datavorous/entrance-exam-dataset huggingface dataset:
The alkali metal with lowest $\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}(\mathrm{V})$ is $\mathrm{X}$ and the alkali metal with highest $\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\circ}(\mathrm{V})$ is $\mathrm{Y}$. Then $\mathrm{X}$ and Y are respectively

Options:
A $\mathrm{Li}, \mathrm{Na}$
B $\mathrm{Li}, \mathrm{Cs}$
C $\mathrm{Na}, \mathrm{Li}$
D $\mathrm{Cs}, \mathrm{Li}$

The correct answer is:
$\mathrm{Li}, \mathrm{Na}$

Solution:
The standard reduction potential of $\mathrm{Li}^{+}$is lowest because of its exceptionally small size and high chargeto-mass ratio. It is highest for $\mathrm{Na}^{+}$to $\mathrm{Na}$ due to its size, electropositivity and hydration enthalpy.

Topic: Chemistry, Electrochemistry, TS EAMCET, TS EAMCET 2023 (14 May Shift 1)
The amine 'A', when treated with nitrous acid, gives a yellow oily substance. The amine A is

Options:
A Triethylamine
B Trimethylamine
C Aniline
D Methylphenylamine

The correct answer is:
Methylphenylamine

Solution:
2o/ Secondary amines âNitrous acid Yellow oily substance.
Out of these given four options, only methyl phenylamine is a secondary amine.

Topic: Chemistry, Amines, MHT CET, MHT CET 2016
The amine which does not react with chloroform and ethanolic potassium hydroxide is

Options:
A
B C6H5-NH2
C
D

The correct answer is:

Solution:
The carbylamine reaction (also known as Hoffmann isocyanide synthesis) is a synthesis of an isocyanide by the reaction of a primary amine, chloroform and a base (ethanolic potassium hydroxide). If a primary amine is present, the isocyanide is formed, which are foul-smelling substances.

Only primary amines give carbylamine test and secondary, tertiary amines do not give this test.

Topic: Chemistry, Amines, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
The amine which will not liberate nitrogen with nitrous acid is

Options:
A ethyl amine
B methyl amine
C dimethyl amine
D 2-amino propane

The correct answer is:
dimethyl amine

Solution:
Only aliphatic primary amines react with nitrous acid to give alcohols with the evolution of $N_{2}$ gas.
$$
R-\mathrm{NH}_{2}+\mathrm{HONO} \stackrel{273-278 \mathrm{~K}}{\longrightarrow} R-\mathrm{OH}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}
$$
$\because$ Dimethyl amine is a secondary amine, it will not give $\mathrm{N}_{2}$ gas.

Topic: Chemistry, Amines, COMEDK, COMEDK 2012
The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis is called:

Options:
A Secondary production
B Primary production
C Gross primary production
D Net primary production

The correct answer is:
Primary production

Solution:
Productivity is the total amount of biomass or organic matter produced per unit area. Over a period of time by plants. This is also called Primary production.
$\rightarrow$ Productivity of consumer is called Secondary production.

Topic: Biology, Ecosystem, NEET, NEET 2022 (Phase 2)
The amount of biomass or organic matter produced per unit area over a time period by the plants during photosynthesis is called

Options:
A Primary productivity
B Secondary productivity
C Standing state
D Standing crop

The correct answer is:
Primary productivity

Solution:
Each trophic level has a certain mass of living material at a particular time called as the standing crop. The standing crop is measured as the mass of living organism (biomass) or the number in a unit area. The biomass of a species is expressed in terms of fresh weight or dry weight. Measurement of biomass in terms of dry weight is more accurate.
Primary production is defined as the amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis.

Topic: Biology, Ecosystem, NEET
The amount of organic matter or biomass that is generated by producers per unit area over a span of time is called ___

Options:
A Primary productivity
B Secondary productivity
C Tertiary productivity
D None of the above

The correct answer is:
Primary productivity

Solution:
The amount of organic matter or biomass that is generated by producers per unit area over a span of time is called primary productivity.

