datavorous/entrance-exam-dataset huggingface dataset:
The angle between the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and the plane $3 x+2 y-3 z=4$ is

Options:
A $45^{\circ}$
B $0^{\circ}$
C $\cos ^{-1}\left(\frac{24}{\sqrt{29} \sqrt{22}}\right)$
D $90^{\circ}$

The correct answer is:
$0^{\circ}$

Solution:
Angle between the plane and line is
$\sin \theta=\frac{a a^{\prime}+b b^{\prime}+c c^{\prime}}{\sqrt{a^2+b^2+c^2} \sqrt{a^{\prime 2}+b^{\prime 2}+c^{\prime 2}}}$
Here, $a a^{\prime}+b b^{\prime}+c c^{\prime}=2 \times 3+3 \times 2-4 \times 3=0$
$\therefore \sin \theta=0 \Rightarrow \theta=0^{\circ} .$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The angle between the line \( \frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{-1} \) and the plane \( x+y+z=4 \) is

Options:
A \( \sin ^{-1}\left(\frac{\sqrt{2}}{3}\right) \)
B \( \cos ^{-1}\left(\frac{\sqrt{2}}{3}\right) \)
C \( \tan ^{-1}\left(\sqrt{\frac{2}{13}}\right) \)
D \( \tan ^{-1}\left(\sqrt{\frac{13}{2}}\right) \)

The correct answer is:
\( \sin ^{-1}\left(\frac{\sqrt{2}}{3}\right) \)

Solution:
Angle θ between line x-x1a=y-y1b=z-z1c and the plane lx+my+nz-d=0

is
θ=sin-1al+am+ana2+b2+c2l2+m2+n2
Here we have line x-21=y-32=z-4-1

and plane x+y+z-4=0
So, θ=sin -11×1+2×1-1×112+22+-1212+12+12=sin-126×3=sin-1218
=sin-123

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The angle between the lines $\frac{x}{1}=\frac{y}{0}=\frac{z}{-1}$ and $\frac{x}{3}=\frac{y}{4}=\frac{z}{5}$ is

Options:
A $\cos ^{-1} \frac{1}{5}$
B $\cos ^{-1} \frac{1}{3}$
C $\cos ^{-1} \frac{1}{2}$
D $\cos ^{-1} \frac{1}{4}$

The correct answer is:
$\cos ^{-1} \frac{1}{5}$

Solution:
$\begin{aligned}
& \theta=\cos ^{-1}\left(\frac{3+0-5}{\sqrt{1+1} \sqrt{9+16+25}}\right) \\
& =\cos ^{-1}\left(\frac{-2}{ \pm 10}\right)=\cos ^{-1}\left(\frac{1}{5}\right)
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The angle between the lines $\frac{x+4}{1}=\frac{y-3}{2}=\frac{z+2}{3}$ and $\frac{x}{3}=\frac{y-1}{-2}=\frac{z}{1}$ is

Options:
A $\sin ^{-1}\left(\frac{1}{7}\right)$
B $\cos ^{-1}\left(\frac{2}{7}\right)$
C $\cos ^{-1}\left(\frac{1}{7}\right)$
D None of these

The correct answer is:
$\cos ^{-1}\left(\frac{1}{7}\right)$

Solution:
Angle between two lines,
$\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$
$\begin{aligned} & \therefore \cos \theta=\frac{1 \times 3+2 \times-2+3 \times 1}{\sqrt{1^2+2^2+3^2} \sqrt{3^2+(-2)^2+1^2}}=\frac{2}{\sqrt{14} \sqrt{14}} \\ & \therefore \theta=\cos ^{-1}\left(\frac{1}{7}\right)\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The angle between the lines $2 x=3 y=-z$ and $6 x=-y=-4 z$, is

Options:
A $0^{\circ}$
B $30^{\circ}$
C $45^{\circ}$
D $90^{\circ}$

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

Solution:
The lines are $\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$ and $\frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$
Since, $a_1 a_2+b_1 b_2+c_1 c_2=6-24+18=0 \Rightarrow \theta=90^{\circ}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The angle between the lines $2 x=3 y=-z$ and $6 x=-y=-4 z$ is

