Given:

• Length of the rope ($AC$) = $20$ m.
• Angle made by the rope with the ground level ($\angle C$) = $30^\circ$.
• The pole is vertical, meaning the angle between the pole and the ground is $90^\circ$ ($\angle B = 90^\circ$).

To Find:

• The height of the pole ($AB$).

Visual Representation:

Step 1: Defining the Variables and Trigonometric Relationship

Let $AB$ be the height of the vertical pole. Let $AC$ be the length of the rope. In the right-angled triangle $\triangle ABC$, where $\angle B = 90^\circ$:

• The side opposite to the angle $\angle C$ is $AB$ (the height of the pole).
• The side opposite to the right angle $\angle B$ is $AC$ (the hypotenuse, which is the rope).

We use the trigonometric ratio that relates the opposite side and the hypotenuse, which is the sine function:

$\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$

Step 2: Setting up the Equation

Substituting the known values into the sine formula:

$\sin(30^\circ) = \frac{AB}{AC}$

Given that $AC = 20$ m and $\sin(30^\circ) = \frac{1}{2}$ [Standard trigonometric value]:

$\frac{1}{2} = \frac{AB}{20}$

Step 3: Solving for $AB$

To isolate $AB$, multiply both sides of the equation by $20$:

$AB = 20 \times \frac{1}{2}$

$AB = 10$

Step 4: Conclusion

The height of the pole is $10$ meters.

Final Answer: The height of the pole is 10 m.

Given:

• Height of the building ($AB$) = $30$ m.
• Height of the boy ($ED = FC$) = $1.5$ m.
• Initial angle of elevation ($\angle AEF$) = $30^\circ$.
• Final angle of elevation ($\angle ADF$) = $60^\circ$.

To find:

The distance the boy walked towards the building, which is the distance $EF$.

Step 1: Determine the height of the triangle above the boy's eye level.

Let the building be $AB$ and the boy's positions be $EF$ and $DC$. The height of the building $AB = 30$ m. Since the boy is $1.5$ m tall, the height of the triangle $AF$ (where $F$ is the point on $AB$ at the eye level of the boy) is:

$AF = AB - FB = 30 - 1.5 = 28.5$ m.

Step 2: Define the variables for the horizontal distances.

Let $DF = x$ and $EF = y$. We need to find $x$.

Step 3: Apply trigonometry in $\triangle ADF$.

In the right-angled triangle $\triangle ADF$, the angle of elevation is $60^\circ$.

Using the tangent ratio: $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$

$\tan(60^\circ) = \frac{AF}{DF}$

Since $\tan(60^\circ) = \sqrt{3}$:

$\sqrt{3} = \frac{28.5}{x}$

$x = \frac{28.5}{\sqrt{3}}$

Rationalizing the denominator: $x = \frac{28.5 \times \sqrt{3}}{3} = 9.5\sqrt{3}$ m.

Step 4: Apply trigonometry in $\triangle AEF$.

In the right-angled triangle $\triangle AEF$, the angle of elevation is $30^\circ$.

$\tan(30^\circ) = \frac{AF}{EF} = \frac{AF}{ED + DF}$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$:

$\frac{1}{\sqrt{3}} = \frac{28.5}{y + x}$

$y + x = 28.5\sqrt{3}$

Step 5: Calculate the distance walked ($y$).

$y = 28.5\sqrt{3} - x$

$y = 28.5\sqrt{3} - 9.5\sqrt{3}$

$y = 19\sqrt{3}$ m.

Final Answer: The distance the boy walked towards the building is $19\sqrt{3}$ m.

Given:

1. The height of the kite above the ground ($AB$) = $60$ m.

2. The angle of inclination of the string with the ground ($\angle ACB$) = $60^\circ$.

3. The string is assumed to be taut (no slack), forming a straight line.

To Find:

The length of the string ($AC$).

Step 1: Defining the Geometric Model

Let the position of the kite be point $A$, the point on the ground directly below the kite be $B$, and the point where the string is tied to the ground be $C$.

This forms a right-angled triangle $\triangle ABC$, where:

• $\angle B = 90^\circ$ (The height is measured perpendicular to the ground).
• $AB = 60$ m (The side opposite to $\angle C$).
• $\angle C = 60^\circ$ (The angle of inclination).
• $AC$ is the length of the string (the hypotenuse of the triangle).

