Heights and Distances Questions in English

(B) Given:
$x = a \cos \theta + b \sin \theta$
$y = b \cos \theta - a \sin \theta$
Squaring both equations:
$x^{2} = (a \cos \theta + b \sin \theta)^{2} = a^{2} \cos^{2} \theta + b^{2} \sin^{2} \theta + 2ab \sin \theta \cos \theta$
$y^{2} = (b \cos \theta - a \sin \theta)^{2} = b^{2} \cos^{2} \theta + a^{2} \sin^{2} \theta - 2ab \cos \theta \sin \theta$
Adding $x^{2}$ and $y^{2}$:
$x^{2} + y^{2} = (a^{2} \cos^{2} \theta + b^{2} \sin^{2} \theta + 2ab \sin \theta \cos \theta) + (b^{2} \cos^{2} \theta + a^{2} \sin^{2} \theta - 2ab \cos \theta \sin \theta)$
$x^{2} + y^{2} = a^{2} (\cos^{2} \theta + \sin^{2} \theta) + b^{2} (\sin^{2} \theta + \cos^{2} \theta)$
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,we get:
$x^{2} + y^{2} = a^{2}(1) + b^{2}(1) = a^{2} + b^{2}$

(A) Let the heights of the two poles be $h_1 = 12 \ m$ and $h_2 = 7 \ m$. The distance between their feet is $d = 12 \ m$.
When we draw a horizontal line from the top of the shorter pole to the taller pole,we form a right-angled triangle.
The base of this triangle is equal to the distance between the poles,which is $12 \ m$.
The vertical height of this triangle is the difference between the heights of the two poles: $12 \ m - 7 \ m = 5 \ m$.
Using the Pythagorean theorem,the distance between the tops $(AC)$ is:
$AC^2 = (12)^2 + (5)^2$
$AC^2 = 144 + 25 = 169$
$AC = \sqrt{169} = 13 \ m$.

(C) Let the ladder be $AC$,where $AC = 10 \ m$. Let the wall be $AB$ and the ground be $BC$. The ladder makes an angle of $30^{\circ}$ with the ground,so $\angle C = 30^{\circ}$.
In the right-angled triangle $ABC$,we need to find the distance of the foot of the ladder from the wall,which is $BC$.
Using the trigonometric ratio $\cos \theta = \frac{\text{base}}{\text{hypotenuse}}$:
$\cos 30^{\circ} = \frac{BC}{AC}$
$\frac{\sqrt{3}}{2} = \frac{BC}{10}$
$BC = 10 \times \frac{\sqrt{3}}{2} = 5 \times \sqrt{3}$
Given $\sqrt{3} = 1.732$,we have:
$BC = 5 \times 1.732 = 8.66 \ m$.

(D) Let the height of the tower be $h$ meters.
Given that the distance from the foot of the tower is $100 \ m$ and the angle of elevation is $30^{\circ}$.
In a right-angled triangle,$\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}$.
Here,$\tan(30^{\circ}) = \frac{h}{100}$.
We know that $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$.
So,$\frac{1}{\sqrt{3}} = \frac{h}{100}$.
Therefore,$h = \frac{100}{\sqrt{3}} \ m$.

(D) Let the height of the person be $CD = 6 \ ft$.
Let the height of the tree be $AB = \frac{26}{3} \ ft$.
The distance between the person and the tree is $BD = \frac{8}{\sqrt{3}} \ ft$.
Let $E$ be a point on $AB$ such that $CE$ is parallel to $BD$. Thus,$BE = CD = 6 \ ft$.
The height of the fruit above the person's eye level is $AE = AB - BE = \frac{26}{3} - 6 = \frac{26 - 18}{3} = \frac{8}{3} \ ft$.
The horizontal distance is $CE = BD = \frac{8}{\sqrt{3}} \ ft$.
In the right-angled triangle $\Delta AEC$,$\tan \theta = \frac{AE}{CE}$.
$\tan \theta = \frac{8/3}{8/\sqrt{3}} = \frac{8}{3} \times \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.
Since $\tan \theta = \frac{1}{\sqrt{3}}$,we have $\theta = 30^{\circ}$.

(B) Let the height of the kite from the ground be $AB = 75 \ m$.
Let the angle made by the string with the ground be $\angle ACB = \Theta$.
Given that $\cot \Theta = 8/15$.
In the right-angled triangle $\triangle ABC$,$\cot \Theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{BC}{AB}$.
Therefore,$\frac{BC}{75} = \frac{8}{15}$.
$BC = \frac{8 \times 75}{15} = 8 \times 5 = 40 \ m$.
The length of the string is the hypotenuse $AC$.
Using the Pythagorean theorem,$AC = \sqrt{AB^2 + BC^2}$.
$AC = \sqrt{75^2 + 40^2} = \sqrt{5625 + 1600} = \sqrt{7225}$.
$AC = 85 \ m$.