Topic: Biology, Ecosystem, NEET
The angle $A$ lies in the third quadrant and it satisfies the equation $4\left(\sin ^{2} x+\cos x\right)=1$. What is the measure of the $\begin{array}{ll}\text { angle } A ?

Options:
A $225^{\circ}$
B $240^{\circ}$
C $210^{\circ}$
D None of these

The correct answer is:
$210^{\circ}$

Solution:
Given equation is $4\left(\sin ^{2} x+\cos x\right)=1$
$\Rightarrow 4 \sin ^{2} x+4 \cos x=1$
$\Rightarrow 4 \sin ^{2} x+4 \cos x-1=0$
$\Rightarrow 4\left(1-\cos ^{2} x\right)+4 \cos x-1=0$
$\Rightarrow 4-4 \cos ^{2} x+4 \cos x-1=0$
$\Rightarrow-4 \cos ^{2} x+4 \cos x+3=0$
$\Rightarrow 4 \cos ^{2} x-4 \cos x-3=0$
This is the quadratic in $\cos x$. $\Rightarrow 4 \cos ^{2} x-6 \cos x+2 \cos x-3=0$
$\Rightarrow(2 \cos x-3)(2 \cos x+1)=0$
$\Rightarrow \cos x=\frac{3}{2}$ and $\cos x=-\frac{1}{2}$
$\cos x=\frac{3}{2}$ is not possible therefore $\cos x=-\frac{1}{2}$
$\Rightarrow \cos A=-\frac{1}{2}=\cos 210^{\circ}$
$\Rightarrow A=210^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 1)
The angle between lines represented by $\left(\sin ^2 \alpha\right) y^2-2 x y\left(\cos ^2 \alpha\right)+\left(\cos ^2 \alpha-1\right) x^2=0$ is

Options:
A $2 \alpha$
B $\alpha$
C $90^{\circ}$
D $45^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
Pair of straight lines
$$
\left(\sin ^2 \alpha\right) y^2-2 x y \cos ^2 \alpha+\left(\cos ^2 \alpha-1\right) x^2=0
$$
Standard form
$$
\begin{aligned}
& a x^2+2 h x y+b y^2=0 \\
& \therefore \quad \begin{array}{l}
a=\cos ^2 \alpha-1 \\
h=-\cos ^2 \alpha \\
b=\sin ^2 \alpha
\end{array} \\
& \because \tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|
\end{aligned}
$$
$\begin{gathered}\tan \theta=\left|\frac{2 \sqrt{\cos ^4 \alpha-\left(\cos ^2 \alpha-1\right) \sin ^2 \alpha}}{\cos ^2 \alpha-1+\sin ^2 \alpha}\right| \\ \because \sin ^2 \alpha+\cos ^2 \alpha=1 \\ \therefore \quad \tan \theta=\left|\frac{2 \sqrt{\cos ^4 \alpha+\sin ^4 \alpha}}{0}\right| \\ \tan \theta=\infty \Rightarrow \theta=90^{\circ}\end{gathered}$

Topic: Mathematics, Pair of Lines, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 1)
The angle between the curves $y=\sin x$ and $y=\cos x 0

MathematicsArea Under CurvesMHT CETMHT CET 2022 (07 Aug Shift 2)

Options:

A $\tan ^{-1}(\sqrt{2})$
B $\tan ^{-1}(3 \sqrt{2})$
C $\tan ^{-1}(2 \sqrt{2})$
D $\tan ^{-1}(3 \sqrt{3})$

Solution:

