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Q1. |
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Ans. |
Given : Length of rope = Hypotenuse = AC = 20 m The angle made by the rope with the ground level = \(\angle BCA = 30^{0}\) We have to find Height of pole = AB = let, h Using trigonometric ratios of \(\sin\theta = \frac {\text {Opposite side to}\angle C}{Hypotenuse}=\frac {AB}{AC}\) Putting value \[\sin(30^\circ) = \frac{h}{20}\] Solving for \(h\): \[ h = 20 \cdot \sin(30^\circ) \] \[ h = 20 \cdot \frac{1}{2} \] \[ h = 10 \, \text{m} \] |
Q2. | A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. |
Ans. |  Construction: By using given instruction, draw a figure as fallow: Here, PR = the broken part of the tree \(\angle R=30^{0}\) QR = Horizontal line = 8m As per figure, Total height of tree = PQ + PR In \(\triangle PQR\) \(\cos 30^{0}=\frac {\text {Adjacent side of }\angle R}{Hypotenuse}=\frac {QR}{PR}\) Putting value \(\frac {\sqrt {3}}{2}=\frac {8}{PR}\) \(PR=\frac {16}{\sqrt {3}}........ (i)\) And also, \(\tan 30^{0}=\frac {\text {Opposite side to} \angle R}{\text {Adjacent side of }\angle R}=\frac {PQ}{QR}\) Putting value \(\frac {1}{\sqrt {3}}=\frac {PQ}{8}\) \(PQ = \frac {8}{\sqrt {3}}........ (ii)\) Hence, total height of tree = PQ+PR \(\frac {16}{\sqrt {3}}+\frac {8}{\sqrt {3}}=\frac {16+8}{\sqrt {3}}=\frac {24}{\sqrt {3}}\) multiply by \(\sqrt {3}\) \(=\frac {24\times \sqrt {3}}{\sqrt {3}\times\sqrt {3}}\) \(=\frac {24\times \sqrt {3}}{3}=8\sqrt {3}\) |
Q3. | A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? |
Ans. |  Case 1: By construction, In \(\triangle ABC\): Height of slide =AB = 1.5 m \(\angle ACB = 30^{0}\) We have to find Length of slide AC. We know that: \(\sin C = \frac {\text {Opposite side to} \angle C}{Hypotenuse}=\frac {AB}{AC}\) Putting values: \(\frac {1}{2}=\frac {1.5}{AC}\) \(1\times AC=1.5\times 2\) \(AC = 3 \text { m}\) So, Length of slide for below 5 year children = 3 m Case 2: By construction, In \(\triangle PQR\): Height of slide = PQ = 3 m \(\angle PRQ = 60^{0}\) We have to find Length of slide PR. We know that: \(\sin R = \frac {\text {Opposite side to} \angle R}{Hypotenuse}=\frac {PQ}{PR}\) Putting values: \(\frac {\sqrt {3}}{2}=\frac {3}{PR}\) \(PR = \frac {3\times 2}{\sqrt {3}}=\frac {6}{\sqrt {3}}\) \(PR =\frac {6\times \sqrt {3}}{\sqrt {3}\times \sqrt {3}}=\frac {6\sqrt {3}}{3}=2\sqrt{3}\) So, Length of slide for Elder children = \(2\sqrt{3}\) |
Q4. | The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. |
Ans. |  Let Tower is AB Let Point of ground be C Distance of point C from the foot of tower = BC = 30m Angle of elevation = \(\angle ACB = 30^{0}\) We have to find height of tower which is AB. In \(\triangle ABC\) \(\tan C = \frac {\text {Opposite side to }\angle C}{\text {Adjacent side to }\angle C}\) \(\tan 30^{0}=\frac {AB}{BC}\) \(\frac {1}{\sqrt {3}}=\frac {AB}{30}\) \(AB = \frac {30}{\sqrt {3}}\) Multiply by \(\sqrt {3}\) \(AB = \frac {30\times \sqrt {3}}{\sqrt {3}\times \sqrt {3}}\) \(AB = \frac {30\sqrt {3}}{3}=10\sqrt {3}\) Hence, Height of tower (AB) = \(10\sqrt {3}\) |
Q5. | A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. |