Step 2: Selecting the Trigonometric Ratio

In a right-angled triangle, the relationship between the side opposite to an angle and the hypotenuse is given by the sine function:

$\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$

Applying this to $\triangle ABC$:

$\sin(60^\circ) = \frac{AB}{AC}$

Step 3: Substituting Known Values

We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$ [Standard trigonometric value].

Substituting the values into the equation:

$\frac{\sqrt{3}}{2} = \frac{60}{AC}$

Step 4: Solving for $AC$

To isolate $AC$, we perform cross-multiplication:

$\sqrt{3} \times AC = 60 \times 2$

$\sqrt{3} \times AC = 120$

$AC = \frac{120}{\sqrt{3}}$

Step 5: Rationalizing the Denominator

To simplify the expression, multiply the numerator and denominator by $\sqrt{3}$:

$AC = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$AC = \frac{120\sqrt{3}}{3}$

$AC = 40\sqrt{3}$

Step 6: Final Calculation

Using the approximation $\sqrt{3} \approx 1.732$:

$AC = 40 \times 1.732 = 69.28$ m

Final Answer: The length of the string is $40\sqrt{3}$ m or approximately $69.28$ m.

Given:

1. A tower $AB$ of height $h$ stands vertically on a horizontal plane.

2. A car is moving towards the foot of the tower ($B$) along a straight highway.

3. At point $C$, the angle of depression from the top of the tower ($A$) is $30^\circ$.

4. After $6$ seconds, the car reaches point $D$, where the angle of depression is $60^\circ$.

To Find:

The time taken by the car to travel from point $D$ to the foot of the tower $B$.

Step 1: Define Variables and Geometric Relationships

Let $AB = h$ be the height of the tower. Let the speed of the car be $v$ (units/sec).

Distance covered by the car from $C$ to $D$ in $6$ seconds is $CD = v \times 6 = 6v$.

Let the time taken to travel from $D$ to $B$ be $t$ seconds. Then, $DB = v \times t = vt$.

Step 2: Apply Trigonometric Ratios in Right-Angled Triangles

In $\triangle ABD$ (where $\angle ADB = 60^\circ$):

$\tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{DB}$

Since $\tan(60^\circ) = \sqrt{3}$, we have:

$\sqrt{3} = \frac{h}{vt} \implies h = vt\sqrt{3}$ --- (Equation 1)

In $\triangle ABC$ (where $\angle ACB = 30^\circ$):

$\tan(30^\circ) = \frac{AB}{CB} = \frac{AB}{CD + DB}$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we have:

$\frac{1}{\sqrt{3}} = \frac{h}{6v + vt}$

$h\sqrt{3} = 6v + vt$ --- (Equation 2)

Step 3: Solve the System of Equations

Substitute the value of $h$ from Equation 1 into Equation 2:

$(vt\sqrt{3})\sqrt{3} = 6v + vt$

$vt(3) = 6v + vt$

$3vt = 6v + vt$

Subtract $vt$ from both sides:

$2vt = 6v$

Divide both sides by $2v$ (assuming $v \neq 0$):

$t = \frac{6v}{2v}$

$t = 3$

Conclusion:

The time taken by the car to travel from point $D$ to the foot of the tower $B$ is $3$ seconds.

Final Answer: 3 seconds

Given:

• Two poles of equal height, let this height be $h$ meters.
• The width of the road is $80$ m.
• A point $P$ is located on the road between the two poles.
• The angle of elevation from point $P$ to the top of the first pole is $60^\circ$.
• The angle of elevation from point $P$ to the top of the second pole is $30^\circ$.

To Find:

• The height of the poles ($h$).
• The distance of point $P$ from each pole.

Step 1: Define variables and set up the geometry.

Let the two poles be $AC$ and $BD$ such that $AC = BD = h$.
Let the width of the road $CD = 80$ m.
Let $P$ be a point on $CD$ such that $CP = x$ meters. Then, $PD = (80 - x)$ meters.

Step 2: Apply trigonometric ratios in the right-angled triangles.

In $\triangle ACP$, using the tangent ratio $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$:

$\tan(60^\circ) = \frac{AC}{CP}$

$\sqrt{3} = \frac{h}{x}$

$h = x\sqrt{3}$ --- (Equation 1)

In $\triangle BDP$:

$\tan(30^\circ) = \frac{BD}{PD}$

$\frac{1}{\sqrt{3}} = \frac{h}{80 - x}$

$h = \frac{80 - x}{\sqrt{3}}$ --- (Equation 2)

Step 3: Solve for $x$.