1233 Upvotes

Verified Answer

The correct answer is:
$\tan ^{-1}(2 \sqrt{2})$

$y=\sin x$ and $y=\cos x$ intersects at $x=\frac{\pi}{4}$
$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for first curve at } x=\frac{\pi}{4}}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for second curve at } x=\frac{\pi}{4}}=-\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}}$
Required angle $=\theta=\tan ^{-1}\left\{\frac{\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)}{1+\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)}\right\}=\tan ^{-1}\left(\frac{\sqrt{2}}{\frac{1}{2}}\right)$
$=\tan ^{-1}(2 \sqrt{2})$

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Options:
A $\tan ^{-1}(\sqrt{2})$
B $\tan ^{-1}(3 \sqrt{2})$
C $\tan ^{-1}(2 \sqrt{2})$
D $\tan ^{-1}(3 \sqrt{3})$

The correct answer is:
$\tan ^{-1}(2 \sqrt{2})$

Solution:
$y=\sin x$ and $y=\cos x$ intersects at $x=\frac{\pi}{4}$
$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for first curve at } x=\frac{\pi}{4}}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for second curve at } x=\frac{\pi}{4}}=-\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}}$
Required angle $=\theta=\tan ^{-1}\left\{\frac{\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)}{1+\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)}\right\}=\tan ^{-1}\left(\frac{\sqrt{2}}{\frac{1}{2}}\right)$
$=\tan ^{-1}(2 \sqrt{2})$

Topic: Mathematics, Area Under Curves, MHT CET, MHT CET 2022 (07 Aug Shift 2)
The angle between the lines with direction ratios $(1,0, \pm \cos \alpha)$ is $60^{\circ} .$ What is the value of ?

Options:
A $\cos ^{-1}(1 / \sqrt{2})$
B $\cos ^{-1}(1 / \sqrt{3})$
C $\cos ^{-1}(1 / 3)$
D $\cos ^{-1}(1 / 2)$

The correct answer is:
$\cos ^{-1}(1 / \sqrt{3})$

Solution:
Let the angle between two lines whose direction ratios are $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ respectively is 0 .
Then, $\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$\cos 60^{\circ}=\frac{1 \times 1+0 \times 0+(\cos \alpha)(-\cos \alpha)}{\sqrt{1^{2}+(0)^{2}+\cos ^{2} \alpha} \sqrt{1^{2}+(0)^{2}+(-\cos \alpha)^{2}}}$
$\Rightarrow \frac{1}{2}=\frac{1-\cos ^{2} \alpha}{\sqrt{1+\cos ^{2}} \alpha \sqrt{1+\cos ^{2} \alpha}}$
$\Rightarrow \frac{1}{2}=\frac{1-\cos ^{2} \alpha}{1+\cos ^{2} \alpha} \Rightarrow \frac{1+2}{1-2}=\frac{2}{-2 \cos ^{2} \alpha}$
$\Rightarrow \frac{3}{-1}=\frac{1}{-\cos ^{2} \alpha}$
$\Rightarrow \cos \alpha=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 2)
The angle between the lines $x+y-3=0$ and $x-y+3=0$ is $\alpha$ and the acute angle between the lines
$\mathrm{x}-\sqrt{3} \mathrm{y}+2 \sqrt{3}=0$ and $\sqrt{3} \mathrm{x}-\mathrm{y}+1=0$ is $\beta .$ Which one of
the following is correct?

Options:
A $\quad \alpha=\beta$
B $\alpha>\beta$
C $\alpha < \beta$
D $\alpha=2 \beta$

The correct answer is:
$\alpha>\beta$

Solution:
$\quad$ The slope of line $x+y-3=0$ is $-1$ The slope of line $x-y+3=0$ is 1 So, these are perpendicular lines and so angle between them is $90^{\circ}$.
$\therefore \alpha=90^{\circ}$
The slope of line $x-\sqrt{3} y+2 \sqrt{3}=0$ is $m_{1}=\frac{1}{\sqrt{3}}$
The slope of line $\sqrt{3} \mathrm{x}-\mathrm{y}+1=0$ is $\mathrm{m}_{2}=\sqrt{3}$.
$\therefore \tan \beta=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+(\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)}\right|=\left|\frac{2}{2 \sqrt{3}}\right|=\frac{1}{\sqrt{3}} \therefore \beta=30^{\circ}$
$\therefore \alpha>\beta$

Topic: Mathematics, Pair of Lines, NDA, NDA 2017 (Phase 2)
The angle between the two lines $1 \mathrm{x}+\mathrm{my}+\mathrm{n}=0$ and $1^{\prime} x+m^{\prime} y+n^{\prime}=0$ is given by $\tan ^{-1} \theta .$ What $\theta$ equal to?