(D) The path taken by the person forms a right-angled triangle where the base is $12 \ km$ (east) and the perpendicular is $5 \ km$ (north).
The shortest distance between the starting point $L$ and the final point $M$ is the hypotenuse of this right-angled triangle.
Using the Pythagorean theorem:
$\text{Shortest distance} = \sqrt{(\text{base})^2 + (\text{perpendicular})^2}$
$= \sqrt{12^2 + 5^2}$
$= \sqrt{144 + 25}$
$= \sqrt{169}$
$= 13 \ km$

(D) Let the building be $BC = 10 \ m$ and the flagstaff be $AB$. Let $P$ be the point on the ground.
In $\Delta BCP$,$\tan 30^{\circ} = \frac{BC}{CP} = \frac{10}{CP}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{10}{CP}$,so $CP = 10\sqrt{3} \ m$.
In $\Delta ACP$,$\tan 45^{\circ} = \frac{AC}{CP} = \frac{AB + BC}{CP}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{AB + 10}{10\sqrt{3}}$.
Therefore,$AC = 10\sqrt{3} \ m$.
Length of flagstaff $AB = AC - BC = 10\sqrt{3} - 10$.
$AB = 10(1.732) - 10 = 17.32 - 10 = 7.32 \ m$.

(D) Let $AD$ be the tower of height $108 \; m$,and $B$ and $C$ be the two objects on either side of the tower.
Given: $AD = 108 \; m$,$\angle ABD = 30^{\circ}$,and $\angle ACD = 45^{\circ}$.
In right-angled $\triangle ABD$:
$\tan 30^{\circ} = \frac{AD}{BD}$
$\frac{1}{\sqrt{3}} = \frac{108}{BD}$
$BD = 108\sqrt{3} \; m$
In right-angled $\triangle ADC$:
$\tan 45^{\circ} = \frac{AD}{DC}$
$1 = \frac{108}{DC}$
$DC = 108 \; m$
The distance between the objects is $BC = BD + DC$.
$BC = 108\sqrt{3} + 108 = 108(\sqrt{3} + 1) \; m$.

(B) Let the height of the tower be $h \ m$.
Let the length of the shadow when the sun's altitude is $60^{\circ}$ be $x \ m$.
In the right-angled triangle formed by the tower and the shadow at $60^{\circ}$,we have $\tan 60^{\circ} = h / x$,so $x = h / \sqrt{3}$.
When the sun's altitude is $30^{\circ}$,the length of the shadow is $(x + 40) \ m$.
In the right-angled triangle formed by the tower and the shadow at $30^{\circ}$,we have $\tan 30^{\circ} = h / (x + 40)$,so $x + 40 = h / \tan 30^{\circ} = h \sqrt{3}$.
Substituting $x = h / \sqrt{3}$ into the equation: $h / \sqrt{3} + 40 = h \sqrt{3}$.
$40 = h \sqrt{3} - h / \sqrt{3} = h (\sqrt{3} - 1 / \sqrt{3}) = h ( (3 - 1) / \sqrt{3} ) = 2h / \sqrt{3}$.
$2h = 40 \sqrt{3}$,which gives $h = 20 \sqrt{3} \ m$.

(B) Let the height of the tower be $h$ and the length of the shadow be $L = 9 \text{ m}$.
Given the angle of elevation (sun's altitude) $\theta = 30^{\circ}$.
In a right-angled triangle formed by the tower and its shadow,we have $\tan \theta = \frac{\text{height}}{\text{shadow length}}$.
Substituting the values: $\tan 30^{\circ} = \frac{h}{9}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we get $\frac{1}{\sqrt{3}} = \frac{h}{9}$.
Solving for $h$: $h = \frac{9}{\sqrt{3}} = \frac{9 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{9 \sqrt{3}}{3} = 3 \sqrt{3} \text{ m}$.

(B) Let $AB = h$ be the height of the pole.
Let $DE = 169 \; cm$ be the height of the man.
Let $CD = 130 \; cm$ be the shadow of the man.
Let $BC = 420 \; cm$ be the shadow of the pole.
Since the sun's rays are parallel,$\Delta CED$ is similar to $\Delta ABC$ by $AA$ similarity criterion.
Therefore,the ratio of their corresponding sides is equal:
$\frac{DE}{AB} = \frac{CD}{BC}$
Substituting the given values:
$\frac{169}{h} = \frac{130}{420}$
$h = \frac{169 \times 420}{130}$
$h = \frac{169 \times 42}{13}$
$h = 13 \times 42$
$h = 546 \; cm$
Thus,the height of the pole is $546 \; cm$.