Options:
A
$0^0$

B
$90^0$

C
$45^0$

D
$30^0$

The correct answer is:
$90^0$

Solution:
Angle between the lines $2 x=3 y=-z \& 6 x=-y=-4 z$ is $90^{\circ}$ Since $a_1 a_2+b_1 b_2+c_1 c_2=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2005
The angle between the lines joining the foci of an ellipse to one particular extremity of the minor axis is $90^{\circ}$. The eccentricity of the ellipse is

Options:
A $\frac{1}{8}$
B $\frac{1}{\sqrt{3}}$
C $\sqrt{\frac{2}{3}}$
D $\sqrt{\frac{1}{2}}$

The correct answer is:
$\sqrt{\frac{1}{2}}$

Solution:
$$
\begin{gathered}
\text { } \frac{b}{a e}=\tan \frac{\pi}{4} \\
\qquad \begin{array}{c}
a e \Rightarrow \frac{b}{a}=e \\
e^2=1-\frac{b^2}{a^2} \\
e^2=1-e^2
\end{array}
\end{gathered}
$$
$$
e^2=\frac{1}{2} \Rightarrow e=\frac{1}{\sqrt{2}}
$$

Topic: Mathematics, Ellipse, JEE Main
The angle between the lines whose direction cosines satisfy the equations l⁡+m+n=0 and l⁡2=m2+n2 is

Options:
A π 6
B π 2
C Ï 3
D π 4

The correct answer is:
π 3

Solution:
l⁡+m+n=0
m+n=-l⁡
∴ m+n2=l⁡2
∴ l⁡2= m2+n2+2mn
But, m2+n2=l⁡2
∴ 2mn=0
∴ m = 0 or n = 0
∴ l⁡=-n or l⁡=-m
∵l⁡2+m2+n2=1
∴ The two direction cosines are 12,0,-12 and 12,-12,0
θ is the angle between them.
∴ cos θ=l1l2+m1m2+n1n2=1212+0+0=12
∴ θ = π 3 .
Note : Taking l⁡ = - 1 2 , will also given the same solution, but the supplementary angle. Both are correct, we choose the one present in the options.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (06 Apr)
The angle between the lines whose intercepts on the axes are $a,-b$ and $b,-a$ respectively, is

Options:
A $\tan ^{-1} \frac{a^2-b^2}{a b}$
B $\tan ^{-1} \frac{b^2-a^2}{2}$
C $\tan ^{-1} \frac{b^2-a^2}{2 a b}$
D None of these

The correct answer is:
$\tan ^{-1} \frac{b^2-a^2}{2 a b}$

Solution:
Equation of lines are $\frac{x}{a}-\frac{y}{b}=1$ and $\frac{x}{b}-\frac{y}{a}=1$
$\begin{aligned}
& \Rightarrow m_1=\frac{b}{a} \text { and } m_2=\frac{a}{b} \\
& \theta=\tan ^{-1} \frac{\frac{b}{a}-\frac{a}{b}}{1+\frac{b}{a} \cdot \frac{a}{b}}=\tan ^{-1} \frac{b^2-a^2}{2 a b} \\
&
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
The angle between the lines $x \cos 30^{\circ}+y \sin 30^{\circ}=3$ and $x \cos 60^{\circ}+y \sin 60^{\circ}=5$ is

Options:
A $90^{\circ}$
B $30^{\circ}$
C $60^{\circ}$
D None of these

The correct answer is:
$30^{\circ}$

Solution:
$\begin{aligned} \theta=\tan ^{-1} & \left(\frac{-\cot 30^{\circ}+\cot 60^{\circ}}{1+\cot 30^{\circ} \cot 60^{\circ}}\right) \\ & =\tan ^{-1}\left(\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 30^{\circ} \tan 60^{\circ}}\right)=30^{\circ}\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
The angle between the lines $y=(2-\sqrt{3}) x+5$ and $y=(2+\sqrt{3}) x-7$ is

Options:
A $30^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $90^{\circ}$

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

Solution:
$\theta=\tan ^{-1}\left(\frac{2-\sqrt{3}-2-\sqrt{3}}{1+4-3}\right)=\tan ^{-1}(-\sqrt{3})=120^{\circ}$
Considering smaller angle $\theta^{\prime}=60^{\circ}$

Topic: Mathematics, Straight Lines, JEE Main
The angle between the minute hand and the hour hand of a clock at “half past 3” is equal to

Options:
A 90o
B 80o
C 75o
D 60o

The correct answer is:
75o

Solution:
The angle subtended by two consecutive marks at the center =30o
Hence, at “half past 3 ” the angle is 30o+30o+15o=75o