Options:
A $\left|\frac{\ell \mathrm{m}^{\prime}-\ell^{\prime} \mathrm{m}}{\ell \ell^{\prime}-\mathrm{mm}^{\prime}}\right|$
B $\left|\frac{\ell \mathrm{m}^{\prime}+\ell^{\prime} \mathrm{m}}{\ell \ell^{\prime}+\mathrm{mm}^{\prime}}\right|$
C $\left|\frac{\varepsilon \mathrm{m}^{\prime}-\ell^{\prime} \mathrm{m}}{\ell \ell^{\prime}+\mathrm{mm}^{\prime}}\right|$
D $\left|\frac{\ell \mathrm{m}^{\prime}+\ell^{\prime} \mathrm{m}}{\ell \ell^{\prime}-\mathrm{mm}^{\prime}}\right|$

The correct answer is:
$\left|\frac{\varepsilon \mathrm{m}^{\prime}-\ell^{\prime} \mathrm{m}}{\ell \ell^{\prime}+\mathrm{mm}^{\prime}}\right|$

Solution:
$\tan ^{-1}(\theta)=\tan ^{-1}\left|\frac{\ell m^{\prime}-\ell^{\prime} m}{\ell \ell^{\prime}+m m^{\prime}}\right|$
$\Rightarrow \theta=\left|\frac{\ell m^{\prime}-\ell^{\prime} m}{\ell \ell^{\prime}+m m^{\prime}}\right|$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 2)
The angle between velocity and acceleration of a particle describing uniform circular motion is

Options:
A 180o
B 90o
C 45o
D 60o

The correct answer is:
90o

Solution:
When an object follows a circular path at a constant speed, the motion of object is called uniform circular motion.
Although the speed does not vary the particle is accelerating because the velocity changes its direction at every point on circular track.

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Topic: Physics, Laws of Motion, NEET
The angle of elevation from a point on the bank of ariver of the top of a temple on the other bank is $45^{\circ}$. Retreating $50 \mathrm{~m}$, the observer finds the new angle of elevation as $30^{\circ} .$ What is the width of the river? $\quad$

Options:
A $50 \mathrm{~m}$
B $50 \sqrt{3} \mathrm{~m}$
C $50 /(\sqrt{3}-1) \mathrm{m}$
D $100 \mathrm{~m}$

The correct answer is:
$50 /(\sqrt{3}-1) \mathrm{m}$

Solution:
$\operatorname{In} \Delta \mathrm{ABC}, \tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$
$\Rightarrow 1=\frac{h}{x}$
$\Rightarrow h=x$
Now, in $\triangle \mathrm{ABD}$.
$\tan 30^{\circ}=\frac{A B}{B D}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+50}$
$\Rightarrow x+50=h \sqrt{3}$
$\Rightarrow h+50=h \sqrt{3}$
[fom Eq. (i)]
$\Rightarrow h=\frac{50}{\sqrt{3}-1} m$

Topic: Mathematics, Heights and Distances, NDA, NDA 2009 (Phase 1)
The angle of elevation of a stationary cloud from a point $25 \mathrm{~m}$ above a lake is $15^{\circ}$ and the angle of depression of its image in the lake is $45^{\circ} .$ The height of the cloud above the lake level is

