Textbook - Some Applications of Trigonometry Questions in English

$(15\sqrt{3})$ First,let us represent the problem with a right-angled triangle $ABC$ (as shown in the figure),where $AB$ is the height of the tower,$BC = 15\, m$ is the distance of the point from the foot of the tower,and $\angle ACB = 60^{\circ}$ is the angle of elevation.
In the right-angled triangle $ABC$,we have:
$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan 60^{\circ} = \frac{AB}{BC}$
Since $\tan 60^{\circ} = \sqrt{3}$,we substitute the values:
$\sqrt{3} = \frac{AB}{15}$
$AB = 15\sqrt{3}\, m$
Thus,the height of the tower is $15\sqrt{3}\, m$.

(N/A) In the figure,the electrician is required to reach the point $B$ on the pole $AD$.
So,
$BD = AD - AB = (5 - 1.3)\, m = 3.7\, m$
Here,$BC$ represents the ladder. We need to find its length,i.e.,the hypotenuse of the right triangle $BDC$.
Now,we consider the trigonometric ratio $\sin 60^{\circ}$.
So,$\frac{BD}{BC} = \sin 60^{\circ}$ or $\frac{3.7}{BC} = \frac{\sqrt{3}}{2}$
Therefore,$BC = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{1.73} \approx 4.28\, m$.
i.e.,the length of the ladder should be $4.28\, m$.
Now,$\frac{DC}{BD} = \cot 60^{\circ} = \frac{1}{\sqrt{3}}$.
i.e.,$DC = \frac{3.7}{\sqrt{3}} = \frac{3.7}{1.73} \approx 2.14\, m$.
Therefore,she should place the foot of the ladder at a distance of $2.14\, m$ from the pole.

(C) Let $AB$ be the chimney,$CD$ be the observer,and $\angle ADE$ be the angle of elevation. In this case,$\triangle ADE$ is a right-angled triangle at $E$,and we need to find the height of the chimney.
We have $AB = AE + BE = AE + 1.5\, m$.
Also,$DE = CB = 28.5\, m$.
To determine $AE$,we use the tangent trigonometric ratio,which involves both $AE$ and $DE$:
$\tan 45^{\circ} = \frac{AE}{DE}$
Since $\tan 45^{\circ} = 1$,we have:
$1 = \frac{AE}{28.5}$
$AE = 28.5\, m$.
Therefore,the height of the chimney $AB = AE + BE = 28.5\, m + 1.5\, m = 30\, m$.

(N/A) In the figure,$AB$ denotes the height of the building,$BD$ the flagstaff,and $P$ the given point. Note that there are two right-angled triangles,$\triangle PAB$ and $\triangle PAD$. We are required to find the length of the flagstaff,i.e.,$DB$,and the distance of the building from the point $P$,i.e.,$PA$.
Since we know the height of the building $AB = 10 \, m$,we will first consider the right $\triangle PAB$.
We have $\tan 30^{\circ} = \frac{AB}{AP}$.
i.e.,$\frac{1}{\sqrt{3}} = \frac{10}{AP}$.
Therefore,$AP = 10\sqrt{3} \, m$.
i.e.,the distance of the building from $P$ is $10 \times 1.732 = 17.32 \, m$.
Next,let us suppose the length of the flagstaff $BD = x \, m$. Then the total height $AD = (10 + x) \, m$.
Now,in right $\triangle PAD$,$\tan 45^{\circ} = \frac{AD}{AP} = \frac{10 + x}{10\sqrt{3}}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{10 + x}{10\sqrt{3}}$.
Therefore,$10\sqrt{3} = 10 + x$.
$x = 10\sqrt{3} - 10 = 10(\sqrt{3} - 1) = 10(1.732 - 1) = 10(0.732) = 7.32 \, m$.
Thus,the length of the flagstaff is $7.32 \, m$ and the distance of the building from $P$ is $17.32 \, m$.

