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...