Some Applications of Trigonometry — Important Questions

Answers

$h=\frac{100}{\sqrt{3}}=\frac{100\sqrt{3}}{3}$ m.
The height of the tower is $\frac{100\sqrt{3}}{3}$ m.

$BC=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}$ m.

$CD=\frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}$ m.
Since AC = CD, AC = $\frac{16\sqrt{3}}{3}$ m.
Total height of the tree, H = AC + BC = $\frac{16\sqrt{3}}{3}+\frac{8\sqrt{3}}{3}=\frac{24\sqrt{3}}{3}=8\sqrt{3}$ m.

$EF=28.5\sqrt{3}$ m.
After walking towards the building, let her position be D. The new distance from the building is DF. The angle of elevation from her eyes at G (at height 1.5 m) to A is $\mathrm{\angle }AGH={60}^{\circ }$. So, AH = 30 - 1.5 = 28.5 m.
In right-angled triangle AGH,

$GH=\frac{28.5}{\sqrt{3}}=\frac{28.5\sqrt{3}}{3}=9.5\sqrt{3}$ m.
The distance walked by the girl is CD = EF - GH = $28.5\sqrt{3}-9.5\sqrt{3}=19\sqrt{3}$ m.

So, AE = x.
Now, the height of the tower CD = CE + ED.
CE = AE = x.
ED = AB = 7 m.
So, CD = x + 7.
In right-angled triangle AEC (this is not a right triangle). Let's reconsider.
Let the height of the building be AB = 7 m. Let the height of the cable tower be CD. Let the distance between the building and the tower be BD = x.
The observer is at A (top of the building).
Angle of elevation of C from A is $\mathrm{\angle }CAE={60}^{\circ }$, where AE is horizontal. AE = BD = x.
Angle of depression of D from A is $\mathrm{\angle }DAF={45}^{\circ }$, where AF is horizontal. AF = BD = x.
In right triangle ABE, where E is a point such that BE = AB = 7m, and AE is the horizontal distance. This is not working.

Let's use standard notation.
Let the height of the building be AB = 7 m. Let the height of the cable tower be CD = H. Let the distance between the base of the building and the base of the tower be BD = x.
The observer is at A (top of the building).
Draw a horizontal line AE parallel to BD. So AE = BD = x.
The angle of elevation of the top of the tower C from A is $\mathrm{\angle }CAE={60}^{\circ }$.
The angle of depression of the foot of the tower D from A is $\mathrm{\angle }DA{E}^{\mathrm{\prime }}={45}^{\circ }$, where AE' is the horizontal line. Since AE and AE' are the same line, let's call it AE.
So, $\mathrm{\angle }DAE={45}^{\circ }$.
In right triangle ABE, where E is a point such that BE = AB = 7m, and AE is the horizontal distance. This is also not working.

Correct approach for Q4:
Let the height of the building be AB = 7 m. Let the height of the cable tower be CD = h. Let the distance between the foot of the building and the foot of the tower be BD = x.
From the top of the building (point A), draw a horizontal line AE.
The angle of elevation of the top of the tower (C) from A is $\mathrm{\angle }CAE={60}^{\circ }$.
The angle of depression of the foot of the tower (D) from A is $\mathrm{\angle }DAE={45}^{\circ }$.
In right triangle ABD, $\mathrm{\angle }ADB$ is not defined directly.
Consider the horizontal line AE through A. AE is parallel to BD.
So, $\mathrm{\angle }ABD$ is not related directly.
In right triangle ADE where E is a point on the line segment CD such that DE = AB = 7m, and AE = BD = x.
In right triangle ADE, $\mathrm{\angle }AED={90}^{\circ }$.

$1=\frac{7}{x}⟹x=7$ m.
Now consider the top of the tower. C is above E. CE = CD - DE = h - 7.
In right triangle ACE, $\mathrm{\angle }AEC={90}^{\circ }$.

$h=7+7\sqrt{3}=7(1+\sqrt{3})$ m.
The height of the tower is $7(1+\sqrt{3})$ m.