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The angle between the pair of tangents drawn to the ellipse 3 x 2 +2 y 2 =5 from the point 1, 2 is tan −1 12 λ , then the value of λ is

Options:

The correct answer is:
5

Solution:
Equation of pair of tangents
3 x 2 +2 y 2 −5 6 = 3x+4y−5 2
9 x 2 −4 y 2 −24xy+30x+40y−55=0
tanθ= 12 5

Topic: Mathematics, Ellipse, JEE Main
The angle between the planes $3 x-4 y+5 z=0$ and $2 x-y-2 z=5$ is

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{2}$
C $\frac{\pi}{6}$
D None of these

The correct answer is:
$\frac{\pi}{2}$

Solution:
$\theta=\cos ^{-1}\left[\frac{6+4-10}{\sqrt{50} \sqrt{9}}\right]=\cos ^{-1}(0)=\frac{\pi}{2}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The angle between the tangents drawn at the end points of the latus rectum of parabola $y^2=4 a x$, is

Options:
A $\frac{\pi}{3}$
B $\frac{2 \pi}{3}$
C $\frac{\pi}{4}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
End points are $(a, \pm 2 a)$.
$\therefore$ Tangents are, $\pm 2 a y=2 a(x+a)$ $m= \pm \frac{2 a}{2 a}= \pm 1$
Hence angle between them is $\frac{\pi}{2}$

Topic: Mathematics, Parabola, JEE Main
The angle between the tangents drawn from the origin to the parabola $y^2=4 a(x-a)$ is

Options:
A $90^{\circ}$
B $30^{\circ}$
C $\tan ^{-1} \frac{1}{2}$
D $45^{\circ}$

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

Solution:
Any line through origin is $y=m x$. Since it is a tangent to $y^2=4 a(x-a)$, it will cut it in two coincident points.
$\therefore$ Roots of $m^2 x^2-4 a x+4 a^2=0$ are equal.
$\therefore 16 a^2-16 a^2 m^2=0$ or $m^2=1$ or $m=1,-1$
Product of slopes $=-1$. Hence it is a right angled triangle.

Topic: Mathematics, Parabola, JEE Main
The angle between the tangents drawn from the point 4,1 to the parabola x2=4y is

Options:
A π6
B π4
C π3
D π2

The correct answer is:
π3

Solution:
Equation of the tangent to the parabola x2=4y having slope m is x=ym+m
This tangent passes through 4,1. So,
4=m+1m or m2-4m+1=0
Now, angle between the tangents is given by
tan⁡θ=m1-m21+m1m2
=m1+m22-4m1m21+m1m2=16-41+1=3
or θ=π3

Topic: Mathematics, Parabola, JEE Main
The angle between the tangents drawn from the point 2,6 to the parabola y2-4y-4x+8=0 is

Options:
A π2
B π4
C π3
D π6

The correct answer is:
π3

Solution:
Given parabola is y-22=4x-1
Equation of tangent in slope form to this parabola is
y-2=mx-1+1m
It passes through 2,6, so
4=m+1m⇒m2-4m+1=0
If m1 and m2 are roots of this equation, then
m1+m2=4, m1·m2=1
The angle between the tangents =tan-1⁡m1-m21+m1m2
=tan-1⁡m1+m22-4m1m21+m1m2=tan-1⁡16-42=tan-13

=π3

Topic: Mathematics, Parabola, JEE Main
The angle between the tangents drawn from the points $(1,4)$ to the parabola $y^2=4 x$ is

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{3}$
C $\frac{\pi}{4}$
D $\frac{\pi}{6}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
Any tangent to $y^2=4 x$ is $y=m x+\frac{1}{m}$
Since it passes throguh $(1,4)$, we have $4=m+\frac{1}{m}$
$\Rightarrow m^2-4 m+1=0 \Rightarrow m_1+m_2=4, m_1 m_2=1$
$\Rightarrow\left|m_1-m_2\right|=2 \sqrt{3}$
If $\theta$ is the required angle, then$\tan \theta=\frac{2 \sqrt{3}}{1+1}=\sqrt{3}$
$\Rightarrow \quad \theta=\frac{\pi}{3}$

Topic: Mathematics, Parabola, JEE Main
The angle between the tangents from $(\alpha, \beta)$ to the circle $x^2+y^2=a^2$, is