Equating Equation 1 and Equation 2:

$x\sqrt{3} = \frac{80 - x}{\sqrt{3}}$

Multiply both sides by $\sqrt{3}$:

$3x = 80 - x$

$3x + x = 80$

$4x = 80$

$x = 20$ m

Step 4: Calculate the height $h$ and the remaining distance.

Substitute $x = 20$ into Equation 1:

$h = 20\sqrt{3}$ m

The distance of point $P$ from the first pole is $CP = x = 20$ m.

The distance of point $P$ from the second pole is $PD = 80 - x = 80 - 20 = 60$ m.

Final Answer: The height of the poles is $20\sqrt{3}$ m (approx. $34.64$ m), and the distances of the point from the poles are $20$ m and $60$ m respectively.

Given:

• Height of the tower ($CD$) = $50$ m.
• Angle of elevation of the top of the building ($AB$) from the foot of the tower ($D$) = $30^\circ$.
• Angle of elevation of the top of the tower ($CD$) from the foot of the building ($B$) = $60^\circ$.

To find:

The height of the building ($AB = h$).

Step 1: Define variables and identify triangles.

Let the height of the building $AB = h$ meters.

Let the distance between the foot of the building ($B$) and the foot of the tower ($D$) be $BD = x$ meters.

We have two right-angled triangles: $\triangle ABD$ and $\triangle CDB$.

Step 2: Analyze $\triangle CDB$ to find the distance $x$.

In $\triangle CDB$, the angle of elevation is $\angle CBD = 60^\circ$.

Using the trigonometric ratio tangent: $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$

$\tan(60^\circ) = \frac{CD}{BD}$

Since $\tan(60^\circ) = \sqrt{3}$ [Trigonometric table value]:

$\sqrt{3} = \frac{50}{x}$

$x = \frac{50}{\sqrt{3}}$ --- (Equation 1)

Step 3: Analyze $\triangle ABD$ to find the height $h$.

In $\triangle ABD$, the angle of elevation is $\angle ADB = 30^\circ$.

Using the trigonometric ratio tangent:

$\tan(30^\circ) = \frac{AB}{BD}$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ [Trigonometric table value]:

$\frac{1}{\sqrt{3}} = \frac{h}{x}$

$h = \frac{x}{\sqrt{3}}$ --- (Equation 2)

Step 4: Substitute Equation 1 into Equation 2.

$h = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}}$

$h = \frac{50}{\sqrt{3} \times \sqrt{3}}$

$h = \frac{50}{3}$

$h = 16.666...$ meters

Final Answer: The height of the building is $\frac{50}{3}$ m or approximately $16.67$ m.

Given:

Case 1 (Children below 5 years):

• Height of the slide ($AB$) = $1.5$ m
• Angle of inclination ($\angle ACB$) = $30^\circ$

Case 2 (Elder children):

• Height of the slide ($PQ$) = $3$ m
• Angle of inclination ($\angle PRQ$) = $60^\circ$

To find:

The length of the slide for each case (i.e., the length of the hypotenuse $AC$ and $PR$).

Step 1: Calculating the length of the slide for children below 5 years.

In the right-angled triangle $\triangle ABC$, where $\angle B = 90^\circ$:

We use the trigonometric ratio that relates the opposite side to the hypotenuse, which is the sine function:

$\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$

Substituting the given values:

$\sin(30^\circ) = \frac{AB}{AC}$

Since $\sin(30^\circ) = \frac{1}{2}$ [Standard trigonometric value]:

$\frac{1}{2} = \frac{1.5}{AC}$

By cross-multiplication:

$AC = 1.5 \times 2$

$AC = 3$ m

Step 2: Calculating the length of the slide for elder children.

In the right-angled triangle $\triangle PQR$, where $\angle Q = 90^\circ$:

Using the sine function again:

$\sin(60^\circ) = \frac{PQ}{PR}$

Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$ [Standard trigonometric value]:

$\frac{\sqrt{3}}{2} = \frac{3}{PR}$

By cross-multiplication:

$PR \times \sqrt{3} = 3 \times 2$

$PR \times \sqrt{3} = 6$

$PR = \frac{6}{\sqrt{3}}$

Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

$PR = \frac{6 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$PR = \frac{6\sqrt{3}}{3}$

$PR = 2\sqrt{3}$ m

Using the approximation $\sqrt{3} \approx 1.732$:

$PR \approx 2 \times 1.732 = 3.464$ m

Final Answer: The length of the slide for children below 5 years is $3$ m, and the length of the slide for elder children is $2\sqrt{3}$ m (approximately $3.46$ m).