Ans. |  C is position of kite Kit flying at height = AB = 60 m Inclination of the string with the ground = \(\angle ACB = 60^{0}\) We have to find Length of string = Hypotenuse = AC In \(\triangle ABC\) \(sin C = \frac {\text {Opposite side to }\angle C}{Hypotenuse}=\frac {AB}{AC}\) Putting value \(\sin 60^{0}=\frac {60}{AC}\) \(\frac {\sqrt {3}}{2}=\frac {60}{AC}\) \(AC = \frac {2}{\sqrt {3}}\times 60 = \frac {120}{\sqrt {3}}\) Multiply by \(\sqrt {3}\) \(AC =\frac {120\times \sqrt {3}}{\sqrt {3}\times \sqrt {3}}\) \(AC = \frac {120\sqrt{3}}{3}=40\sqrt {3}\text { m}\) Hence, Length of string \(=AC=40\sqrt {3}\text { m}\) |
Q6. | A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. |
Ans. |  is 1.5 m tall so, PQ = 1.5 m and building is 30 m tall so, AB = 30 m Angle of elevation from start point (Q) to top of building \(=30^{0}\) so, \(\angle APC = 30^{0}\) When boy moves from point Q to Point R, angle of elevation changes \(30^{0}\) to \(60^{0}\) So, \(\angle ASC = 60^{0}\) and \(PQ\parallel CB\), so, we can say that \(CQ=CB=1.5\text { m}\) Now, \(AC = AB-CB\) \(AC = 30-1.5=28.5\text { m}\) And also, \(PC\parallel QB\) so, We can say that \(PS=QR \text { and } SC = RB\) Since tower is vertical so \(\angle ACP = 90^{0}\) Now, In \(\triangle APC\) \(\tan P = \frac {\text {Opposite side to}\angle P}{\text {Adjacent side to }\angle P}=\frac {AC}{PC}\) Putting value \(\tan 30^{0}=\frac {28.5}{PC}\) \(\frac {1}{\sqrt {3}}=\frac {28.5}{PC}\) \(PC = 28.5\sqrt {3}\) In \(\triangle ASC\) \(\tan S = \frac {\text {Opposite side to}\angle S}{\text {Adjacent side to }\angle S}=\frac {AC}{SC}\) Putting value \(\tan 60^{0}=\frac {AC}{PC}\) \(\sqrt {3}=\frac {28.5}{SC}\) \(SC = \frac {28.5}{\sqrt {3}}\) Because \(PC = PS+SC\) putting value \(28.5\sqrt{3}=PS+\frac {28.5}{\sqrt{3}}\) \(PS = 28.5\sqrt{3}-\frac {28.5}{\sqrt{3}}\) \(PS=\frac {28.5\sqrt{3}\times\sqrt{3}-28.5}{\sqrt{3}}\) \(PS=\frac {28.5\times 3-28.5}{\sqrt{3}}\) \(PS=\frac {28.5(3-1)}{\sqrt {3}}\) \(PS=\frac {28.5\times 2}{\sqrt {3}}\) Multiply by \(\sqrt {3}\) \(PS=\frac {28.5\times 2}{\sqrt {3}}\times \frac {\sqrt{3}}{\sqrt{3}}\) \(PS=\frac {28.5\times \sqrt {3}\times 2}{3}\) \(PS=\frac {57\times \sqrt {3}}{3}\) \(PS=19\sqrt {3}\) Since \(QR=PS\), so, \(QR=19\sqrt {3}\) Hence,Walking distance toward the building is \(19\sqrt {3}\). |
Q7. | From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. |
Ans. |  Let Building be AB and Tower be CA Height of building \(= AB = 20\text { m}\) Let point on the ground \(O\). Angle of elevation from point \(O\) to top of building\( = \angle AOB = 45^{0}\) Angle of elevation from point \(O\) to top of Tower \( = \angle COB = 60^{0}\) We have to find \(AC\). In \(\triangle AOB\), \(\tan O = \frac {\text {Opposite side to angle O}}{\text {Adjacent side to angle O}}=\frac {AB}{OB}\) Putting value \(\tan 45^{0}=\frac {20}{OB}\) \(1=\frac {20}{OB}\) \(OB = 20 \text { m}\) and In \(\triangle COB\), \(\tan O = \frac {\text {Opposite side to angle O}}{\text {Adjacent side to angle O}}=\frac {BC}{OB}\) \(\tan 60^{0}=\frac {BC}{OB}\) Putting value \(\sqrt {3}=\frac {BC}{20}\) \(BC = 20\sqrt {3}\) So, \(AB+AC=20\sqrt {3}\) \(20+AC=20\sqrt {3}\) \(AC=20\sqrt {3}-20\) \(AC = 20(\sqrt {3}-1)\text { m}\) Hence, Height of the tower is \(20(\sqrt {3}-1)\text { m}\). |
Q8. | A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. |
