Last Updated On [03-11-2023]
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
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
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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