MCA NIMCET Heights and Distances Previous Year Question Paper and Solution with PDF

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MCA NIMCET Heights And Distances PYQ

Qus : 1 MCA NIMCET PYQ

From the top of a viewpoint at a height of 80 m, the angles of depression of the top and bottom of a flag standing on the same plane are $30^\circ$ and $45^\circ$. Find the height of the flag.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

Let the distance between viewpoint and flag be $x$.

For bottom of flag:

$\tan45^\circ=\frac{80}{x}$

$x=80$

Let height of flag be $h$.

For top of flag: $\tan30^\circ=\frac{80-h}{x}$

$\frac{1}{\sqrt{3}}=\frac{80-h}{80}$

$80-h=\frac{80}{\sqrt{3}}$

$h=80-\frac{80}{\sqrt{3}}$

$h=80\left(1-\frac{1}{\sqrt{3}}\right)$

Answer: $(c) 80\left(1-\frac{1}{\sqrt{3}}\right)$

Qus : 2 MCA NIMCET PYQ

A line segment AB of length 10 meters is passing through the foot of the perpendicular of a pillar, which is standing at right angle to the ground. Top of the pillar subtends angles $tan^{–1}$ 3 and $tan^{–1} 2$ at A and B respectively. Which of the following choice represents the height of the pillar?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 3 MCA NIMCET PYQ

Car travels half distance with $v_1$, half with $v_2$. Average speed is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

For equal distances: Average velocity $=\dfrac{2v_1v_2}{v_1+v_2}$

Qus : 4 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 5 MCA NIMCET PYQ

A man observes the angle of elevation of the top of mountain to be 30^o. He walks 1000 feet nearer and finds the angle of elevation to be $45^{o}$. What is the distance of the first point of observation from the foot of the mountain?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 6 MCA NIMCET PYQ

Angles of elevation of the top of a tower from three points (collinear) A, B and C on a road leading to the foot of the tower are 30°, 45° and 60° respectively. The ratio of AB and BC is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 7 MCA NIMCET PYQ

If the angle of elevation of the top of a hill from each of the vertices A, B and C of a horizontal triangle is $\alpha$, then the height of the hill is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 8 MCA NIMCET PYQ

A tower subtends angles $\alpha, 2\alpha$ and $3\alpha$ respectively at points A, B and C which are lying on a horizontal line through the foot of the tower. Then $\frac{AB}{BC}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Qus : 9 MCA NIMCET PYQ

The angles of depression of the top and bottom of an 8m tall building from the top of a multi storied building are 30° and 45°, respectively. What is the height of the multistoried building and the distance between the two buildings?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

The height of the multistoried building is $H$ and the horizontal distance between the buildings is $D$.

The angles of depression to the top and bottom of the $8\text{ m}$ building are $30^\circ$ and $45^\circ$ respectively.

From the $45^\circ$ angle:

$\tan 45^\circ = \dfrac{H}{D}$

$1 = \dfrac{H}{D}$

$H = D$

From the $30^\circ$ angle:

$\tan 30^\circ = \dfrac{H - 8}{D}$

$\dfrac{1}{\sqrt{3}} = \dfrac{H - 8}{H}$

$H - 8 = \dfrac{H}{\sqrt{3}}$

$H\left(1 - \dfrac{1}{\sqrt{3}}\right) = 8$

Solve for $H$:

$H = \dfrac{8}{1 - \dfrac{1}{\sqrt{3}}}$

$H = 4(3 + \sqrt{3})$

Since $H = D$:

$D = 4(3 + \sqrt{3})$

Final answers:

$\boxed{H = 4(3 + \sqrt{3})\ \text{m}}$

$\boxed{D = 4(3 + \sqrt{3})\ \text{m}}$

Qus : 10 MCA NIMCET PYQ

A person stands at a point $A$ due south of a tower and observes elevation $60^\circ$. He walks west to $B$, elevation becomes $45^\circ$. At point $C$ on $AB$ extended, elevation becomes $30^\circ$. Find $\dfrac{AB}{BC}$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Let height of tower = $h$

From A (angle $60^\circ$):

$\tan 60^\circ = \frac{h}{OA} \Rightarrow OA = \frac{h}{\sqrt{3}}$

From B (angle $45^\circ$):

$\tan 45^\circ = \frac{h}{OB} \Rightarrow OB = h$

$AB = OB - OA = h - \frac{h}{\sqrt{3}}$

From C (angle $30^\circ$):

$\tan 30^\circ = \frac{h}{OC} \Rightarrow OC = \sqrt{3}h$

$BC = OC - OB = \sqrt{3}h - h$

Now ratio:

$\frac{AB}{BC} = \frac{h - \frac{h}{\sqrt{3}}}{\sqrt{3}h - h}$

$= \frac{h\left(1 - \frac{1}{\sqrt{3}}\right)}{h(\sqrt{3} - 1)}$

$= \frac{1 - \frac{1}{\sqrt{3}}}{\sqrt{3} - 1}$

Multiply numerator & denominator by $\sqrt{3}$:

$= \frac{\sqrt{3} - 1}{\sqrt{3}(\sqrt{3} - 1)}$

$= \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{1} = 1$

Final Answer: $$\boxed{1}$$

Qus : 11 MCA NIMCET PYQ

If $BC=a$, $AC=b$, and $AB=c$ are the sides of a triangle $ABC$, and $\angle C\ne \frac{\pi}{2}$, then which one of the following is not correct?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

By sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

So,

$\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A+\sin B}$

Now,

$\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

and

$\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

Therefore,

$\frac{\sin A-\sin B}{\sin A+\sin B}=\cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right)$

Since,

$A+B=\pi-C$

So,

$\cot\left(\frac{A+B}{2}\right)=\cot\left(\frac{\pi-C}{2}\right)$

$=\cot\left(\frac{\pi}{2}-\frac{C}{2}\right)$

$=\tan\left(\frac{C}{2}\right)$

Thus,

$\frac{a-b}{a+b}=\tan\left(\frac{C}{2}\right)\tan\left(\frac{A-B}{2}\right)$

But option $4$ gives

$\frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{C}{2}\right)}$

Qus : 12 MCA NIMCET PYQ

An airplane, when 4000m high from the ground, passes vertically above another airplane at an instant when the angles of elevation of the two airplanes from the same point on the ground are 60° and 30°, respectively. Find the vertical distance between the two airplanes.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Let the higher airplane be at a height of $4000\,\text{m}$.

From a point on the ground, the angles of elevation to the two airplanes are $60^\circ$ (upper plane) and $30^\circ$ (lower plane).

Let the horizontal distance from the observer to the airplanes be $x$.

For the upper airplane:

$\tan 60^\circ = \dfrac{4000}{x}$

$\sqrt{3} = \dfrac{4000}{x}$

$\Rightarrow x = \dfrac{4000}{\sqrt{3}}$.

For the lower airplane with height $h$:

$\tan 30^\circ = \dfrac{h}{x}$

$\dfrac{1}{\sqrt{3}} = \dfrac{h}{4000/\sqrt{3}}$

Thus, $h = \dfrac{4000}{3}$.

Now the vertical distance between the two airplanes: $4000 - \dfrac{4000}{3} = \dfrac{8000}{3}.$

Final Answer: $\displaystyle \frac{8000}{3}\text{ m}$.

Qus : 13 MCA NIMCET PYQ

A tower subtends an angle of 30° at a point on the same level as the foot of the tower. At a second point h meters above the first, the depression of the foot of the tower is 60°. What is the horizontal distance of the tower from the point?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Qus : 14 MCA NIMCET PYQ

From three collinear points A, B and C on a level ground, which are on the same side of a tower, the angles of elevation of the top of the tower are 30°, 45° and 60° respectively. If BC = 60 m, then AB is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 15 MCA NIMCET PYQ

A circus artist is climbing a $20$ m long rope, which is tightly streached and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^\circ$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 16 MCA NIMCET PYQ

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° degree and 45° repectively. If the lighthouse is 100 m high, the distance between the two ships is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Let distances from the lighthouse be $x$ (for $30^\circ$) and $y$ (for $45^\circ$). With height $h=100$ m:

$\tan 30^\circ=\dfrac{100}{x}\ \Rightarrow\ x=\dfrac{100}{\tan30^\circ}=100\sqrt{3}$, $\tan 45^\circ=\dfrac{100}{y}\ \Rightarrow\ y=\dfrac{100}{\tan45^\circ}=100$.

Ships are on opposite sides ⇒ distance $=x+y=100\sqrt{3}+100=100(\sqrt{3}+1)\approx \boxed{273.2\ \text{m}}$.

Qus : 17 MCA NIMCET PYQ

Two towers face each other separated by a distance of 25 meters. As seen from the top of the first tower, the angle of depression of the second tower’s base is 60° and that of the top is 30°. The height (in meters) of the second tower is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Two Towers – Height of the Second Tower

Distance between towers’ bases = 25 m

Let the first tower be $AB$ (top $A$, base $B$) and the second tower be $CD$ (top $C$, base $D$). The bases $B$ and $D$ are 25 m apart.

From the top $A$: angle of depression to base $D$ is $60^\circ$ and to top $C$ is $30^\circ$.

From right $\triangle ABD$: \[ \tan 60^\circ=\frac{AB}{BD}=\frac{AB}{25}\;\Rightarrow\; AB=25\sqrt{3}. \]
From right $\triangle ACD$: vertical difference $=AB-CD$ and horizontal $=25$. \[ \tan 30^\circ=\frac{AB-CD}{25}=\frac{1}{\sqrt{3}} \;\Rightarrow\; AB-CD=\frac{25}{\sqrt{3}}. \]
Substitute $AB=25\sqrt{3}$: \[ 25\sqrt{3}-CD=\frac{25}{\sqrt{3}} \;\Rightarrow\; CD=25\!\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) =\frac{50}{\sqrt{3}}\;\text{m}. \]

Answer: $ \displaystyle CD=\frac{50}{\sqrt{3}} \approx 28.87\ \text{m} $

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