(N/A) Let $AB$ be the tower of height $h \, m$ and $BC$ be the length of the shadow when the Sun's altitude is $60^{\circ}$. Let $BC = x \, m$.
When the Sun's altitude is $30^{\circ}$,the length of the shadow is $DB = (x + 40) \, m$.
In $\triangle ABC$,$\tan 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \dots (1)$.
In $\triangle ABD$,$\tan 30^{\circ} = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \implies h = \frac{x + 40}{\sqrt{3}} \dots (2)$.
Equating $(1)$ and $(2)$,we get $x\sqrt{3} = \frac{x + 40}{\sqrt{3}}$.
$3x = x + 40 \implies 2x = 40 \implies x = 20 \, m$.
Substituting $x = 20$ in $(1)$,$h = 20\sqrt{3} \, m$.
Thus,the height of the tower is $20\sqrt{3} \, m$ (or approximately $34.64 \, m$).

(N/A) Let $PC$ be the multi-storeyed building and $AB$ be the $8 \, m$ tall building. We need to find the height of the multi-storeyed building $(PC)$ and the distance between the two buildings $(AC)$.
From the figure,$PQ$ is the horizontal line from the top of the building $P$. Since $PQ \parallel BD$,the alternate interior angles are equal.
Therefore,$\angle PBD = 30^{\circ}$ and $\angle PAC = 45^{\circ}$.
In right-angled $\triangle PBD$:
$\tan 30^{\circ} = \frac{PD}{BD} \implies \frac{1}{\sqrt{3}} = \frac{PD}{BD} \implies BD = PD\sqrt{3}$.
In right-angled $\triangle PAC$:
$\tan 45^{\circ} = \frac{PC}{AC} \implies 1 = \frac{PC}{AC} \implies PC = AC$.
Since $AC = BD$ and $PC = PD + DC$,where $DC = AB = 8 \, m$:
$PD + 8 = AC = BD = PD\sqrt{3}$.
Solving for $PD$:
$PD\sqrt{3} - PD = 8 \implies PD(\sqrt{3} - 1) = 8 \implies PD = \frac{8}{\sqrt{3} - 1}$.
Rationalizing the denominator:
$PD = \frac{8(\sqrt{3} + 1)}{3 - 1} = \frac{8(\sqrt{3} + 1)}{2} = 4(\sqrt{3} + 1) \, m$.
Height of the multi-storeyed building $PC = PD + DC = 4\sqrt{3} + 4 + 8 = 4\sqrt{3} + 12 = 4(3 + \sqrt{3}) \, m$.
Distance between the buildings $AC = PC = 4(3 + \sqrt{3}) \, m$.

(N/A) Let $A$ and $B$ represent points on the bank on opposite sides of the river,so that $AB$ is the width of the river. Let $P$ be a point on the bridge at a height of $3 \, m$ from the river level,i.e.,$PD = 3 \, m$,where $D$ is the point on the river surface directly below $P$.
In right-angled $\triangle APD$,$\angle PAD = 30^{\circ}$.
Using trigonometry,$\tan 30^{\circ} = \frac{PD}{AD}$.
$\frac{1}{\sqrt{3}} = \frac{3}{AD} \implies AD = 3\sqrt{3} \, m$.
In right-angled $\triangle PBD$,$\angle PBD = 45^{\circ}$.
Using trigonometry,$\tan 45^{\circ} = \frac{PD}{BD}$.
$1 = \frac{3}{BD} \implies BD = 3 \, m$.
The total width of the river is $AB = AD + BD = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1) \, m$.

(D) From the figure,let $AB$ be the vertical pole and $AC$ be the rope of length $20\, m$.
In the right-angled triangle $\triangle ABC$,the angle made by the rope with the ground is $\angle C = 30^{\circ}$.
We need to find the height of the pole,which is $AB$.
Using the trigonometric ratio $\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}$,we have:
$\sin 30^{\circ} = \frac{AB}{AC}$
Since $\sin 30^{\circ} = \frac{1}{2}$ and $AC = 20\, m$,we get:
$\frac{1}{2} = \frac{AB}{20}$
$AB = \frac{20}{2} = 10\, m$.
Thus,the height of the pole is $10\, m$.