Correct approach for Q5:
Let the height of the multi-storeyed building be PC = h. Let the height of the other building be AB = 8 m. Let the horizontal distance between the two buildings be AC = x.
From the top of the multi-storeyed building (P), draw a horizontal line PQ.
The angle of depression of the top of building A is 30°, so $\mathrm{\angle }QPA={30}^{\circ }$.
The angle of depression of the bottom of building B is 45°, so $\mathrm{\angle }QPB={45}^{\circ }$.
Draw a line from P parallel to AC, meeting AB at D. Then PD = AC = x.
And AD = PB = h - 8.
In right triangle APD, $\mathrm{\angle }ADP={90}^{\circ }$.

$\frac{1}{\sqrt{3}}=\frac{h-8}{x}⟹x=\sqrt{3}(h-8)$ (Equation 1)
Now consider the angle of depression to B.
Draw a line from P parallel to AC, meeting the line extended from AB at D. This is confusing.

Let's use the diagram provided in NCERT, Fig 9.9.
PC is the multi-storeyed building, AB is the 8m tall building. AC is the distance.
Draw a horizontal line through P. Let it be PQ.
$\mathrm{\angle }QPA={30}^{\circ }$. $\mathrm{\angle }QPB={45}^{\circ }$.
Since PQ is parallel to AB, $\mathrm{\angle }PBD={30}^{\circ }$ (alternate interior angles).
In right triangle PBD, $\tan {30}^{\circ }=\frac{PD}{BD}$. Here PD is the horizontal distance, which is AC = x. BD is the height difference, which is PC - BC = h - 8.
So, $\tan {30}^{\circ }=\frac{x}{h-8}$.
$\frac{1}{\sqrt{3}}=\frac{x}{h-8}⟹x=\frac{h-8}{\sqrt{3}}$ (Equation 1)
Now consider the angle of depression to B.
$\mathrm{\angle }QPB={45}^{\circ }$.
Draw a horizontal line through P.
Consider the right triangle formed by P, B, and the point on the line PB at the same height as B.
This is also confusing. Let's re-read the NCERT solution for Example 6.
"In Fig. 9.9, PC denotes the multi- storyed building and AB denotes the 8 m tall building. We are interested to determine the height of the multi-storeyed building, i.e., PC and the distance between the two buildings, i.e., AC. Look at the figure carefully. Observe that PB is a transversal to the parallel lines PQ and BD. Therefore, $\mathrm{\angle }QPB$ and $\mathrm{\angle }PBD$ are alternate angles, and so are equal. So $\mathrm{\angle }PBD={30}^{\circ }$."
This statement in the book is incorrect based on the diagram. The diagram shows the angle of depression of the top and bottom of the 8m building.
Let's assume PC is the multi-storeyed building and AB is the 8m building.
Let PC = h and AC = x.
Observer is at P.
Angle of depression of A is 30°.
Angle of depression of B is 45°.
Draw a horizontal line through P, say PX.
$\mathrm{\angle }XPA={30}^{\circ }$.
$\mathrm{\angle }XPB={45}^{\circ }$.
Since PX is parallel to AC, and AC is perpendicular to AB, PX is perpendicular to AB line.
In $\mathrm{△}PAX$, AX = PC - AC = h - 8. This is not correct. AX = height difference.
Let's assume A is the top of the 8m building and B is the bottom.
Let C be the top of the multi-storeyed building and D be the bottom.
Let the height of the multi-storeyed building be CD = h. Let the height of the 8m building be AB = 8. Let the distance between them be BD = x.
Observer is at C.
Angle of depression of A is 30°.
Angle of depression of B is 45°.
Draw a horizontal line through C, say CE.
$\mathrm{\angle }ECA={30}^{\circ }$.
$\mathrm{\angle }ECB={45}^{\circ }$.
Since CE is parallel to BD, $\mathrm{\angle }CAB={30}^{\circ }$ (alternate interior angles). This is wrong.
Let's go back to the NCERT text's interpretation.
PC = multi-storeyed building (height h). AB = 8 m building. AC = x (distance).
Observer at P.
Angle of depression of A is 30°. $\mathrm{\angle }XPA={30}^{\circ }$.
Angle of depression of B is 45°. $\mathrm{\angle }XPB={45}^{\circ }$.
Draw horizontal line PQ.
In right triangle formed by P, A and the point vertically below P on AB line (let's call it D). PD = x. AD = h - 8.