Given:

• Height of the statue ($AB$) = $1.6$ m.
• The statue stands on a pedestal ($BC$).
• Angle of elevation from a point $D$ on the ground to the top of the statue ($A$) is $\angle ADB = 60^\circ$.
• Angle of elevation from the same point $D$ to the top of the pedestal ($B$) is $\angle CDB = 45^\circ$.

To Find:

The height of the pedestal ($BC = h$ meters).

Step 1: Define Variables and Assumptions

Let $BC = h$ be the height of the pedestal in meters.

Let $CD = x$ be the distance from the point $D$ to the base of the pedestal $C$ in meters.

The total height of the statue and pedestal is $AC = AB + BC = 1.6 + h$.

Step 2: Analyze the smaller triangle $\triangle BCD$

In the right-angled triangle $\triangle BCD$, the angle $\angle CDB = 45^\circ$.

Using the trigonometric ratio $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$:

$\tan(45^\circ) = \frac{BC}{CD}$

Since $\tan(45^\circ) = 1$:

$1 = \frac{h}{x} \implies x = h$ --- (Equation 1)

Step 3: Analyze the larger triangle $\triangle ACD$

In the right-angled triangle $\triangle ACD$, the angle $\angle ADB = 60^\circ$.

Using the trigonometric ratio $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$:

$\tan(60^\circ) = \frac{AC}{CD}$

Substitute $AC = 1.6 + h$ and $CD = x$:

$\sqrt{3} = \frac{1.6 + h}{x}$ --- (Equation 2)

Step 4: Solve for $h$

Substitute $x = h$ (from Equation 1) into Equation 2:

$\sqrt{3} = \frac{1.6 + h}{h}$

Multiply both sides by $h$:

$h\sqrt{3} = 1.6 + h$

Rearrange the terms to isolate $h$:

$h\sqrt{3} - h = 1.6$

$h(\sqrt{3} - 1) = 1.6$

$h = \frac{1.6}{\sqrt{3} - 1}$

Step 5: Rationalize the denominator

To simplify, multiply the numerator and denominator by the conjugate $(\sqrt{3} + 1)$:

$h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$

Using the identity $(a-b)(a+b) = a^2 - b^2$:

$h = \frac{1.6(\sqrt{3} + 1)}{3 - 1}$

$h = \frac{1.6(\sqrt{3} + 1)}{2}$

$h = 0.8(\sqrt{3} + 1)$

Using $\sqrt{3} \approx 1.732$:

$h = 0.8(1.732 + 1) = 0.8(2.732) = 2.1856$ m.

Final Answer: The height of the pedestal is $0.8(\sqrt{3} + 1)$ m or approximately $2.186$ m.

Given:

• The distance from the foot of the tower to the point on the ground ($BC$) = $30$ m.
• The angle of elevation of the top of the tower from the point on the ground ($\angle ACB$) = $30^\circ$.

To Find:

• The height of the tower ($AB$).

Step 1: Defining the Variables and Assumptions

Let $AB$ be the tower of height $h$ meters. Let $C$ be the point on the ground at a distance of $30$ m from the foot of the tower $B$. Thus, $BC = 30$ m. The angle of elevation $\angle ACB = 30^\circ$. We assume the tower stands vertically on the ground, making $\triangle ABC$ a right-angled triangle at $B$.

Step 2: Selecting the Trigonometric Ratio

In the right-angled triangle $\triangle ABC$, we have the side adjacent to the angle $\angle C$ (which is $BC$) and we need to find the side opposite to the angle $\angle C$ (which is $AB$). The trigonometric ratio that relates the opposite side and the adjacent side is the tangent function:

$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$

Step 3: Substituting the Values

Substitute $\theta = 30^\circ$, $BC = 30$ m, and $AB = h$ into the formula:

$\tan(30^\circ) = \frac{h}{30}$

Step 4: Solving for $h$

We know from trigonometric standard values that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

$\frac{1}{\sqrt{3}} = \frac{h}{30}$

Multiply both sides by $30$ to isolate $h$:

$h = \frac{30}{\sqrt{3}}$

Step 5: Rationalizing the Denominator

To simplify the expression, multiply the numerator and the denominator by $\sqrt{3}$:

$h = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$h = \frac{30\sqrt{3}}{3}$

$h = 10\sqrt{3}$

Step 6: Final Calculation

Using the approximate value of $\sqrt{3} \approx 1.732$:

$h = 10 \times 1.732 = 17.32$ m

Final Answer: The height of the tower is $10\sqrt{3}$ m or approximately $17.32$ m.

