Kaliyuga Ekalavya

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 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. Given: The tree breaks at a certain point and the broken part touches the ground, making an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 8 m. To Find: The original height of the tree. Formula: The tangent of an angle in a right triangle is given by: $$ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$ Solution: Step 1: Let the tree break at a point such that the unbroken part is of height \( h \) meters and the broken part is of length \( x \) meters. The broken part touches the ground at a distance of 8 m from the foot of the tree, making...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Height of the boy = 1.5 m Height of the building = 30 m Initial angle of elevation = 30° Final angle of elevation = 60° To Find: The distance the boy walked towards the building. Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the initial distance from the boy to the building be \( x \) metres, and the final distance be \( y \) metres. The effective height from the boy's eyes to the top of the building is: \[ \text{Effective height} = 30\,\text{m} - 1.5\,\text{m} = 28.5\,\text{m} \] Step 2: Using the tangent ratio for the initial position (angle of elevati...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Height of the kite above ground = 60 m Inclination of string with ground = 60° To Find: Length of the string Formula: \[ \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \] Solution: Step 1: Let the length of the string be \( l \) meters. Step 2: In the right triangle formed, the height of the kite is the perpendicular, and the string is the hypotenuse. Using the sine ratio: $$ \sin 60^\circ = \frac{60}{l} $$ Step 3: We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). $$ \frac{\sqrt{3}}{2} = \frac{60}{l} $$ Step 4: Cross-multiplying to solve for \( l \): $$ l = \f...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? Given: Height of slide for children below 5 years, \( h_1 = 1.5\,\text{m} \) Angle of inclination for children below 5 years, \( \theta_1 = 30^\circ \) Height of slide for elder children, \( h_2 = 3\,\text{m} \) Angle of inclination for elder children, \( \theta_2 = 60^\circ \) To Find: Length of the slide for children below 5 years (\( l_1 \)) Length of the slide for elder children (\( l_2 \)) Formula: \[ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} \] ...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Angle of elevation, \( \theta = 30^\circ \) Distance from the point to the foot of the tower, \( = 30\,\text{m} \) To Find: Height of the tower (\( h \)) Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the height of the tower be \( h \) m. The distance from the point to the foot of the tower is 30 m. Step 2: Using the definition of tangent for the angle of elevation, $$ \tan 30^\circ = \frac{h}{30} $$ Step 3: We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). Substitute this value: $$ \frac{1}{\sqrt{3}} = \frac{h}{30} $$ Step 4: Cross-multiply to solve for \( h \): $$ h = 30 \times \frac{1}{\sqrt{3}} = \frac{30}{\sqrt{3}} $$ ...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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°. Find the distance travelled by the balloon during the interval. Given: Height of the girl = 1.2 m Height of the balloon from the ground = 88.2 m Height of the balloon above the girl's eyes = 88.2 - 1.2 = 87 m Initial angle of elevation (\( \theta_1 \)) = \( 60^\circ \) Final angle of elevation (\( \theta_2 \)) = \( 30^\circ \) To Find: Distance travelled by the balloon during the interval. Formula: The tangent of an angle in a right triangle is given by: $$ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$ Solution: Step 1: Let the initial horizontal distance from the girl to the balloon be \( x_1 ...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. Given: Length of the rope = 20 m Angle made by the rope with the ground = \(30^\circ\) To Find: Height of the pole Formula: \[ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} \] Solution: Step 1: Let the height of the pole be \( h \) meters. Step 2: In the right-angled triangle formed by the pole (vertical), rope (hypotenuse), and ground (base), apply the sine formula: \[ \sin 30^\circ = \frac{h}{20} \] Step 3: Substitute the value \(\sin 30^\circ = \frac{1}{2}\): \[ \frac{1}{2} = \frac{h}{20} \] Step 4: Solve for \( h \): \[ h = 20 \times \frac{1}{2} = 10 \] Result: The height of the pole is \( 10 \) m. Next question solu...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Height of the building = 7 m Angle of elevation to top of tower = 60° Angle of depression to foot of tower = 45° To Find: Height of the tower Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the height of the building be \( AB = 7\,\text{m} \). Let the height of the tower be \( CD = h\,\text{m} \). Let the horizontal distance between the building and the tower be \( x\,\text{m} \). Step 2: From the top of the building, the angle of depression to the foot of the tower is 45°. In right triangle \( ABE \): $$ \tan 45^\circ = \frac{AB}{BE} $$ $$ 1 = \frac{7}{x} \implies x = 7\,\text{m} $$ Step 3: The angle of elevation to...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Angle of elevation of the top of the building from the foot of the tower = \(30^\circ\) Angle of elevation of the top of the tower from the foot of the building = \(60^\circ\) Height of the tower = \(50\,\text{m}\) To Find: Height of the building. Formula: Trigonometric ratio: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the height of the building be \(h\) meters. Let the distance between the building and the tower be \(x\) meters. Step 2: From the foot of the tower, the angle of elevation to the top of the building is \(30^\circ\): \[ \tan 30^\circ = \frac{h}{x} \] \[ \frac{1}{\sqrt{3}} = \frac{h}...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Angle of depression at first observation = \(30^\circ\) Angle of depression at second observation = \(60^\circ\) Time between observations = 6 seconds The car moves with uniform speed To Find: Time taken by the car to reach the foot of the tower from the second observation point. Formula: \(\tan \theta = \dfrac{\text{Opposite side}}{\text{Adjacent side}}\) Uniform speed: \(\dfrac{\text{Distance}_1}{\text{Time}_1} = \dfrac{\text{Distance}_2}{\text{Time}_2}\) Solution: Step 1: Let the height of the tower be \( h \...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Height of lighthouse = 75 m Angle of depression of first ship = 45° Angle of depression of second ship = 30° Both ships are on the same side and one is exactly behind the other To Find: Distance between the two ships. Formula: The tangent of an angle in a right triangle is given by: $$ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$ Solution: Step 1: Let the distance of the first ship from the base of the lighthouse be \( x_1 \) and the distance of the second ship be \( x_2 \), where the second ship is farther from the lighthouse than the first. Step 2: For the first ship (angle of depression = 45°), ...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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 joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. Given: Let the height of the tower be \( h \) meters and the width of the canal be \( x \) meters. The angle of elevation from the point on the other bank directly opposite the tower is \( 60^\circ \). From another point 20 m farther on the same line, the angle of elevation is \( 30^\circ \). To Find: Height of the tower (\( h \)) and width of the canal (\( x \)). Formula: For a right triangle, \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the height of the tower b...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Width of the road = 80 m Angles of elevation of the tops of the poles from a point between them are \(60^\circ\) and \(30^\circ\) The poles are of equal height To Find: The height of each pole The distances of the point from each pole Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the height of each pole be \( h \) meters. Let the distances from the point to the two poles be \( x \) meters and \( y \) meters, respectively. The road is 80 m wide, so: \[ x + y = 80 \] Step 2: Using the angle of elevation \(60^\c...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Height of the statue = 1.6 m Angle of elevation of the top of the statue = 60° Angle of elevation of the top of the pedestal = 45° To Find: Height of the pedestal Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the height of the pedestal be \( h \) meters. The total height of the statue and pedestal is \( h + 1.6 \) meters. Let the distance from the point on the ground to the base of the pedestal be \( x \) meters. Step 2: Using the angle of elevation to the top of the pedestal (45°): \[ \tan 45^\circ = \frac{h}{x} \] \[ 1 = \frac{h}{x} \implies h = x \] Step 3: ...