Ans. |  Let Pedestal be AB and Statue be CA Height of Statue \(= AC = 1.6\text { m}\) Let point on the ground \(O\). Angle of elevation from point \(O\) to top of Pedestal\( = \angle AOB = 45^{0}\) Angle of elevation from point \(O\) to top of Statue \( = \angle COB = 60^{0}\) We have to find \(AB\). In \(\triangle AOB\), \(\tan O = \frac {\text {Opposite side to angle O}}{\text {Adjacent side to angle O}}=\frac {AB}{OB}\) Putting value \(\tan 45^{0}=\frac {AB}{OB}\) \(1=\frac {AB}{OB}\) \(OB = AB ...................... (i)\) and In \(\triangle COB\), \(\tan O = \frac {\text {Opposite side to angle O}}{\text {Adjacent side to angle O}}=\frac {BC}{OB}\) \(\tan 60^{0}=\frac {BC}{OB}\) Putting value \(\sqrt {3}=\frac {BC}{OB}\) \(\sqrt {3} OB = BC\) From equation (i) \(\sqrt {3} AB = BC\) \(\sqrt {3} AB = AB+AC\) \([since BC = AB+AC]\) Putting \(AC\) value \(\sqrt {3} AB = AB+1.6\) \(\sqrt {3} AB -AB = 1.6\) \((\sqrt {3} -1) AB = 1.6\) \(AB = \frac {1.6}{\sqrt {3} -1}\) Multiply by \(\sqrt {3} +1\) \(AB = \frac {1.6}{\sqrt {3} -1}\times \frac {\sqrt {3} +1}{\sqrt {3} +1}\) \(AB=\frac {1.6\times (\sqrt {3} +1)}{(\sqrt {3})^{2}-(1)^{2}}\) \(AB=\frac {1.6\times (\sqrt {3} +1)}{3-1}\) \(AB=\frac {1.6\times (\sqrt {3} +1)}{2}\) \(AB = 0.8(\sqrt {3} +1)\) Hence, Height of the Pedestal is \(0.8(\sqrt {3} +1)\text { m}\). |
Q9. | The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. |
Ans. |  Given: Height of Tower = 50 m = Let = H Height of building, Let = h Distance between tower and building, Let = d Now, In \(\triangle ABC\) \(\tan 60^{0}=\frac {\text {Opposite side to angle A}}{\text {Adjacent side to angle A}}=\frac {BC}{AB}=\frac {H}{d}\) Putting value \(\tan 60^{0}=\frac {H}{d}\) \(\sqrt {3}=\frac {50}{d}\) \(d=\frac {50}{\sqrt {3}}\) And , In \(\triangle ABD\) \(\tan 30^{0}=\frac {\text {Opposite side to angle B}}{\text {Adjacent side to angle B}}=\frac {AD}{AB}=\frac {h}{d}\) Putting value \(\tan 30^{0}=\frac {h}{d}\) \(\frac {1}{\sqrt {3}}=\frac {h}{\frac {50}{\sqrt {3}}}\) \(\frac {1}{\sqrt {3}}=\frac {\sqrt {3}h}{50}\) \(3h = 50\) \(h = \frac {50}{3} = 16.7 \text { m}\) |
Q10. | Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. |
Ans. |  \begin{align*}
\text{Construction :} \quad & \text{Construct a line diagram as per direction given in question} \\
\text{Given:} \quad & \angle APB = 60^\circ, \quad \angle CPD = 30^\circ, \quad AC = 80 \, \text{m} \\
\text{To find:} \quad & \text{The height of the pole} = AB = CD \\
\text{Solution:} \quad & \text{Let } AB \text{ and } CD \text{ be the two poles of equal height, and }\\
& P \text{ be the point on the road between the poles.} \\
& \text{In } \triangle APB,\\
& \tan 60^\circ = \frac{AB}{AP} \\
& \text{or, } AP = AB \cdot \frac{1}{\tan 60^\circ}\\
& \text{or, } AP = \frac {AB} {\sqrt{3}} ........ \quad \text{(i)} \\
& \text{In } \triangle CPD,\\
& \tan 30^\circ = \frac{CD}{CP} \\
& \text{or, } CP = CD \cdot \frac{1}{\tan 30^\circ}\\
& \text{or, } \quad CP = \sqrt{3} CD = \sqrt{3} AB \quad ...........\quad \text{(ii)} \\
& \text{Adding equation (i) and (ii) we get,} \\
& AP + CP = \frac {AB} {\sqrt{3}} + \sqrt{3} AB \\
& \text{or, } AC = AB \left(\sqrt{3} + \frac {1}{\sqrt{3}}\right) \\
& \text{or, } 80 \, \text{m} = 4\cdot \frac {AB} {\sqrt{3}}\\
& \text{or, } \quad AB = 20 \sqrt{3} \, \text{m} \\