Options:
A $25 \mathrm{~m}$
B $25 \sqrt{3} \mathrm{~m}$
C $50 \mathrm{~m}$
D $50 \sqrt{3} \mathrm{~m}$

The correct answer is:
$25 \sqrt{3} \mathrm{~m}$

Solution:
PQ is the lake and $\mathrm{A}$ is the point of observation. Given, $\mathrm{AP}=25 \mathrm{~m}, \angle \mathrm{BAD}=15^{\circ}$ and $\angle \mathrm{DAC}=45^{\circ}$
Let $\mathrm{BD}=\mathrm{x}$
Then, $\mathrm{DQ}=\mathrm{AP}=25 \mathrm{~m}$ and $\mathrm{QC}=\mathrm{BD}+\mathrm{DQ}=\mathrm{x}+25$
Also, $\mathrm{DC}=\mathrm{DQ}+\mathrm{QC}=25+\mathrm{x}+25=\mathrm{x}+50$
In $\triangle \mathrm{ADC}$
$\tan 45^{\circ}=\frac{\mathrm{DC}}{\mathrm{AD}}=\frac{\mathrm{x}+50}{\mathrm{AD}}$
$\Rightarrow 1=\frac{\mathrm{x}+50}{\mathrm{AD}} \Rightarrow \mathrm{AD}=\mathrm{x}+50$
Also, $\tan 15^{\circ}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$.
In $\Delta \mathrm{ABD}, \tan 15^{\circ}=\frac{\mathrm{BD}}{\mathrm{AD}}$
$\Rightarrow \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{x}{x+50} \quad($ from $(1))$
$\Rightarrow(x+50)(\sqrt{3}-1)=x(\sqrt{3}+1)$
$\Rightarrow \sqrt{3} x-x+50 \sqrt{3}-50=\sqrt{3} x+x$
$\Rightarrow 2 \mathrm{x}=50 \sqrt{3}-50=50(\sqrt{3}-1)$
$\Rightarrow x=25(\sqrt{3}-1)$
Now, $B Q=B D+D Q=x+25$
$=25(\sqrt{3}-1)+25$
$=25 \sqrt{3}-25+25$
$=25 \sqrt{3}$
$\therefore$ Height of the cloud above the lake $=25 \sqrt{3} \mathrm{~m}$.

Topic: Mathematics, Heights and Distances, NDA, NDA 2017 (Phase 2)
The angle of elevation of a tower at a level ground is $30^{\circ}$. The angle of elevation becomes $\theta$ when $10 \mathrm{~m}$ moved towards the tower. If the height of tower is $5 \sqrt{3} \mathrm{~m}$, then what is $\theta$ equal to? $\quad$

Options:
A $45^{\circ}$
B $60^{\circ}$
C $75^{\circ}$
D None of the above

The correct answer is:
$60^{\circ}$

Solution:
Let $\mathrm{BC}$ be the tower of height $5 \sqrt{3} \mathrm{~m}$ Let $\mathrm{AD}=10 \mathrm{~m}$
In $\Delta \mathrm{ABC}$
$\frac{\mathrm{BC}}{\mathrm{AB}}=\tan 30^{\circ}$
$\Rightarrow A B=\frac{B C}{\tan 30^{\circ}}=\frac{5 \sqrt{3}}{1 / \sqrt{3}}=15$
Given $\mathrm{AD}=10 \mathrm{~m}$
$\therefore \mathrm{DB}=(15-10) \mathrm{m}=5 \mathrm{~m}$
In $\Delta \mathrm{BCD}$
$\frac{\mathrm{BC}}{\mathrm{BD}}=\tan \theta$
$\frac{5 \sqrt{3}}{5}=\tan \theta \Rightarrow \tan \theta=\sqrt{3}$
$\Rightarrow \theta=60^{\circ}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2012 (Phase 1)
The angle of elevation of a tower of height h from a point $A$ due South of it is $\mathrm{x}$ and from a point $\mathrm{B}$ due East of $\mathrm{A}$ is $\mathrm{y}$. If $\mathrm{AB}=\mathrm{z}$, then which one of the following is correct?