(B) Let $AB$ be the height of the tower and $C$ be the point on the ground.
Given that the distance from the foot of the tower $B$ to the point $C$ is $BC = 20 \; m$.
The angle of elevation $\angle ACB = 30^{\circ}$.
In the right-angled triangle $\triangle ABC$,we have:
$\tan(30^{\circ}) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$,we get:
$\frac{1}{\sqrt{3}} = \frac{AB}{20}$
$AB = \frac{20}{\sqrt{3}} \; m$.
Therefore,the height of the tower is $\frac{20}{\sqrt{3}} \; m$.

(A) Let $AB$ be the tower of height $h \; m$.
Let $C$ be the initial point such that $BC = 160 \; m$ and $\angle ACB = \theta$.
After moving $100 \; m$ towards the tower,we reach point $D$,such that $CD = 100 \; m$. Thus,$BD = BC - CD = 160 - 100 = 60 \; m$.
At point $D$,the angle subtended is $2\theta$,so $\angle ADB = 2\theta$.
In $\triangle ABC$,$\tan \theta = \frac{AB}{BC} = \frac{h}{160}$.
In $\triangle ABD$,$\tan 2\theta = \frac{AB}{BD} = \frac{h}{60}$.
Using the formula $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$,we have:
$\frac{h}{60} = \frac{2(h/160)}{1 - (h/160)^2}$.
$\frac{h}{60} = \frac{h/80}{1 - h^2/25600}$.
Dividing by $h$ (since $h \neq 0$):
$\frac{1}{60} = \frac{1/80}{(25600 - h^2)/25600} = \frac{25600}{80(25600 - h^2)} = \frac{320}{25600 - h^2}$.
$25600 - h^2 = 320 \times 60 = 19200$.
$h^2 = 25600 - 19200 = 6400$.
$h = \sqrt{6400} = 80 \; m$.

(B) Let $AB$ be the height of the tower,$h \; m$.
Let $C$ be the point at a distance of $50 \; m$ from the foot of the tower $B$.
Thus,$BC = 50 \; m$.
The angle of elevation is given as $\angle ACB = 30^{\circ}$.
In the right-angled triangle $\triangle ABC$,we have:
$\tan 30^{\circ} = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$
Substituting the values,we get:
$\frac{1}{\sqrt{3}} = \frac{h}{50}$
$h = \frac{50}{\sqrt{3}} \; m$.
Therefore,the height of the tower is $\frac{50}{\sqrt{3}} \; m$.

(D) Let $AB = 3x$ units and $BC = 2x$ units.
Since $P$ is the midpoint of $AB$,$PB = \frac{AB}{2} = \frac{3x}{2}$ units.
In the right-angled triangle $\triangle CPB$,by the Pythagorean theorem:
$CP = \sqrt{PB^2 + BC^2} = \sqrt{(\frac{3x}{2})^2 + (2x)^2}$
$CP = \sqrt{\frac{9x^2}{4} + 4x^2} = \sqrt{\frac{9x^2 + 16x^2}{4}} = \sqrt{\frac{25x^2}{4}} = \frac{5x}{2}$ units.
Now,in $\triangle CPB$,$\sin(\angle CPB) = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{CP}$.
$\sin(\angle CPB) = \frac{2x}{\frac{5x}{2}} = 2x \cdot \frac{2}{5x} = \frac{4}{5}$.

(A) To simplify the expression $\frac{\sin A}{1+\cos A}+\frac{\sin A}{1-\cos A}$,we find a common denominator:
$= \frac{\sin A(1-\cos A) + \sin A(1+\cos A)}{(1+\cos A)(1-\cos A)}$
$= \frac{\sin A - \sin A \cos A + \sin A + \sin A \cos A}{1 - \cos^2 A}$
$= \frac{2 \sin A}{\sin^2 A}$
$= \frac{2}{\sin A} = 2 \operatorname{cosec} A$

(D) Given equations are:
$r \sin \theta = 1$ --- $(1)$
$r \cos \theta = \sqrt{3}$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{r \sin \theta}{r \cos \theta} = \frac{1}{\sqrt{3}}$
$\tan \theta = \frac{1}{\sqrt{3}}$
Now,substitute the value of $\tan \theta$ into the expression $(\sqrt{3} \tan \theta + 1)$:
$= \sqrt{3} \times \left( \frac{1}{\sqrt{3}} \right) + 1$
$= 1 + 1 = 2$
Thus,the correct option is $(d)$.