Options:
A $\tan ^{-1}\left(\frac{a}{\sqrt{\alpha^2+\beta^2-a^2}}\right)$
B $\tan ^{-1}\left(\frac{\sqrt{\alpha^2+\beta^2-a^2}}{a}\right)$
C $2 \tan ^{-1}\left(\frac{a}{\sqrt{\alpha^2+\beta^2-a^2}}\right)$
D None of these

The correct answer is:
$2 \tan ^{-1}\left(\frac{a}{\sqrt{\alpha^2+\beta^2-a^2}}\right)$

Solution:
$\tan \frac{\theta}{2}=\frac{C T_1}{P T_1}=\frac{a}{\sqrt{\alpha^2+\beta^2-a^2}}$
$\frac{\theta}{2}=\tan ^{-1} \frac{a}{\sqrt{\alpha^2+\beta^2-a^2}} \Rightarrow \theta=2 \tan ^{-1} \frac{a}{\sqrt{\alpha^2+\beta^2-a^2}}$

Topic: Mathematics, Circle, JEE Main
The angle between the tangents to the curves $y=2 x^2$ and $x=2 y^2$ at $(1,1)$ is

Options:
A $\tan ^{-1}\left(\frac{15}{8}\right)$
B $\tan ^{-1}\left(\frac{7}{8}\right)$
C $\tan ^{-1}\left(\frac{3}{4}\right)$
D $\tan ^{-1}\left(\frac{1}{4}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{15}{8}\right)$

Solution:
$$
y=2 x^2
$$
$\therefore \quad$ Slope of the tangent to this curve is $\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{m}_1=4 x$
$\therefore \quad$ at $(1,1), \mathrm{m}_1=4$
$$
x=2 y^2
$$
$\therefore \quad$ Slope of the tangent to this curve is $\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{m}_2=\frac{1}{4 y}$
$\therefore \quad$ at $(1,1), \mathrm{m}_2=\frac{1}{4}$
Let $\theta$ be the angle between two tangents.
$$
\begin{aligned}
& \therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{4-\frac{1}{4}}{1+4 \times \frac{1}{4}}\right|=\frac{15}{8} \\
& \therefore \quad \theta=\tan ^{-1}\left(\frac{15}{8}\right)
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
The angle between the two sides of the triangle having length $6 \mathrm{~cm}$ and $8 \mathrm{~cm}$ is $60^{\circ}$, find the third side of the triangle.

Options:
A $\sqrt{26}\mathrm{~cm}$
B $12 \mathrm{~cm}$
C $\sqrt{52}\mathrm{~cm}$
D $13 \mathrm{~cm}$

The correct answer is:
$\sqrt{52}\mathrm{~cm}$

Solution:
Given $a=6, b=8$ and $\theta=60^{\circ}$
So according to the cosine rule,
$\cos 60=\frac{6^2+8^2-c^2}{2(6)(8)}$
$\frac{1}{2}=\frac{36+64-c^2}{96}$
$48=100-c^2$
$c^2=52$
$\therefore c=\sqrt{52}$

Topic: Mathematics, Properties of Triangles, JEE Main
The angle between the vectors $3 \mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}$ is

Options:
A $\cos ^{-1} \frac{2}{\sqrt{7}}$
B $\sin ^{-1} \frac{2}{\sqrt{7}}$
C $\cos ^{-1} \frac{2}{\sqrt{5}}$
D $\sin ^{-1} \frac{2}{\sqrt{5}}$

The correct answer is:
$\sin ^{-1} \frac{2}{\sqrt{7}}$

Solution:
$\begin{aligned} \cos \theta=\frac{3(2)+(1)(-2)+2(4)}{\sqrt{9+1+4} \sqrt{4+4+16}} & =\frac{12}{\sqrt{14} \sqrt{24}}=\frac{6}{\sqrt{14} \sqrt{6}} \\ \Rightarrow \cos \theta & =\frac{\sqrt{3}}{\sqrt{7}} \Rightarrow \sin \theta=\frac{2}{\sqrt{7}} \Rightarrow \theta=\sin ^{-1}\left(\frac{2}{\sqrt{7}}\right) .\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
The angle between two linear transmembrane domains is defined by following vectors $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

Options:
A $\cos ^{-1}\left(\frac{1}{3}\right)$
B $\cos ^{-1}\left(\frac{-1}{3}\right)$
C $\sin ^{-1}\left(\frac{-1}{3}\right)$
D $\sin ^{-1}\left(\frac{1}{3}\right)$

