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(B) From the geometry of the problem,we have a right-angled triangle $ABC$ where $\angle BAC = 90^{\circ}$ and $\angle ABC = 	heta = 40^{\circ}$.Usin

Vedclass · Accepted Answer

$119$

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(B) From the geometry of the problem,we have a right-angled triangle $ABC$ where $\angle BAC = 90^{\circ}$ and $\angle ABC = 	heta = 40^{\circ}$.Using the trigonometric relation for the triangle $ABC$:$	an 	heta = \frac{AC}{AB}$Given that $AB = 100 \; m$ and $	heta = 40^{\circ}$,we can rearrange the formula to solve for $AC$:$AC = AB 	imes 	an 	heta$Wait,looking at the provided diagram,the angle $	heta$ is at vertex $B$ (the angle between $BC$ and $AB$ is $90^{\circ} - 	heta$,but the angle given as parallax $	heta$ is at $B$ relative to the line $BC$ and the vertical). Actually,in the triangle $ABC$,$	an 	heta = \frac{AC}{AB}$ is incorrect. Based on the diagram,$	an 	heta = \frac{AB}{AC}$.Therefore,$AC = \frac{AB}{	an 	heta} = \frac{100 \; m}{	an 40^{\circ}}$.Using $	an 40^{\circ} \approx 0.8391$:$AC = \frac{100}{0.8391} \approx 119.17 \; m$.Rounding to the nearest whole number,the distance is $119 \; m$.

List-$I$	List-$II$
$(1)$ Rotation period of earth	$(i)$ $10^5\, s$
$(2)$ Revolution period of earth	$(ii)$ $10^7\, s$
$(3)$ Period of light wave	$(iii)$ $10^{-15}\, s$
$(4)$ Period of sound wave	$(iv)$ $10^{-3}\, s$

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