Options:
A $\mathrm{h}^{2}\left(\cot ^{2} \mathrm{y}-\cot ^{2} \mathrm{x}\right)=\mathrm{z}^{2}$
B $z^{2}\left(\cot ^{2} y-\cot ^{2} x\right)=h^{2}$
C $h^{2}\left(\tan ^{2} y-\tan ^{2} x\right)=z^{2}$
D $z^{2}\left(\tan ^{2} y-\tan ^{2} x\right)=h^{2}$

The correct answer is:
$\mathrm{h}^{2}\left(\cot ^{2} \mathrm{y}-\cot ^{2} \mathrm{x}\right)=\mathrm{z}^{2}$

Solution:
from figure (iii), $\mathrm{h}^{2} \cot ^{2} \mathrm{y}=\mathrm{z}^{2}+\mathrm{h}^{2} \cot ^{2} \mathrm{x}$
$\Rightarrow \mathrm{z}^{2}=\mathrm{h}^{2}\left(\cot ^{2} \mathrm{y}-\cot ^{2} \mathrm{x}\right)$

Topic: Mathematics, Heights and Distances, NDA, NDA 2019 (Phase 1)
The angle of elevation of the tip of a flag staff from a point 10 $\mathrm{m}$ due South of its base is $60^{\circ}$. What is the height of the flag staff correct to the nearest meter? $\quad$

Options:
A $15 \mathrm{~m}$
B $16 \mathrm{~m}$
C $17 \mathrm{~m}$
D $18 \mathrm{~m}$

The correct answer is:
$17 \mathrm{~m}$

Solution:
Let ' $\mathrm{A}$ ' be the top of the flag staff. Let ' $\mathrm{x}$ ' be the height of the flag staff
$\operatorname{In} \Delta \mathrm{ABC}, \tan 60^{\circ}=\frac{\mathrm{x}}{10}$
$\sqrt{3}=\frac{\mathrm{x}}{10}$
$\mathrm{x}=10 \sqrt{3}=10 \times 1.732$
$=17.32 \cong 17$

Topic: Mathematics, Heights and Distances, NDA, NDA 2012 (Phase 1)
The angle of elevation of the top of a flag post from a point 5 m away fromits base is $75^{\circ}$. What is the approximate height of the flag post?

Options:
A $15 \mathrm{~m}$
B $17 \mathrm{~m}$
C $19 \mathrm{~m}$
D $21 \mathrm{~m}$

The correct answer is:
$19 \mathrm{~m}$

Solution:
Let $h$ be the height of the flag post where $\mathrm{BC}$ is the base. In $\Delta \mathrm{ABC}$
$\tan 75^{\circ}=\frac{A B}{B C}=\frac{h}{5}$
$\Rightarrow \tan \left(45^{\circ}+30^{\circ}\right)=\frac{h}{5}$
$\Rightarrow \frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}=\frac{h}{5}$
$\begin{aligned}(\because \tan (A+B)=& \left.\frac{\tan A+\tan B}{1-\tan A \tan B}\right) \\ \Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{h}{5} \Rightarrow h=\frac{(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2}-(1)^{2}} \times 5 \\ \Rightarrow \quad h=5\left(\frac{3+1+2 \sqrt{3}}{3-1}\right)=5(2+\sqrt{3}) \\=5 \times 3.732=18.660 \end{aligned}$
$=19 \mathrm{~m}$ (approx.)

Topic: Mathematics, Heights and Distances, NDA, NDA 2010 (Phase 1)
The angle of elevation of the top of a pillar of height $\mathrm{h}$ at a point on the ground at a distance $\mathrm{x}$ from the pillar is $30^{\circ} . \mathrm{On}$ walking a distance'd towards the pillar the angle of elevation becomes $60^{\circ}$. Then, which one of the following is correct?