Given:

• Height of the building ($BC$) = $20$ m.
• The transmission tower ($CD$) is fixed on top of the building.
• Angle of elevation from a point $A$ on the ground to the bottom of the tower (point $C$) = $45^\circ$.
• Angle of elevation from point $A$ on the ground to the top of the tower (point $D$) = $60^\circ$.

To find:

The height of the transmission tower ($h = CD$).

Step 1: Define variables and identify triangles.

Let $AB = x$ be the distance from the point on the ground to the base of the building. Let $CD = h$ be the height of the tower. The total height of the building and tower is $BD = BC + CD = 20 + h$.

Step 2: Analyze $\triangle ABC$ (Right-angled at $B$).

In $\triangle ABC$, the angle of elevation $\angle BAC = 45^\circ$.

Using the trigonometric ratio: $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$

$\tan(45^\circ) = \frac{BC}{AB}$

Since $\tan(45^\circ) = 1$ [Trigonometric table value]:

$1 = \frac{20}{x}$

$x = 20$ m [Equation 1]

Step 3: Analyze $\triangle ABD$ (Right-angled at $B$).

In $\triangle ABD$, the angle of elevation $\angle BAD = 60^\circ$.

$\tan(60^\circ) = \frac{BD}{AB}$

Since $\tan(60^\circ) = \sqrt{3}$ [Trigonometric table value]:

$\sqrt{3} = \frac{20 + h}{x}$

Step 4: Solve for $h$.

Substitute $x = 20$ from Equation 1 into the equation from Step 3:

$\sqrt{3} = \frac{20 + h}{20}$

$20\sqrt{3} = 20 + h$ [Multiplying both sides by 20]

$h = 20\sqrt{3} - 20$

$h = 20(\sqrt{3} - 1)$

Using $\sqrt{3} \approx 1.732$:

$h = 20(1.732 - 1)$

$h = 20(0.732)$

$h = 14.64$ m

Final Answer: The height of the transmission tower is $20(\sqrt{3} - 1)$ m or approximately $14.64$ m.

Given:

• Height of the girl ($h_g$) = $1.2$ m.
• Height of the balloon from the ground ($H$) = $88.2$ m.
• Initial angle of elevation ($\theta_1$) = $60^\circ$.
• Final angle of elevation ($\theta_2$) = $30^\circ$.

To Find:

The distance travelled by the balloon during the interval, let this be $d$.

Step 1: Determine the effective height of the balloon from the girl's eye level.

Since the girl is $1.2$ m tall, the height of the balloon from her eye level ($h$) is the total height minus the girl's height.

$h = 88.2\text{ m} - 1.2\text{ m} = 87\text{ m}$.

Step 2: Analyze the first position of the balloon (Triangle 1).

Let the first position of the balloon be $C$ and the girl's eye position be $E$. Let the point on the ground directly below the balloon be $F$.

In the right-angled triangle formed by the eye level, the balloon, and the vertical line:

$\tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{x_1}$

$\sqrt{3} = \frac{87}{x_1}$

$x_1 = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3}\text{ m}$.

Step 3: Analyze the second position of the balloon (Triangle 2).

Let the second position of the balloon be $D$.

In the right-angled triangle formed by the eye level, the new balloon position, and the vertical line:

$\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{x_2}$

$\frac{1}{\sqrt{3}} = \frac{87}{x_2}$

$x_2 = 87\sqrt{3}\text{ m}$.

Step 4: Calculate the distance travelled by the balloon.

The distance travelled ($d$) is the difference between the two horizontal distances $x_2$ and $x_1$.

$d = x_2 - x_1$

$d = 87\sqrt{3} - 29\sqrt{3}$

$d = (87 - 29)\sqrt{3}$

$d = 58\sqrt{3}\text{ m}$.