& \text{Therefore, height of the pole} = 20 \sqrt{3} \, \text{m} \approx 34.64 \, \text{m}
\end{align*} |
Q11. | A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.  |
Ans. | \begin{align*}
\text{Given that: } & AB \text{ Height of tower and }\\
& AC \text{ width of canal.}\\
& C \text{ is the point on the other side} \\
& \text {of the canal directly opposite the tower.} \\
& \text{Let } AB = h, \quad CA = x, \quad CD = 20 \\
\text{Solution: } & \text{In } \triangle BAC:\\
& \cot 60^\circ = \frac{x}{h} \\
& \text{or, } \frac{1}{\sqrt{3}} = \frac{x}{h}\\
& \text{or, } x = \frac{h}{\sqrt{3}} ........... \text{(i)} \\
& \text{In } \triangle BAD:\\
& \cot 30^\circ = \frac{AD}{AB} \\
& \text{or, } \sqrt{3} = \frac{x+20}{h} \\
& \text{or, } x = h\sqrt{3} - 20 ........... \text{(2)} \\
& h = 10\sqrt{3} \\
& \text{From equation } (1):\\
& x = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \\
& \text{Therefore, the width of the canal = } 10\text{m}.\\
& \text{and the height of the canal = } 10\sqrt{3} \, \text{m}.
\end{align*}
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Q12. | From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. |
Ans. |  \begin{align*}
\text {Construction: }& \text {Construct a line diagram as per question}\\
& \text{Let, the height of the tower = } CE\\
& \text{ the height of the building = } AB\\
& \text{The angle of elevation } \angle EAD = 60^\circ \\
& \text{the angle of depression} \angle ACB = 45^\circ \\
& \text{Draw } AD \parallel BC\\
& \text{Then, } \angle DAC = \angle ACB = 45^\circ \text{ (alternate interior angles)} \\
\text {Solution :} & \text{In } \triangle ABC:\\
& \tan 45^\circ = \frac{AB}{BC}\\
& \text{or, } 1 = \frac{7}{BC} \\
& \text{or, } \quad BC = 7 \\
& ABCD \text{ is a square, so: } BC = AD = 7 \quad \text{and} \quad AB = CD = 7 \\
& \text{In } \triangle ADE:\\
& \tan 60^\circ = \frac{ED}{AD} \\
& \text{or, } \sqrt{3} = \frac{ED}{7}\\
& \text{or, } ED = 7\sqrt{3} \\
& \text{As per construction, Height of the tower } CE = ED + CD\\
& CE = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \\
& \text{Therefore, height of the tower }= 7 (\sqrt{3} + 1) \, \text{m.}
\end{align*} |
Q13. | As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. |
Ans. |  \begin{align*}
\text {Given :} & \text { Height of Lighthouse = } 75 \text {m}\\
& \text {Let, Two ships be at point C and D, so }\\
& \text {Angle of depression with ship D =} 30^\circ \\
& \text {Angle of depression with ship C =} 45^\circ \\
\text {Construction :} & \text {Construct line diagram as per question, Where}\\
& AB = \text {Lighthouse height =} 75 \text {m}\\
\text {Solution :} & \text {In} \triangle ABC,\\
& \Rightarrow \tan 45^\circ = \frac {AB}{BC}\\
& \Rightarrow 1=\frac {75}{BC}\\
& \Rightarrow BC = 75 \text { m}\\
& \text {In} \triangle ABD,\\
& \Rightarrow \tan 30^\circ = \frac {AB}{BD}\\
& \text {As per construction BD = BC + CD, So}\\
& \Rightarrow \frac {1}{\sqrt {3}}=\frac {75}{BC + CD}\\
& \Rightarrow \frac {1}{\sqrt {3}}=\frac {75}{75 + CD}\\
& \Rightarrow 75\sqrt {3} = 75 + CD\\
& \Rightarrow CD = 75\sqrt {3} - 75\\
& \Rightarrow CD = 75 (\sqrt {3} - 1)\\
& \text {By taking } \sqrt {3} = 1.73, We get \\
& CD = 54.75 \text { m}\\
& \text {Hence, the distance between two ship = 54.75 m}.