$\frac{1}{\sqrt{3}}=\frac{h-8}{x}⟹x=\sqrt{3}(h-8)$ (1)
Now consider the angle of depression to B.
In right triangle formed by P, B and the point vertically below P on the ground (which is C itself). This is not right.

Let's restart Q5 with a clear diagram concept:
Let the height of the taller building be $H$ and the shorter building be $h=8$ m. Let the distance between them be $d$.
From the top of the taller building, the angle of depression to the top of the shorter building is ${30}^{\circ }$.
From the top of the taller building, the angle of depression to the bottom of the shorter building is ${45}^{\circ }$.
Let the top of the taller building be $T$. Let the top of the shorter building be $S$. Let the bottom of the shorter building be $B$.
Let the point directly below $T$ at the level of $S$ be $P$. Then $TP=d$. $TS=H-h$. $\mathrm{\angle }TPS={30}^{\circ }$.
In right-angled triangle $TPS$:

$\frac{1}{\sqrt{3}}=\frac{H-8}{d}⟹d=\sqrt{3}(H-8)$ (Eq. 1)

Let the point directly below $T$ at the level of $B$ be $Q$. Then $TQ=d$. $TB=H$.
The angle of depression to $B$ is ${45}^{\circ }$.
In right-angled triangle $TBQ$:

$1=\frac{H}{d}⟹d=H$ (Eq. 2)

From Eq. 1 and Eq. 2:

$H=\frac{8\sqrt{3}}{\sqrt{3}-1}=\frac{8\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{8(3+\sqrt{3})}{3-1}=\frac{8(3+\sqrt{3})}{2}=4(3+\sqrt{3})$ m.
$H=12+4\sqrt{3}$ m.
The height of the multistoried building is $12+4\sqrt{3}$ m.
The distance between the two buildings is $d=H=12+4\sqrt{3}$ m.

Correct approach for Q6:
Let the height of the building be AB = 60 m. Let the height of the lamp post be CD = h. Let the horizontal distance between them be AC = x.
From the top of the building (A), draw a horizontal line AE.
The angle of depression of the top of the lamp post (C) is 30°. So, $\mathrm{\angle }EAC={30}^{\circ }$.
The angle of depression of the bottom of the lamp post (D) is 60°. So, $\mathrm{\angle }EAD={60}^{\circ }$.
Since AE is parallel to BD, $\mathrm{\angle }ACD={30}^{\circ }$ is wrong.
In $\mathrm{△}ACE$, CE = AB - CD = 60 - h. AE = AC = x.

$\frac{1}{\sqrt{3}}=\frac{60-h}{x}⟹x=\sqrt{3}(60-h)$ (1)
In $\mathrm{△}ADE$, DE = AB = 60 m. AE = AC = x.

$\sqrt{3}=\frac{60}{x}⟹x=\frac{60}{\sqrt{3}}=20\sqrt{3}$ m.
Substitute x in (1):

$h=60-20=40$ m.
(i) The horizontal distance between the building and the lamp post is $x=20\sqrt{3}$ m.
(ii) The distance between the tops of the building and the lamp post is AC. We need to find AC.
This is not correct. The question asks for the distance between the tops, which is the length of the line segment AC.
AC is the hypotenuse in triangle ACE.

$AC=\sqrt{1600}=40$ m.
The distance between the tops of the building and the lamp post is 40 m.

$h=\frac{80}{2}=40$ m.
The height of the kite above the ground is 40 m.

$1=\frac{h_p}{x}⟹x=h_p$.
The total height of the pedestal and statue is AC = $h_p+h_s=h_p+1.6$.
In right-angled triangle CPB:

$\sqrt{3}=\frac{h_p+1.6}{h_p}$ (since $x=h_p$)

$h_p=\frac{1.6}{\sqrt{3}-1}=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{1.6(\sqrt{3}+1)}{3-1}=\frac{1.6(\sqrt{3}+1)}{2}=0.8(\sqrt{3}+1)$ m.
The height of the pedestal is $0.8(\sqrt{3}+1)$ m.

$h=\frac{20}{2}=10$ m.
The height of the pole is 10 m.

$h=\frac{30}{\sqrt{3}}=\frac{30\sqrt{3}}{3}=10\sqrt{3}$ m.
The height of the tower is $10\sqrt{3}$ m.

