Question

A person standing at point $A$ on the ground saw an object at point $B$ on the ground $600$ m away. The object started flying towards him at an angle of $30^\circ$ with the ground. The person saw it the second time at point $C$ when the angle of elevation was $30^\circ$. At $C$, the object changed direction and continued flying upwards. The person saw it the third time when it was directly above him. The object flew at a constant speed of $10$ kmph. Find the angle (with the ground) at which it flew after the second sighting. You may use additional statement(s) if required.

Statement I: After changing direction the object took $3$ more minutes than it had taken before.
Statement II: After changing direction the object travelled an additional $200\sqrt{3}$ metres.
Which of the following is correct?

• A. Statement I alone is sufficient to find the angle but Statement II is not.
• B. Statement II alone is sufficient to find the angle but Statement I is not.
• C. Statement I and Statement II are consistent with each other.
• D. Statement I and Statement II are inconsistent with each other.
• E. Neither Statement I nor Statement II is sufficient to find the angle.

Hint:

Fix the geometry first: the second sighting at $30^\circ$ pins down point $C$ ($x=300$, $h=100\sqrt{3}$). After that, $s_2=\dfrac{300}{\cos\theta}$. Translate each statement into an equation for $s_2$ and compare—if they yield different $\theta$, the statements are inconsistent.

Answer 1

The Correct Option is D

Step 1: Geometry up to the second sighting.
Let $A$ be at $x=0$ and $B$ at $x=600$ on the ground. The object flies from $B$ towards $A$ at $30^\circ$ to the ground. At the second sighting, angle of elevation from $A$ is $30^\circ$. If the horizontal distance from $A$ then is $x$, the height is $x/\sqrt{3}$. Along the first $30^\circ$ path, height also equals $(600-x)/\sqrt{3}$. Equating gives $x=300$ and height $=100\sqrt{3}$. Thus $BC$ has horizontal projection $300$ m, so \[ s_1=\frac{300}{\cos 30^\circ}=\frac{300}{\sqrt{3}/2}=\frac{600}{\sqrt{3}}=200\sqrt{3}\ \text{m}. \] Step 2: Motion after the second sighting.
From $C$, the object heads in a straight line at angle $\theta$ to the ground and passes vertically above $A$. Horizontal gap to cover is $300$ m, hence the second-leg distance \[ s_2=\frac{300}{\cos\theta}. \] Step 3: Use each statement to obtain $\theta$.
Speed $v=10$ kmph $=\dfrac{10{,}000}{60}=166.\overline{6}$ m/min. {Statement I (time $3$ minutes more):} \[ t_2=t_1+3 \ \Rightarrow\ \frac{s_2}{v}=\frac{s_1}{v}+3 \ \Rightarrow\ s_2=s_1+3v=200\sqrt{3}+500. \] Hence \[ \cos\theta=\frac{300}{s_2}=\frac{300}{200\sqrt{3}+500} \ \Rightarrow\ \theta\approx 69.2^\circ. \] {Statement II (distance $200\sqrt{3}$ m more):} \[ s_2=s_1+200\sqrt{3}=400\sqrt{3}\ \Rightarrow\ \cos\theta=\frac{300}{400\sqrt{3}}\ \Rightarrow\ \theta\approx 64.4^\circ. \] Step 4: Conclusion.
Statements I and II give {different} values of $\theta$ for the same situation, i.e., they cannot both hold simultaneously. Therefore, the two statements are inconsistent. \[ \boxed{\text{Answer (D): Statements I and II are inconsistent with each other.}} \]