(N/A) Let $AB$ be the original height of the tree. Let the tree break at point $B$,and the top $A$ touches the ground at point $A^{\prime}$.
In the right-angled triangle $\triangle BCA^{\prime}$,where $\angle BCA^{\prime} = 90^{\circ}$ and $\angle BA^{\prime}C = 30^{\circ}$:
Given $CA^{\prime} = 8\, m$.
Using $\tan 30^{\circ} = \frac{BC}{CA^{\prime}}$:
$\frac{BC}{8} = \frac{1}{\sqrt{3}} \implies BC = \frac{8}{\sqrt{3}}\, m$.
Using $\cos 30^{\circ} = \frac{CA^{\prime}}{BA^{\prime}}$:
$\frac{8}{BA^{\prime}} = \frac{\sqrt{3}}{2} \implies BA^{\prime} = \frac{16}{\sqrt{3}}\, m$.
The total height of the tree is $AB = BC + BA^{\prime}$ (since $BA^{\prime} = BA$):
$AB = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}}\, m$.
Rationalizing the denominator:
$AB = \frac{24 \sqrt{3}}{3} = 8\sqrt{3}\, m$.
Thus,the height of the tree is $8\sqrt{3}\, m$.

(N/A) It can be observed that $AC$ and $PR$ are the slides for younger and elder children respectively.
In $\triangle ABC$:
$\sin 30^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC}$
$\frac{1}{2} = \frac{1.5}{AC}$
$AC = 3 \, m$
In $\triangle PQR$:
$\sin 60^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{PR}$
$\frac{\sqrt{3}}{2} = \frac{3}{PR}$
$PR = \frac{6}{\sqrt{3}} = 2\sqrt{3} \, m \approx 3.46 \, m$

(N/A) Let $AB$ be the tower and the angle of elevation from point $C$ (on the ground) is $30^{\circ}$.
In $\triangle ABC$,
$\frac{AB}{BC} = \tan 30^{\circ}$
$\frac{AB}{30} = \frac{1}{\sqrt{3}}$
$AB = \frac{30}{\sqrt{3}} = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}\, m$
Therefore,the height of the tower is $10\sqrt{3}\, m$.

$(40\sqrt{3}\, m) $ Let $K$ be the position of the kite and $P$ be the point on the ground where the string is tied. Let $L$ be the point on the ground directly below the kite.
In the right-angled triangle $\triangle KLP$:
$KL = 60\, m$ (height of the kite)
$\angle KPL = 60^{\circ}$ (angle of inclination)
We need to find the length of the string,which is the hypotenuse $KP$.
Using the trigonometric ratio $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$:
$\sin 60^{\circ} = \frac{KL}{KP}$
$\frac{\sqrt{3}}{2} = \frac{60}{KP}$
$KP = \frac{60 \times 2}{\sqrt{3}}$
$KP = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$KP = \frac{120\sqrt{3}}{3} = 40\sqrt{3}\, m$
Thus,the length of the string is $40\sqrt{3}\, m$ (approximately $69.28\, m$).

(N/A) Let the boy be standing at point $A$ initially. He walks towards the building and reaches point $B$.
Let $PQ$ be the building of height $30 \, m$. The boy's height is $1.5 \, m$.
Therefore,the height of the triangle $PR = PQ - RQ = 30 - 1.5 = 28.5 \, m = \frac{57}{2} \, m$.
In $\triangle PAR$,$\tan 30^{\circ} = \frac{PR}{AR} \implies \frac{1}{\sqrt{3}} = \frac{57/2}{AR} \implies AR = \frac{57\sqrt{3}}{2} \, m$.
In $\triangle PBR$,$\tan 60^{\circ} = \frac{PR}{BR} \implies \sqrt{3} = \frac{57/2}{BR} \implies BR = \frac{57}{2\sqrt{3}} = \frac{19\sqrt{3}}{2} \, m$.
The distance walked towards the building is $AB = AR - BR$.
$AB = \frac{57\sqrt{3}}{2} - \frac{19\sqrt{3}}{2} = \frac{38\sqrt{3}}{2} = 19\sqrt{3} \, m$.
Thus,the boy walked $19\sqrt{3} \, m$ towards the building.

(N/A) Let $BC$ be the building,$AB$ be the transmission tower,and $D$ be the point on the ground from where the elevation angles are measured.
In $\triangle BCD$:
$\frac{BC}{CD} = \tan 45^{\circ}$
$\frac{20}{CD} = 1$
$CD = 20 \, m$
In $\triangle ACD$:
$\frac{AC}{CD} = \tan 60^{\circ}$
$\frac{AB + BC}{CD} = \sqrt{3}$
$\frac{AB + 20}{20} = \sqrt{3}$
$AB + 20 = 20\sqrt{3}$
$AB = 20\sqrt{3} - 20$
$AB = 20(\sqrt{3} - 1) \, m$
Therefore,the height of the transmission tower is $20(\sqrt{3} - 1) \, m$.