$\frac{1}{2}=\frac{1.5}{l_1}⟹l_1=2\times 1.5=3$ m.
For elder children: height $h_2=3$ m, angle ${\theta }_2={60}^{\circ }$. Let the length of the slide be $l_2$.
In right-angled triangle:

$\frac{\sqrt{3}}{2}=\frac{3}{l_2}⟹l_2=\frac{3\times 2}{\sqrt{3}}=\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}$ m.
The length of the slide for children below 5 years is 3 m and for elder children is $2\sqrt{3}$ m.

$\sqrt{3}=\frac{h}{x}⟹h=x\sqrt{3}$ (1)
In right-angled triangle CDP:

$\frac{1}{\sqrt{3}}=\frac{h}{80-x}⟹h=\frac{80-x}{\sqrt{3}}$ (2)
Equating (1) and (2):

$4x=80⟹x=20$ m.
Now find the height of the poles:
$h=x\sqrt{3}=20\sqrt{3}$ m.
The distances of the point from the poles are 20 m and $80-20=60$ m.
The height of the poles is $20\sqrt{3}$ m.

$\sqrt{3}=\frac{h}{x}⟹h=x\sqrt{3}$ (1)
In right-angled triangle ABD:

$\frac{1}{\sqrt{3}}=\frac{h}{x+20}⟹h=\frac{x+20}{\sqrt{3}}$ (2)
Equating (1) and (2):

$2x=20⟹x=10$ m.
Now find the height of the tower:
$h=x\sqrt{3}=10\sqrt{3}$ m.
The height of the tower is $10\sqrt{3}$ m and the width of the canal is 10 m.

Correct approach for Q14:
Let the height of the bridge be h = 3 m. Let the width of the river be W. Let the point on the bridge be P. Let the banks on opposite sides be A and B. So, AB = W.
Let the point directly below P on the ground be D. So, PD = 3 m.
The angle of depression of bank A is 30°. So, $\mathrm{\angle }APD$ is not the angle.
Let's consider the horizontal line through P. Let it be PX.
$\mathrm{\angle }XPA={30}^{\circ }$.
$\mathrm{\angle }XPB={45}^{\circ }$.
Since PX is parallel to AB, $\mathrm{\angle }PAB={30}^{\circ }$ and $\mathrm{\angle }PBA={45}^{\circ }$.
In right-angled triangle PDA, $\mathrm{\angle }PDA={90}^{\circ }$.

$\frac{1}{\sqrt{3}}=\frac{3}{AD}⟹AD=3\sqrt{3}$ m.
In right-angled triangle PDB, $\mathrm{\angle }PDB={90}^{\circ }$.

$1=\frac{3}{BD}⟹BD=3$ m.
The width of the river AB = AD + BD = $3\sqrt{3}+3=3(\sqrt{3}+1)$ m.

$\sqrt{3}=\frac{87}{CD}⟹CD=\frac{87}{\sqrt{3}}=\frac{87\sqrt{3}}{3}=29\sqrt{3}$ m.
After some time, the angle of elevation reduces to 30°. Let the new position of the balloon be B.

$\frac{1}{\sqrt{3}}=\frac{87}{CE}⟹CE=87\sqrt{3}$ m.
The distance travelled by the balloon is AB = CE - CD = $87\sqrt{3}-29\sqrt{3}=58\sqrt{3}$ m.

$\frac{1}{\sqrt{3}}=\frac{75}{AC}⟹AC=75\sqrt{3}$ m.
In right-angled triangle BBC:

$\sqrt{3}=\frac{75}{BC}⟹BC=\frac{75}{\sqrt{3}}=\frac{75\sqrt{3}}{3}=25\sqrt{3}$ m.
The distance travelled by the car in 6 seconds is AB = AC - BC = $75\sqrt{3}-25\sqrt{3}=50\sqrt{3}$ m.
The speed of the car is $\frac{50\sqrt{3}}{6}=\frac{25\sqrt{3}}{3}$ m/s.
The time taken by the car to reach the foot of the tower from point B is $\frac{BC}{\text{speed}}=\frac{25\sqrt{3}}{\frac{25\sqrt{3}}{3}}=3$ seconds.
The time taken by the car to reach the foot of the tower from point A is $\frac{AC}{\text{speed}}=\frac{75\sqrt{3}}{\frac{25\sqrt{3}}{3}}=75\sqrt{3}\times \frac{3}{25\sqrt{3}}=9$ seconds.
The question asks for the time taken from the point where the angle of depression was 60°. This is the point B. The time taken from point B to C is 3 seconds.