NCERT Class X Chapter 9: Some Applications of Trigonometry Question: 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. Given: Angle of elevation of the bottom of the tower = \(45^\circ\) Angle of elevation of the top of the tower = \(60^\circ\) Height of the building = \(20\,\text{m}\) To Find: Height of the transmission tower (\(h\)) Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the distance from the observation point to the foot of the building be \(x\) meters. Step 2: Using the angle of elevation to the bottom of the tower (top of the building): \[ \tan 45^\circ = \frac{20}{x} \] Since \(\tan 45^\circ = 1\): \[ 1 = \frac{20}{x} \implies x = 20 \] Step 3: Let the height of the tower be \(h\) meters. The to...

NCERT Class X Chapter 9: Some Applications of Trigonometry Example 7 Question: From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river. Given: Angles of depression to the two banks: \(30^\circ\) and \(45^\circ\) Height of the bridge above the banks: \(3\,\text{m}\) To Find: The width of the river. Formula: \[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \] Solution: Step 1: Let the points where the lines of sight from the bridge meet the two banks be \(A\) and \(B\). Let \(P\) be the point on the bridge directly above the river at height \(3\,\text{m}\). Let the distances from \(P\) vertically down to \(A\) and \(B\) along the river banks be \(x\) and \(y\) respectively. The total width of the river is \(x + y\). Step 2: For the bank with angle of depression \(45^\circ\)...

NCERT Class X Chapter 9: Some Applications of Trigonometry Example 6 Question: The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings. Given: Angle of depression of the top of the building = \(30^\circ\) Angle of depression of the bottom of the building = \(45^\circ\) Height of the smaller building = \(8\,\text{m}\) To Find: Height of the multi-storeyed building Distance between the two buildings Formula: \[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \] Solution: Step 1: Let the height of the multi-storeyed building be \(h\) meters and the distance between the two buildings be \(x\) meters. Step 2: Consider the angle of depression to the bottom of the 8 m building (\(45^\circ\)). From the top of the multi-storeyed building, the vertica...