Options:
A $\mathrm{x}=\mathrm{d}+\mathrm{h}$
B $x=\frac{3 d}{2}$
C $\mathrm{x}=\frac{5 \mathrm{~d}}{4} \quad$
D $\mathrm{x}=2 \mathrm{~d} \quad\mathrm{}$

The correct answer is:
$x=\frac{3 d}{2}$

Solution:
Let DC be the pillar of height $\mathrm{h}$ and $\mathrm{A}$ be the point at a distance $\mathrm{x}$ from pillar such that $\angle \mathrm{CAD}=30^{\circ} .$ On walking a distance $\mathrm{d}$ towards pillar (point B) $\angle \mathrm{CBD}$ $=60^{\circ} .$ So, in $\Delta \mathrm{BCD}$
$\tan 60^{\circ}=\frac{\mathrm{CD}}{\mathrm{BC}}$
$\Rightarrow \sqrt{3}=\frac{\mathrm{h}}{\mathrm{x}-\mathrm{d}}$
$\Rightarrow \mathrm{h}=\sqrt{3}(\mathrm{x}-\mathrm{d})$
and in $\triangle \mathrm{ACD}$,
$$
\begin{array}{l}
\tan 30^{\circ}=\frac{\mathrm{CD}}{\mathrm{AC}} \\
\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{\mathrm{x}} \\
\Rightarrow \mathrm{x}=\mathrm{h} \sqrt{3} \\
\Rightarrow \mathrm{x}=3(\mathrm{x}-\mathrm{d})
\end{array}
$$
(using Eq. (i))
$$
\Rightarrow x=3 x-3 d
$$
$$
\begin{array}{l}
\Rightarrow 2 \mathrm{x}=3 \mathrm{~d} \\
\Rightarrow \mathrm{x}=\frac{3 \mathrm{~d}}{2}
\end{array}
$$

Topic: Mathematics, Heights and Distances, NDA, NDA 2006 (Phase 2)
The angle of elevation of the top of a tower EF (F being the foot of the tower) as seen from a point $A$ which is on the same level as $\mathrm{F}$, is $\alpha$. On advancing towards the foot of the tower the angle of elevation of the top of the tower as seen from a point $B$ such that $A B=x$, is $\beta .$ If $B F=y$, his the height of the tower and $\alpha+\beta=\frac{\pi}{2}$, then which one of the following is correct?

Options:
A $\mathrm{h}^{2}=\mathrm{x}^{2}+\mathrm{xy}$
B $\mathrm{h}=\mathrm{y}^{2}+\mathrm{xy}^{2}$
C $\mathrm{h}^{2}=\mathrm{y}^{2}+\mathrm{x} y$
D $\mathrm{h}=\mathrm{y}+\mathrm{x}^{2} \mathrm{y} \quad$

The correct answer is:
$\mathrm{h}^{2}=\mathrm{y}^{2}+\mathrm{x} y$

Solution:
Let EF be the height of the tower, and $\mathrm{FB}=\mathrm{Y}, \mathrm{AB}=\mathrm{x}$ and $\mathrm{EF}=\mathrm{h}$.
$\operatorname{In} \Delta \mathrm{BEF}$,
$$
\begin{array}{l}
\tan \beta=\frac{\mathrm{EF}}{\mathrm{BF}} \\
\tan \beta=\frac{\mathrm{h}}{\mathrm{y}}
\end{array}
$$
and in $\triangle \mathrm{AFE}$,
$\begin{aligned} & \tan \alpha=\frac{E F}{A F} \\ \Rightarrow & \tan \alpha=\frac{\mathrm{h}}{\mathrm{x}+\mathrm{y}}, \\ \Rightarrow & \tan \left(\frac{\pi}{2}-\beta\right)=\frac{\mathrm{h}}{\mathrm{x}+\mathrm{y}} \quad\left(\text { Given that } \alpha+\beta=\frac{\pi}{2}\right) \\ \Rightarrow & \cot \beta=\frac{\mathrm{h}}{\mathrm{x}+\mathrm{y}} \end{aligned}$
From Eqs. (i) and (ii), we get
$\begin{aligned} & \tan \beta \cdot \cot \beta=\frac{\mathrm{h}}{\mathrm{y}} \cdot \frac{\mathrm{h}}{(\mathrm{x}+\mathrm{y})} \quad(\because \tan \beta \cdot \cot \beta=1) \\ \Rightarrow \mathrm{xy}+\mathrm{y}^{2}=\mathrm{h}^{2} \end{aligned}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2006 (Phase 2)
The angle of elevation of the top of a tower from a point $20 \mathrm{~m}$ away from its base is $45^{\circ}$. What is the height of the tower?