$\frac{1}{\sqrt{3}}=\frac{75}{PX}⟹PX=75\sqrt{3}$ m.
In right-angled triangle BPX, where PX is the horizontal distance from the lighthouse to B. $\mathrm{\angle }PBX={45}^{\circ }$.

$1=\frac{75}{PB}⟹PB=75$ m.
The distance between the two ships is AB = PB - PX = $75-75\sqrt{3}$. This is incorrect as B is behind A.
The angle of depression of the second ship (B) is 45°. This means B is closer to the lighthouse.
So, the angle of depression of the ship farther away is 30°, and the angle of depression of the ship nearer is 45°.
Let the distance of the farther ship be $d_1$ and the nearer ship be $d_2$.
$\tan {30}^{\circ }=\frac{75}{d_1}⟹d_1=75\sqrt{3}$ m.
$\tan {45}^{\circ }=\frac{75}{d_2}⟹d_2=75$ m.
The distance between the two ships is $d_1-d_2=75\sqrt{3}-75=75(\sqrt{3}-1)$ m.

$1=\frac{10}{d_2}⟹d_2=10$ m.
The first boy is at a distance $d_1$ from the point directly below the kite.

$\frac{1}{\sqrt{3}}=\frac{30}{d_1}⟹d_1=30\sqrt{3}$ m.
The boys are on opposite sides of the kite.
The distance of the first boy from the base of the building is the sum of the horizontal distance of the first boy from the point below the kite and the horizontal distance of the building from the point below the kite.
The horizontal distance of the building from the point below the kite is $d_2=10$ m.
So, the distance of the first boy from the base of the building = $d_1+d_2=30\sqrt{3}+10$ m.
Height of the kite from the ground is 30 m.

$\frac{1}{\sqrt{3}}=\frac{10}{x}⟹x=10\sqrt{3}$ m.
The total height of the pole and the flag is AC = AB + BC = 10 + length of flag. Let the length of the flag be l. So AC = 10 + l.
In right-angled triangle ACP:

$l=10\sqrt{3}-10=10(\sqrt{3}-1)$ m.
The length of the flag is $10(\sqrt{3}-1)$ m.

Let's interpret "subtends an angle of 60°". This usually means the angle between the lines of sight to the extreme points of the object. For a sphere, this would mean the angle between the tangents from the observer to the sphere.
If the angle subtended by the diameter of the balloon at the eye is 60°, then in the triangle formed by the observer and the diameter, the angle at the observer is 60°.
Let the observer be at O. Let the centre of the balloon be C. Let the radius be r.
The angle of elevation of the centre C from O is 45°. Let the height of the centre be $h_c$.
Let the point directly below C be D. So, OD is the horizontal distance. CD = $h_c$.
In right triangle ODC:
$\tan {45}^{\circ }=\frac{CD}{OD}=\frac{h_c}{OD}⟹OD=h_c$.
Now, consider the angle subtended by the balloon.
This implies that the angle formed by the tangents from O to the sphere is 60°.
Let the points of tangency be T1 and T2. Then $\mathrm{\angle }T1OT2={60}^{\circ }$.
Consider the triangle OT1C. This is a right triangle at T1. OC is the hypotenuse.
The line OC bisects $\mathrm{\angle }T1OT2$. So $\mathrm{\angle }T1OC={30}^{\circ }$.
In right triangle OT1C:

$\frac{1}{2}=\frac{r}{OC}⟹OC=2r$.
Now consider the height of the centre.

$4r^2=O{D}^2+h_c^2$.
Since $OD=h_c$,
$4r^2=h_c^2+h_c^2=2h_c^2$.
$h_c^2=\frac{4r^2}{2}=2r^2$.
$h_c=\sqrt{2r^2}=r\sqrt{2}$.
Thus, the height of the centre of the balloon is $\sqrt{2}$ times its radius.