The correct answer is:
$\cos ^{-1}\left(\frac{-1}{3}\right)$

Solution:
The angle between two vectors is given by
$\begin{array}{rlrl} & \cos \theta & =\frac{\mathbf{a} \cdot \mathbf{b}}{a b}=\frac{1-1-1}{\sqrt{3} \times \sqrt{3}} \\ \Rightarrow & \cos \theta & =-\frac{1}{3} \\ & \therefore & \theta & =\cos ^{-1}\left(-\frac{1}{3}\right)\end{array}$

Topic: Physics, Mathematics in Physics, JIPMER, JIPMER 2015
The angle between vector (A→) and (A→-B→) is :

Options:
A tan-1BcosθA-Bsinθ
B tan-13 B2 A-B
C tan-1-B2A-B32
D tan-1A0.7B

The correct answer is:
tan-13 B2 A-B

Solution:
tanϕ=Bsin60°A-Bcos60°

ϕ=tan-1Bsin60°A-Bcos60°=tan-13B2A-B

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The angle of dip at a place is $37^{\circ}$ and the vertical component of the earth's magnetic field is $6 \times 10^{-5} \mathrm{~T}$. The earth's magnetic field at this place is $\left(\tan 37^{\circ}=\frac{3}{4}\right)$

Options:
A $7 \times 10^{-5} \mathrm{~T}$
B $6 \times 10^{-5} \mathrm{~T}$
C $5 \times 10^{-5} \mathrm{~T}$
D $10^{-4} \mathrm{~T}$

The correct answer is:
$10^{-4} \mathrm{~T}$

Solution:
$\tan \theta=\frac{\mathrm{V}}{\mathrm{H}}=\frac{3}{4}\left[\because \tan 37^{\circ}=\frac{3}{4}\right]$

$\therefore \mathrm{V}=\frac{3}{4} \mathrm{H}$

$\mathrm{V}=6 \times 10^{-5} \mathrm{~T}$

$\mathrm{H}=\frac{4}{3} \times 6 \times 10^{-5} \mathrm{~T}=8 \times 10^{-5} \mathrm{~T}$

$\therefore \mathrm{B}_{\text {total }}=\sqrt{\mathrm{V}^{2}+\mathrm{H}^{2}}=\sqrt{(36+64)} \times 10^{-5}$

$=10 \times 10^{-5}=10^{-4} \mathrm{~T} .$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The angle of dip at a point is $30^{\circ}$ and the magnetic field of earth is $16 \mathrm{~T}$. find the vertical component of earth's magnetic field at that point.

Options:
A $32 \mathrm{~T}$
B $10 \mathrm{~T}$
C $12 \mathrm{~T}$
D $8 \mathrm{~T}$

The correct answer is:
$8 \mathrm{~T}$

Solution:
Given that: Angle of dip $(\Phi)=30^{\circ}$ and magnetic field $(B)=16 \mathrm{~T}$
The Vertical component $\left(\mathrm{B}_{\mathrm{V}}\right)$ is $\mathrm{B}_{\mathrm{V}}=B \operatorname{Sin} \Phi$
$B_V=16 \times \operatorname{Sin} 30^{\circ}=16 \times 1 / 2=8 \mathrm{~T}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The angle of elevation of a cloud C from a point P,200 m above a still take is 30o. If the angle of depression of the image of C in the lake from the point P is 60o, then PC (in m) is equal to

Options:
A 100
B 2003
C 400
D 4003

The correct answer is:
400

Solution:
tan30°=xy=13⇒y=3x and

tan60°=x+400y

So, x+400=3y=3x

2x=400, x=200

sin30°=200PC⇒PC=400

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The angle of elevation of a cloud from a point 10 meters above the surface of a lake is 30o and the angle of depression of its reflection from that point is 60o. Then the height of the cloud above the lake is

Options:
A 20 meters
B 203 meters
C 203-13 meters
D 203-1 meters

The correct answer is:
20 meters

Solution:
Let C be the cloud, D its reflection and PB is the surface of the lake
Then,
PC=PD=H,
QA=H-10cot⁡30o=H+10cot⁡60o
H=10cot⁡30o+cot⁡60ocot⁡30o-cot⁡60o=10sin⁡90o/sin30o=20 meters

