{4 R^{2}+D^{2}}$
(C) $\frac{D^{2}}{R^{2}+D^{2}}$
(D) $\frac{D^{2}}{4 R^{2}+D^{2}}$

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