(N/A) Let $AB$ be the statue,$BC$ be the pedestal,and $D$ be the point on the ground from where the angles of elevation are measured.
In $\triangle BCD$,we have:
$\frac{BC}{CD} = \tan 45^{\circ}$
$\frac{BC}{CD} = 1$
$BC = CD$
In $\triangle ACD$,we have:
$\frac{AC}{CD} = \tan 60^{\circ}$
$\frac{AB + BC}{CD} = \sqrt{3}$
Since $CD = BC$,we can write:
$\frac{1.6 + BC}{BC} = \sqrt{3}$
$1.6 + BC = BC \sqrt{3}$
$1.6 = BC(\sqrt{3} - 1)$
$BC = \frac{1.6}{\sqrt{3} - 1}$
Rationalizing the denominator:
$BC = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$
$BC = \frac{1.6(\sqrt{3} + 1)}{3 - 1}$
$BC = \frac{1.6(\sqrt{3} + 1)}{2}$
$BC = 0.8(\sqrt{3} + 1) \, m$
Thus,the height of the pedestal is $0.8(\sqrt{3} + 1) \, m$.

(N/A) Let $AB$ be the building and $CD$ be the tower.
In $\triangle CDB$,
$\frac{CD}{BD} = \tan 60^{\circ}$
$\frac{50}{BD} = \sqrt{3}$
$BD = \frac{50}{\sqrt{3}}$
In $\triangle ABD$,
$\frac{AB}{BD} = \tan 30^{\circ}$
$AB = BD \times \tan 30^{\circ} = \frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{50}{3} = 16 \frac{2}{3} \, m$
Therefore,the height of the building is $16 \frac{2}{3} \, m$.

(N/A) Let $AB$ and $CD$ be the poles of equal height $h$,and $O$ be the point on the road $BD$ such that $BD = 80\, m$.
Let $BO = x\, m$,then $OD = (80 - x)\, m$.
In $\triangle ABO$,$\tan 60^{\circ} = \frac{AB}{BO} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \dots(1)$.
In $\triangle CDO$,$\tan 30^{\circ} = \frac{CD}{OD} \implies \frac{1}{\sqrt{3}} = \frac{h}{80 - x} \implies h = \frac{80 - x}{\sqrt{3}} \quad \dots(2)$.
Equating $(1)$ and $(2)$:
$x\sqrt{3} = \frac{80 - x}{\sqrt{3}}$
$3x = 80 - x$
$4x = 80 \implies x = 20\, m$.
So,$BO = 20\, m$ and $OD = 80 - 20 = 60\, m$.
Substituting $x$ in $(1)$:
$h = 20\sqrt{3}\, m$.
Thus,the height of the poles is $20\sqrt{3}\, m$ and the distances of the point from the poles are $20\, m$ and $60\, m$.

(N/A) Let the height of the $TV$ tower be $AB = h$ and the width of the canal be $BC = x$.
In $\triangle ABC$,we have:
$\frac{AB}{BC} = \tan 60^{\circ}$
$\frac{h}{x} = \sqrt{3} \implies h = x\sqrt{3} \quad \dots(1)$
In $\triangle ABD$,we have:
$\frac{AB}{BD} = \tan 30^{\circ}$
$\frac{h}{x + 20} = \frac{1}{\sqrt{3}} \implies h\sqrt{3} = x + 20 \quad \dots(2)$
Substituting the value of $h$ from equation $(1)$ into equation $(2)$:
$(x\sqrt{3})\sqrt{3} = x + 20$
$3x = x + 20$
$2x = 20 \implies x = 10 \, m$
Now,substituting $x = 10$ into equation $(1)$:
$h = 10\sqrt{3} \, m$
Therefore,the height of the tower is $10\sqrt{3} \, m$ and the width of the canal is $10 \, m$.