\end{align*}
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Q14. | A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.  |
Ans. |  \begin{align*}
\text{Given:} \quad & \text{Height of girl = AG = 1.2 m} \\
& \text{Also, } AG \parallel BF \\
& \text {So,}BF = AG = 1.2 \, \text{m}\\
& \text{height of Baloon}= EF = 88.2 \, \text{m} \\
& \text{Girl sees the balloon first at } 60^\circ \\
& \text{After change, the angle of elevation becomes } 30^\circ \\
& \text{So, Distance traveled by the balloon} = BC \\
& \text{Now, } BE = EF - BF = 88.2 - 1.2 = 87\, \text{m} \\
& \text{Also, } BE = DC = 87 \, \text{m} \\
& \text{Here, } \angle ABE = 90^\circ \text{ and } \angle ACD = 90^\circ \\
\text {Solution :} & \text{In } \triangle EBA\\
& \tan A = \frac{BE}{AB}\\
& \text {Putting value}\\
& \text{or, }\tan 60^\circ = \frac{87}{AB} \\
& \Rightarrow \sqrt{3} = \frac{87}{AB}\\
& \Rightarrow AB = \frac {87}{\sqrt{3}} \, \text{m} \\
& \text{In } \triangle DAC:\\
& \tan A = \frac{CD}{AC} \\
& \text{or, } \tan 30^\circ = \frac{87}{AC} \\
& \Rightarrow \frac{1}{\sqrt{3}} = \frac{87}{AC}\\
& \Rightarrow AC = 87\sqrt{3} \, \text{m} \\
& \text{Now, as per construction } AC = AB + BC \\
& \text {Putting value}\\
& \Rightarrow 87\sqrt{3} = \frac {87}{\sqrt{3}} + BC \\
& \Rightarrow BC = 87\sqrt{3} - \frac {87}{\sqrt{3}} \\
& \Rightarrow BC = 87\left(\sqrt{3} - \frac {1}{\sqrt{3}}\right) \\
& \Rightarrow BC = 87\left(\frac {3-1}{\sqrt {3}}\right) \\
& \Rightarrow BC = 87\times \frac {2}{\sqrt {3}} \\
& \text {Multiply by } \sqrt {3}\\
& \Rightarrow BC = 87\times \frac {2}{\sqrt {3}}\times \frac {\sqrt {3}}{\sqrt {3}} \\
& \Rightarrow BC = 87\times \frac {2\sqrt {3}}{3} \\
& \Rightarrow BC = 29\times 2\sqrt{3} \\
& \Rightarrow BC = 58\sqrt{3} \\
& \text{Hence, the distance traveled by the balloon} = 58\sqrt{3} \, \text{m}
\end{align*} |
Q15. | A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. |
Ans. |  \begin{align*}
\text {Construction :} & \text{Let, the tower be } AB \\
& \text{Initial position of Car = D} \\
& \text { Final position of the car = C} \\
& \text{Angles of depression are measured from A.}\\
& \text {Distance between the foot of the tower to the car = BC} \\
\text {Solution :} & \text{In } \triangle ABC:\\
& \tan 60^\circ = \frac{AB}{BC}\\
& \text{or, } \sqrt{3} = \frac{AB}{BC}\\
& \text{or, } BC = \frac{AB}{\sqrt{3}}....... (i) \\
& \text{Also, In } \triangle ABD:\\
& \tan 30^\circ = \frac{AB}{BD}\\
& \text{or, } \quad \frac{1}{\sqrt{3}} = \frac{AB}{BC + CD} \\
& \text{or, } \quad AB\sqrt{3} = BC + CD \\
& \text{or, } \quad CD = AB\sqrt{3}-BC\\
& \text {Putting BC value from equation (i)}\\
& \text{or, } \quad CD = AB\sqrt{3}-\frac {AB}{\sqrt{3}} \\
& \text{or, } \quad CD = AB(\sqrt{3} - \frac{1}{\sqrt{3}}) \\
& \text{or, } \quad CD = \frac {2AB}{\sqrt{3}}\\
& \text{or, } \quad AB = \frac{\sqrt{3}CD}{2} ...... (ii)\\
& \text{Substitute this } AB \text{ value frome equ. (ii) to equ. (i) } :\\
& BC = \frac{\sqrt{3}CD}{2\sqrt{3}} \quad \text{or, } \quad BC = \frac{CD}{2} \\
& \text{Here, the distance of } BC \text{ is half of } CD.\\
& \text{Thus, the time taken is also half.} \\
& \text{Time taken by car to travel distance } CD \text{ is 6 seconds.} \\
& \text{Time taken by car to travel BC is } \frac{6}{2} = 3 \, \text{seconds}
\end{align*}
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