(A) Let $AB$ be the tower of height $h$ units.
Let the points $P$ and $Q$ be at distances $PB = a$ and $QB = b$ from the base $B$ of the tower.
Given that the angles of elevation are complementary,let $\angle AQB = \theta$ and $\angle APB = 90^{\circ} - \theta$.
In $\triangle AQB$,$\tan \theta = \frac{AB}{QB} = \frac{h}{b}$.
In $\triangle APB$,$\tan(90^{\circ} - \theta) = \frac{AB}{PB} = \frac{h}{a}$.
Since $\tan(90^{\circ} - \theta) = \cot \theta$,we have $\cot \theta = \frac{h}{a}$.
Multiplying the two equations: $\tan \theta \cdot \cot \theta = \left(\frac{h}{b}\right) \cdot \left(\frac{h}{a}\right)$.
Since $\tan \theta \cdot \cot \theta = 1$,we get $1 = \frac{h^2}{ab}$.
Therefore,$h^2 = ab$,which implies $h = \sqrt{ab}$.

(B) Let $AB$ be the tower of height $100 \ m$.
Let $C$ be the initial position of the car and $D$ be the final position after some time.
Let the distance travelled by the car be $CD = x \ m$.
In $\Delta ABD$,the angle of elevation is $60^{\circ}$.
$\tan 60^{\circ} = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{100}{BD} \Rightarrow BD = \frac{100}{\sqrt{3}} \ m$.
In $\Delta ABC$,the angle of elevation is $30^{\circ}$.
$\tan 30^{\circ} = \frac{AB}{BC} = \frac{AB}{BD + CD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{\frac{100}{\sqrt{3}} + x}$.
$\frac{100}{\sqrt{3}} + x = 100 \sqrt{3}$.
$x = 100 \sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} \ m$.
Rationalizing the denominator,$x = \frac{200 \sqrt{3}}{3} \ m$.

(A) Let the height of the shorter post be $h \text{ m}$ and the height of the taller post be $2h \text{ m}$.
Let the posts be $CD$ and $AB$ respectively,with $CD = h$ and $AB = 2h$.
The distance between their feet $B$ and $D$ is $x \text{ m}$.
Let $O$ be the midpoint of $BD$,so $OB = OD = \frac{x}{2} \text{ m}$.
From $\triangle OCD$,$\tan \theta = \frac{CD}{OD} = \frac{h}{x/2} = \frac{2h}{x}$.
From $\triangle OAB$,the angle of elevation is $(90^{\circ} - \theta)$,so $\tan(90^{\circ} - \theta) = \frac{AB}{OB} = \frac{2h}{x/2} = \frac{4h}{x}$.
Since $\tan(90^{\circ} - \theta) = \cot \theta$,we have $\cot \theta = \frac{4h}{x}$.
Multiplying the two equations: $\tan \theta \cdot \cot \theta = \left(\frac{2h}{x}\right) \cdot \left(\frac{4h}{x}\right)$.
Since $\tan \theta \cdot \cot \theta = 1$,we get $1 = \frac{8h^2}{x^2}$.
Thus,$h^2 = \frac{x^2}{8}$,which gives $h = \sqrt{\frac{x^2}{8}} = \frac{x}{2\sqrt{2}} \text{ m}$.

(C) Let the two aeroplanes be at points $A$ and $D$. Let the point on the ground be $C$. The height of the first aeroplane $A$ is $AB = 5000 \; m$.
Let the height of the second aeroplane $D$ be $DB = h$.
The distance of the point $C$ from the point $B$ (directly below the aeroplanes) is $BC = x$.
In $\Delta ABC$,$\tan 60^{\circ} = \frac{AB}{BC} = \frac{5000}{x}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $x = \frac{5000}{\sqrt{3}} \; m$.
In $\Delta DBC$,$\tan 45^{\circ} = \frac{DB}{BC} = \frac{h}{x}$.
Since $\tan 45^{\circ} = 1$,we have $h = x = \frac{5000}{\sqrt{3}} \; m$.
The vertical distance between the aeroplanes is $AD = AB - DB = 5000 - h$.
$AD = 5000 - \frac{5000}{\sqrt{3}} = 5000\left(1 - \frac{1}{\sqrt{3}}\right) \; m$.

(C) Let $AB$ be the tower and $BC$ be its shadow.
Let the height of the tower $AB = x$.
According to the problem,the length of the shadow $BC = \frac{1}{\sqrt{3}} \times AB = \frac{x}{\sqrt{3}}$.
In the right-angled triangle $ABC$,the angle of elevation $\theta$ is given by:
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$
$\tan \theta = \frac{x}{\frac{x}{\sqrt{3}}} = \sqrt{3}$
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\theta = 60^{\circ}$.
Therefore,the angle of elevation of the Sun is $60^{\circ}$.