Using the approximation $\sqrt{3} \approx 1.732$:

$d = 58 \times 1.732 = 100.456\text{ m}$.

Final Answer: The distance travelled by the balloon is $58\sqrt{3}$ m (or approximately $100.46$ m).

Given:

• Height of the lighthouse ($AB$) = $75$ m.
• Angle of depression of the first ship ($C$) = $45^\circ$.
• Angle of depression of the second ship ($D$) = $30^\circ$.
• The ships are on the same side of the lighthouse and lie on the same straight line.

To Find:

The distance between the two ships, i.e., the length of segment $CD$.

Step 1: Defining Variables and Geometric Relationships

Let $AB$ be the lighthouse of height $75$ m. Let $C$ and $D$ be the positions of the two ships. Since the angles of depression are $45^\circ$ and $30^\circ$, the angles of elevation from the ships to the top of the lighthouse are also $45^\circ$ and $30^\circ$ respectively [By the property of alternate interior angles].

Let $BC = x$ and $BD = y$. The distance between the ships is $CD = y - x$.

Step 2: Calculating distance $BC$ in $\triangle ABC$

In right-angled triangle $\triangle ABC$:

$\tan(45^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$

We know $\tan(45^\circ) = 1$ and $AB = 75$ m.

$1 = \frac{75}{x}$

$x = 75$ m

Step 3: Calculating distance $BD$ in $\triangle ABD$

In right-angled triangle $\triangle ABD$:

$\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BD}$

We know $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ and $AB = 75$ m.

$\frac{1}{\sqrt{3}} = \frac{75}{y}$

$y = 75\sqrt{3}$ m

Step 4: Finding the distance between the ships ($CD$)

The distance between the two ships is $CD = BD - BC$.

$CD = y - x$

$CD = 75\sqrt{3} - 75$

$CD = 75(\sqrt{3} - 1)$

Using the approximation $\sqrt{3} \approx 1.732$:

$CD = 75(1.732 - 1)$

$CD = 75(0.732)$

$CD = 54.9$ m

Final Answer: The distance between the two ships is $75(\sqrt{3} - 1)$ m or approximately $54.9$ m.

Given:

1. Height of the building ($AB$) = $7$ m.

2. Angle of elevation of the top of the tower ($EC$) from the top of the building ($A$) = $60^\circ$.

3. Angle of depression of the foot of the tower ($D$) from the top of the building ($A$) = $45^\circ$.

To find:

The total height of the cable tower ($CD$).

Step 1: Define the variables and geometric relationships.

Let $AB$ be the building of height $7$ m. Let $CD$ be the cable tower. Let $A$ be the top of the building and $B$ be its foot. Let $C$ be the top of the tower and $D$ be its foot. Let $E$ be a point on $CD$ such that $AE$ is horizontal. Thus, $AE = BD$ and $ED = AB = 7$ m.

Step 2: Analyze triangle $\triangle ABD$.

In the right-angled triangle $\triangle ABD$, the angle of depression from $A$ to $D$ is $45^\circ$. Therefore, the angle of elevation from $D$ to $A$ is also $45^\circ$ (alternate interior angles).

Using the trigonometric ratio tangent:

$\tan(45^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BD}$

Since $\tan(45^\circ) = 1$:

$1 = \frac{7}{BD}$

$BD = 7$ m

[Since $AE$ is parallel to $BD$, $AE = BD = 7$ m].

Step 3: Analyze triangle $\triangle AEC$.

In the right-angled triangle $\triangle AEC$, the angle of elevation is $60^\circ$.

$\tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CE}{AE}$

We know $\tan(60^\circ) = \sqrt{3}$ and $AE = 7$ m:

$\sqrt{3} = \frac{CE}{7}$

$CE = 7\sqrt{3}$ m

Step 4: Calculate the total height of the tower.

The total height of the tower $CD = CE + ED$.

We know $ED = AB = 7$ m.

$CD = 7\sqrt{3} + 7$

$CD = 7(\sqrt{3} + 1)$ m

Using the approximation $\sqrt{3} \approx 1.732$:

$CD \approx 7(1.732 + 1) = 7(2.732) = 19.124$ m.

Final Answer: The height of the tower is $7(\sqrt{3} + 1)$ m or approximately $19.12$ m.