Topic: Mathematics, Heights and Distances, JEE Main
The angle of elevation of a cloud from a point 250 m above a lake is 15o and angle of depression of its reflection in the lake is 45o . The height of the cloud is

Options:
A 2503 m
B 250 m
C 2503 m
D None of these

The correct answer is:
2503 m

Solution:
h=250
Î±=15o
Î¸=45o
x=hsinâ¡Î¸+Î±sinâ¡Î¸-Î±
=250sinâ¡60osinâ¡30o
=250 3

Topic: Mathematics, Heights and Distances, JEE Main
The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is:

Options:
A 12003 m
B 24003 m
C 18003 m
D 36003 m

The correct answer is:
12003 m

Solution:
tan60°=hy

3=hy⇒h=3y …1

tan30°=hx+y

13=hx+y⇒3h=x+y …2

Speed 432 km h-1⇒432×2060×60⇒125 km

3h=125+y

3h-125=y

from 1

h=33h-125

h=3 h-1235

h=635 km

h=12003 m

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The angle of elevation of an object on a hill is observed from a certain point in the horizontal plane through its base, to be 30°. After walking 120 m towards it on a level ground, the angle of elevation is found to be 60°. Then the height of the object (in metres) is

Options:
A 120
B 603
C 1203
D 60

The correct answer is:
603

Solution:
In âCAD, tan â¡30Â°=CDAC
â 13=h120+x
â 3h=120+x ...(i)
and in âCBD, tan â¡60Â°=CDBC
â 3=hx â h=3x ...(ii)
From Eqs. (i) and (ii), we get, x=60 m
On putting x=60 in Eq.(i), we get
h=603 m

Topic: Mathematics, Heights and Distances, JEE Main
The angle of elevation of the summit of a mountain from a point on the ground is 45°. After climbing up one km towards the summit at an inclination of 30° from the ground, the angle of elevation of the summit is found to be 60°. Then the height (in km) of the summit from the ground is :

Options:
A 3-13+1
B 3+13-1
C 13-1
D 13+1

The correct answer is:
13-1

Solution:
In∆CDF

sin30°=z1CD=1 km

(given)

z=12 ................(1)

cos30°=y1⇒32

now in

∆ABC

tan45°=hx+y

⇒h=x+y

⇒x=h-32 .............(2)

now in

∆BDE,

tan60°=h-zx

3x=h-12⇒3h-32=h-12⇒3-1h=1

h=13-1km

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be 45°. After walking a distance of 80 meters towards the top, up a slope inclined at angle of 30° to the horizontal plane the angle of elevation of the top of the hill becomes 75°. Then the height of the hill (in meters) is _____.

Options:

The correct answer is:
80

Solution:
x=80cos30°=403

y=80sin30°=40

tan45°=1

⇒ x+z=h

⇒ 403+z=h

tan75°=h-yz

⇒ 2+3=h-40z

⇒ z=h-402+3

⇒ h-403=h-402+3

(1+3)h=80(1+3)

h=80

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The angle of elevation of the top of a tower from a point \( 20 m \) away from its base is \( 45^{\circ} . \) Then find the height of the tower.

Options:

The correct answer is:
20.00

Solution:
We have, AC=20 m, θ=45°

Draw the diagram,

∴ tan45°=BCAC=x20

⇒1=x20⇒x=20 m

Topic: Mathematics, Heights and Distances, JEE Main
The angle of elevation of the top of a tower from a point A due north of it is α and from a point B at a distance of 9 units due west of A is cos-1313. If the distance of the point B from the tower is 15 units, then cotα is equal to

Options:
A 65
B 95
C 43
D 73

The correct answer is:
65

Solution:
Given,

The angle of elevation of the top of a tower from a point A due north of it is α and from a point B at a distance of 9 units due west of A is cos-1313. If the distance of the point B from the tower is 15 units, then cotα is equal to

Now making the diagram of given value we have,

Given OB=15

cosβ=313

So, tanβ=23

Now using the value of tanβ in △OPB we get,

⇒tanβ=h15

⇒23=h15, so h=10

Now in △OAB

OA2+AB2=225

⇒OA2+81=225

⇒OA=12

Now in triangle △OAP

tanα=1012

⇒cotα=1210=65

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2022 (29 Jul Shift 1)
The angle of elevation of the top of a vertical tower from a point A, due east of it is 45o . The angle of elevation of the top of the same tower from a point B, due south of A is 30o . If the distance between A and B is 542m , then the height of the tower (in meters), is:

Options:
A 108
B 363
C 543
D 54

The correct answer is:
54

Solution:
Let height of the tower be 'H'
Let AP=x
BP=y
In ∆APQ⇒tan⁡45o=Hx ⇒ H=x
In ∆BPQ⇒tan⁡30o=Hy ⇒ y= 3 H
Since, ∆APB is a right-angled triangle.
⇒x2+5422=y2 (Pythagoras Theorem )
⇒H2+5422=3H2
⇒5422=2H2
⇒542= 2H
⇒54=H

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2016 (10 Apr Online)
The angle of elevation of the top of a vertical tower from a point $\mathrm{P}$ on the horizontal ground was observed to be $\alpha$. After moving a distance 2 metres from $\mathrm{P}$ towards the foot of the tower, the angle of elevation changes to $\beta$. Then the height (in metres) of the tower is:

Options:
A
$\frac{2 \sin \alpha \sin \beta}{\sin (\beta-\alpha)}$

B
$\frac{\sin \alpha \sin \beta}{\cos (\beta-\alpha)}$

C
$\frac{2 \sin (\beta-\alpha)}{\sin \alpha \sin \beta}$

D
$\frac{\cos (\beta-\alpha)}{\sin \alpha \sin \beta}$

The correct answer is:
$\frac{2 \sin \alpha \sin \beta}{\sin (\beta-\alpha)}$

Solution:
$$
\text { Let } \mathrm{AB} \text { be the tower of height ' } \mathrm{h} \text { '. }
$$
Given : In $\triangle \mathrm{ABP}$
$\tan \alpha=\frac{\mathrm{AB}}{\mathrm{PB}}$
or $\frac{\sin \alpha}{\cos \alpha}=\frac{h}{x+2}$
$\Rightarrow(x+2) \sin \alpha=\mathrm{h} \cos \alpha$
$\Rightarrow h=\frac{x \sin \alpha+2 \sin \alpha}{\cos \alpha}$
Now, In $\Delta \mathrm{ABC}, \tan \beta=\frac{\mathrm{AB}}{\mathrm{BC}}$
$$
\Rightarrow \frac{\sin \beta}{\cos \beta}=\frac{h}{x} \Rightarrow x=\frac{h \cos \beta}{\sin \beta}
$$
Putting the value of $x$ in eq. (2) to eq. (1), we get
$$
\begin{aligned}
&h=\frac{\frac{h \cos \beta \sin \alpha}{\sin \beta}+\frac{2 \sin \alpha}{1}}{\cos \alpha} \\
&\Rightarrow h=\frac{h \cos \beta \cdot \sin \alpha+2 \sin \alpha \sin \beta}{\sin \beta \cdot \cos \alpha} \\
&
\end{aligned}
$$
$\Rightarrow h(\sin \beta \cdot \cos \alpha-\cos \beta \cdot \sin \alpha)=2 \sin \alpha \cdot \sin \beta$
$\Rightarrow h[\sin (\beta-\alpha)]=2 \sin \alpha \cdot \sin \beta$
$\Rightarrow h=\frac{2 \sin \alpha \cdot \sin \beta}{\sin (\beta-\alpha)}$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2014 (11 Apr Online)
The angle of elevation of the top P of a tower from the feet of one person standing due south of the tower is 45° and from the feet of another person standing due west of the tower is 30°. If the height of the tower is 5 meters, then the distance (in meters) between the two persons is equal to

Options:
A 525
B 10
C 5
D 55

The correct answer is:
10

Solution:
Given that the elevation to the top of the tower from the feet of one person due south is =45°

Similarly the angle of elevation to the top of tower from the feet of another person to the west of tower is =30°

The given information can be represented in the form of a diagram as below:

We need to find the distance AB.

Let the distance of A,B from the tower be x1,x2.
⇒tan 45°=5x1, tan 30°=5x2
⇒x1=5, x2=53.

⇒AB=x12+x22=52+532=10 m

Hence this is the correct option.

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refraction index $n$ ) is

Options:
A $\sin ^{-1}(n)$
B $\sin ^{-1}\left(\frac{1}{n}\right)$
C $\tan ^{-1}\left(\frac{1}{n}\right)$
D $\tan ^{-1}(n)$

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

Solution:
$\mu=\tan \theta_e \Rightarrow \theta_e=\tan ^{-1} n$

Topic: Physics, Wave Optics, JEE Main