(N/A) Let $AB$ be the building of height $7 \,m$ and $CD$ be the cable tower.
In $\triangle ABD$,$\angle ABD = 90^{\circ}$ and $\angle ADB = 45^{\circ}$.
$\tan 45^{\circ} = \frac{AB}{BD}$
$1 = \frac{7}{BD}$
$BD = 7 \,m$.
Since $ABED$ is a rectangle,$AE = BD = 7 \,m$ and $ED = AB = 7 \,m$.
In $\triangle ACE$,$\angle AEC = 90^{\circ}$ and $\angle CAE = 60^{\circ}$.
$\tan 60^{\circ} = \frac{CE}{AE}$
$\sqrt{3} = \frac{CE}{7}$
$CE = 7\sqrt{3} \,m$.
The total height of the tower $CD = CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \,m$.
Thus,the height of the tower is $7(\sqrt{3} + 1) \,m$.

(N/A) Let $AB$ be the lighthouse and the two ships be at points $C$ and $D$ respectively.
In $\triangle ABC$,
$\frac{AB}{BC} = \tan 45^{\circ}$
$\frac{75}{BC} = 1$
$BC = 75 \, m$
In $\triangle ABD$,
$\frac{AB}{BD} = \tan 30^{\circ}$
$\frac{75}{BC + CD} = \frac{1}{\sqrt{3}}$
$\frac{75}{75 + CD} = \frac{1}{\sqrt{3}}$
$75\sqrt{3} = 75 + CD$
$CD = 75(\sqrt{3} - 1) \, m$
Therefore,the distance between the two ships is $75(\sqrt{3} - 1) \, m$.

(N/A) Let the initial position of the balloon be $A$ and its final position be $B$. Let $CD$ be the girl,where $CD = 1.2\, m$.
The height of the balloon from the ground is $88.2\, m$. The height of the balloon from the eye level of the girl is $88.2\, m - 1.2\, m = 87\, m$.
Let the eye level of the girl be the horizontal line $CG$. Thus,$AE = BG = 87\, m$.
In $\triangle ACE$,$\tan 60^{\circ} = \frac{AE}{CE} \implies \sqrt{3} = \frac{87}{CE} \implies CE = \frac{87}{\sqrt{3}} = 29\sqrt{3}\, m$.
In $\triangle BCG$,$\tan 30^{\circ} = \frac{BG}{CG} \implies \frac{1}{\sqrt{3}} = \frac{87}{CG} \implies CG = 87\sqrt{3}\, m$.
The distance travelled by the balloon is $EG = CG - CE = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\, m$.

(B) Let $AB$ be the tower of height $h$ and $C$ be the initial position of the car. After $6$ seconds,the car reaches point $D$.
In $\triangle ADB$,$\tan 60^{\circ} = \frac{AB}{DB} \implies \sqrt{3} = \frac{h}{DB} \implies DB = \frac{h}{\sqrt{3}}$.
In $\triangle ABC$,$\tan 30^{\circ} = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{h}{BC} \implies BC = h\sqrt{3}$.
The distance covered in $6$ seconds is $CD = BC - DB = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$.
Since the speed is uniform,the time taken to cover distance $CD$ is $6$ seconds.
Therefore,time taken to cover distance $DB = \frac{h}{\sqrt{3}}$ is $\frac{6 \times (h/\sqrt{3})}{2h/\sqrt{3}} = \frac{6}{2} = 3$ seconds.

(N/A) Let $AQ$ be the tower and $R, S$ be the points at a distance of $4 \,m$ and $9 \,m$ from the base of the tower $(Q)$ respectively.
The angles of elevation are complementary. Therefore,if one angle is $\theta$,the other will be $(90^\circ - \theta)$.
In $\triangle AQR$,
$\tan \theta = \frac{AQ}{QR} = \frac{AQ}{4} \quad \dots(i)$
In $\triangle AQS$,
$\tan(90^\circ - \theta) = \frac{AQ}{SQ}$
$\cot \theta = \frac{AQ}{9} \quad \dots(ii)$
Multiplying equations $(i)$ and $(ii)$:
$\tan \theta \cdot \cot \theta = \left(\frac{AQ}{4}\right) \cdot \left(\frac{AQ}{9}\right)$
$1 = \frac{AQ^2}{36}$
$AQ^2 = 36$
$AQ = \sqrt{36} = 6 \,m$ (Since height cannot be negative).
Hence,the height of the tower is $6 \,m$.