Given:

A TV tower $AB$ stands vertically on the bank of a canal. Let $BC$ be the width of the canal. From a point $C$ on the other bank, the angle of elevation of the top of the tower $A$ is $60^\circ$. From another point $D$, which is $20\text{ m}$ away from $C$ on the line joining $C$ to the foot of the tower $B$, the angle of elevation of the top of the tower $A$ is $30^\circ$.

To Find:

1. The height of the tower ($h = AB$).

2. The width of the canal ($x = BC$).

Step 1: Defining Variables and Assumptions

Let the height of the tower $AB = h$ meters.

Let the width of the canal $BC = x$ meters.

The distance $BD = BC + CD = x + 20$ meters.

Step 2: Analyzing Triangle ABC

In the right-angled triangle $\triangle ABC$, the angle of elevation at $C$ is $60^\circ$.

Using the trigonometric ratio $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$:

$\tan(60^\circ) = \frac{AB}{BC}$

Since $\tan(60^\circ) = \sqrt{3}$:

$\sqrt{3} = \frac{h}{x}$

$h = x\sqrt{3}$ --- (Equation 1)

Step 3: Analyzing Triangle ABD

In the right-angled triangle $\triangle ABD$, the angle of elevation at $D$ is $30^\circ$.

Using the trigonometric ratio $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$:

$\tan(30^\circ) = \frac{AB}{BD}$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ and $BD = x + 20$:

$\frac{1}{\sqrt{3}} = \frac{h}{x + 20}$

$x + 20 = h\sqrt{3}$ --- (Equation 2)

Step 4: Solving the System of Equations

Substitute the value of $h$ from Equation 1 into Equation 2:

$x + 20 = (x\sqrt{3})\sqrt{3}$

$x + 20 = 3x$ [Since $\sqrt{3} \times \sqrt{3} = 3$]

$20 = 3x - x$

$20 = 2x$

$x = 10$

Thus, the width of the canal is $10\text{ m}$.

Step 5: Calculating the Height of the Tower

Substitute $x = 10$ back into Equation 1:

$h = 10\sqrt{3}$

Using $\sqrt{3} \approx 1.732$:

$h = 10 \times 1.732 = 17.32\text{ m}$

Final Answer: The height of the tower is $10\sqrt{3}\text{ m}$ (or approximately $17.32\text{ m}$) and the width of the canal is $10\text{ m}$.

Given:

• A tree breaks at a certain point, and the top touches the ground.
• The angle of elevation of the top of the tree from the point where it touches the ground is $\theta = 30^\circ$.
• The distance between the foot of the tree and the point where the top touches the ground is $BC = 8$ m.

To Find:

The total height of the tree ($H = AB + AC$, where $A$ is the top, $B$ is the foot, and $C$ is the point where the top touches the ground, with the break occurring at point $D$).

Step 1: Define the variables and model the triangle.

Let the original tree be $AD$. Let it break at point $D$. The part $AD$ bends such that the top $A$ touches the ground at point $C$. Let $B$ be the foot of the tree. Thus, $BD$ is the vertical part remaining standing, and $DC$ is the broken part that touches the ground. The total height of the tree is $H = BD + DC$.

In the right-angled triangle $\triangle DBC$ (right-angled at $B$):

• Base $BC = 8$ m
• Angle $\angle DCB = 30^\circ$

Step 2: Calculate the height of the standing part ($BD$).

Using the trigonometric ratio for tangent: $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$

$\tan(30^\circ) = \frac{BD}{BC}$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ [Standard trigonometric value]:

$\frac{1}{\sqrt{3}} = \frac{BD}{8}$

$BD = \frac{8}{\sqrt{3}}$ m

Step 3: Calculate the length of the broken part ($DC$).

Using the trigonometric ratio for cosine: $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

$\cos(30^\circ) = \frac{BC}{DC}$

Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ [Standard trigonometric value]:

$\frac{\sqrt{3}}{2} = \frac{8}{DC}$

$DC = \frac{16}{\sqrt{3}}$ m

Step 4: Calculate the total height of the tree ($H$).

$H = BD + DC$

$H = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}}$

$H = \frac{24}{\sqrt{3}}$

Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

$H = \frac{24 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3}$ m

Using $\sqrt{3} \approx 1.732$:

$H = 8 \times 1.732 = 13.856$ m

Final Answer: The height of the tree is $8\sqrt{3}$ m (or approximately $13.86$ m).