Options:
A $10 \mathrm{~m}$
B $20 \mathrm{~m}$
C $30 \mathrm{~m}$
D $40 \mathrm{~m}$

The correct answer is:
$20 \mathrm{~m}$

Solution:
$\operatorname{In} \Delta \mathrm{ABC}$,
$\mathrm{AB}=\mathrm{h}$
$\mathrm{BC}=20 \mathrm{~m}$
$\angle \mathrm{C}=45^{\circ}$
$\therefore \quad \tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$
$1=\frac{\mathrm{h}}{20}$
$\mathrm{h}=20 \mathrm{~m}$
$\therefore \quad$ Height of the tower $=20 \mathrm{~m}$
$\therefore$ Option (b) is correct.

Topic: Mathematics, Heights and Distances, NDA, NDA 2015 (Phase 1)
The angle of elevation of the top of a tower from two places situated at distances $21 \mathrm{~m}$. and $x \mathrm{~m}$. from the base of the tower are $45^{\circ}$ and $60^{\circ}$ respectively. What is the value of $x$ ?

Options:
A $7 \sqrt{3}$
B $7-\sqrt{3}$
C $7+\sqrt{3}$
D 14

The correct answer is:
$7 \sqrt{3}$

Solution:
$\operatorname{In} \triangle \mathrm{ADB}, \tan 45^{\circ}=\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AD}}{21}$
$\mathrm{AD}=21 \mathrm{~m}$
$\operatorname{In} \triangle \mathrm{ADC}, \tan 60^{\circ}=\frac{\mathrm{AD}}{\mathrm{x}}$
$\mathrm{x}=\frac{21}{\sqrt{3}}=7 \sqrt{3} \mathrm{~m}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2013 (Phase 2)
The angle of elevation of the top of a tower of height $\mathrm{H}$ from the foot of another tower in the same plane is $60^{\circ}$ and the angle of elevation of the top of the second tower from the foot of the first tower is $30^{\circ}$. If $\mathrm{h}$ is the height of the other tower, then which one of the following is correct?

Options:
A $\mathrm{H}=2 \mathrm{~h}$
B $\mathrm{H}=\sqrt{3} \mathrm{~h} \quad$
C $\quad \mathrm{H}=3 \mathrm{~h}$
D None of the above

The correct answer is:
$\quad \mathrm{H}=3 \mathrm{~h}$

Solution:
Let $\mathrm{AB}$ be the tower of height $\mathrm{H}$. Let $\mathrm{CD}$ be the tower of height $\mathrm{h}$. Let $\mathrm{x}$ be the distance between the $\mathrm{AB}$ and $\mathrm{CD}$.
from $\Delta \mathrm{ABC}, \frac{\mathrm{H}}{\mathrm{x}}=\sqrt{3}$ and from $\Delta \mathrm{ACD}, \frac{\mathrm{h}}{\mathrm{x}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{\mathrm{H}}{\mathrm{h}}=3 \Rightarrow \mathrm{H}=3 \mathrm{~h} .$

Topic: Mathematics, Heights and Distances, NDA, NDA 2013 (Phase 1)