10th class is one of the most crucial stages in a student life. ‘The Central Board of Secondary Education’ (CBSE) is the most relevant board for engineering and medical aspirants today as both NEET (AIPMT) and the JEE Main exams are directly based on the CBSE board syllabus. It determines the future of a student to a very large extent. If a student gets good marks in 10th, then there is very much possibility that he will get good marks in 12th board too and so on.

Height and distance are a very important section under trigonometry of Quantitative Aptitude. Usually, all government competitive examinations ask questions based on Trigonometry. In general, the concept and questions based on trigonometry are very easy to solve but sometimes it becomes a little bit tricky in nature for its analytical nature.

In this article we are going to cover the key concepts of application of trigonometry particularly problems on height and distance, along with various types of questions, and tips and tricks. We have also added a few solved examples, and some unsolved questions which candidates will find beneficial in their exam preparation.

For Class 10 students who love Mathematics (and for those who don’t), here are some study tips that can help them in the CBSE Class X board exam or any other competitive examination perfectly to the full marks in solving questions on height and distance. This is an important topic from competitive examination point of view.

Heights and Distances is an application of Trigonometry. It helps us in solving complex real-life issues like calculating the distance between two or more objects, the height of mountains, etc. Heights and Distances are helpful in finding the distance, height, slope, etc., of the objects in the constructions. It is helpful in calculating the height of the towers, buildings, celestial bodies, etc. Furthermore, it is also helpful to calculate the angle of inclination of the object without the height and distance of a particular object.

So basically, you should know how to draw the diagram where to keep the angle and all. For this just practice more and more questions. Se firstly just practices to make diagrams and check solutions for the respective question. If your diagram is right then be happy lol. Once you know how to draw the triangle, you just have to practice questions on Trigonometry. It should be clear to you.

    Heights and Distances – Tricks

    There are a few things to be noted while solving the problems of heights and distances. Some of the important points are listed below:

     1. In the problems of heights and distances, if the height of the observer is not given, then the observer is denoted by a point.

    2. The height of the objects like towers, buildings, celestial bodies, etc., are represented by a line segment.

    3. A horizontal line is a line obtained by drawing parallel to the ground.

    1. Angle of elevation:   

    The term angle of elevation denotes the angle from the horizontal upward to an object. Angle of elevation implies that an object is upwards from the position of the observer. An observer’s line of sight would be above the horizontal. If the observer moves towards the object, then the angle of elevation increases.

    5. Angle of depression:  

    If a person looks straight ahead, their line of sight is on a horizontal plane at eye level. The angle of depression implies that an object is downwards from the position of the observer If that person looks to a point on the ground, they are now looking down at another horizontal plane, one that is below eye level. Their line of sight is now aimed at a downward angle. The angle between that line and their horizontal line of sight is the angle of depression. If the observer moves away from the object, then the angle of depression decreases.

    6.  Angle of elevation and angle of depression, so formed are (congruent) equal.

    7. An increase in the angle of elevation of the sun results in a decrease in the length of the shadow of an object.

    1. A decrease in the angle of elevation of the sun results in an increase in the length of the shadow of an object.

    Solution to questions on heights and distances are provided here with simple step-by-step explanations. These solutions for Heights and Distances will be extremely to Class 10 students. 

    Our objective is to provide you a smooth learning experience.

    While solving word problems understanding the question is very important. So:

    • Read the problem carefully.
    • Draw the problem with the sketch.
    • Heights and distance problems are solved by using trigonometric ratios therefore, refresh your trigonometric formulas.

    Also, always remember that paper presentation is an important aspect and you must attempt the paper in a way that not only make an impact on the checker but also makes him read your paper easily. The best thing about step wise solution is that you will get marks for all the correct steps even if your final solution is wrong.

    All these tips if followed will surely help you write a perfect exam within the allotted time. Ultimately, focus on giving your best without being worried about the result. Have confidence in your preparation and you will come with an overwhelming triumph.

    Remember:

    Math is not something that you can learn by heart. You have to practice it daily to master it. Hence, solve at least some Maths questions everyday – without fail. Good preparation along with exam paper writing techniques is necessary to score good marks in CBSE exams. Thus, two things are must: studies and answer writing practice.

    The proverb ‘Practice makes Perfect’ holds very true for Maths too.

    Practise makes a man perfect. Genius is one percent inspiration and ninety percent perspiration: This is a saying of the American inventor Thomas Edison.

    “One ounce of practise is worth a thousand pounds of theory. Practise makes us what we shall be.” Swami Vivekananda.

    In the Bhagavad Gita Arjuna asked the Lord, “how to control the mind which by its very nature is turbulent, obstinate and restless?” Shri Krishna’s simple reply is that only practise and non-attachment can help control the mind. 

    In this article we have solved some previous years question papers in an easy to understand and detailed way for your study and practice.

    Question 1: 

    An aeroplane is flying at a height of 300 m above the ground. Flying at the height 

    The angle of depression from the aeroplane of two points on both banks of the river in opposite directions are 45 deg. and 60 deg. respectively. Find the width of the river. [use √3 = 1.732]

    Solution:

    Let the aeroplane is at point P and A and B are the two points

    on both banks of the river. Based on the information the given figure has been drawn.

    Given:

    Height of the plane PM = 300 m

    Angle of depression from point P to A = XPA = 45 deg.

    Angle of depression from point P to B = YPB = 60 deg.

    To find: Width of the river i.e., length AB

    From the figure it is clear that XPA = MAP = 45 deg. [alternate angles]

    Similarly, YPB = MBP= 60 deg. [alternate angles]

    In right ΔPMA, tan MAP = PM/AM

    tan 450 = 300/AM

    1 = 300/AM

    AM =300 m …………………………… (i)

    Again, in right ΔPMB, tan MBP = PM/BM

    tan 600 = 300/MB

    √3 = 300/MB

    MB = 300/√3 …………………(i) 

    From equation (i) and (ii) we get,

    (AM + MB) = 300√3 + 300/√3

    AB = 300 + 100√3

    AB = 300 + 100*1.732

    AB = 300 + 173.2

    AB = 473.2 m

    Hence, width of the river = 473.2 m

    Question 2 :

    Two poles of equal height are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of one pole is 600 and the angle of depression from the top of another pole at P is 300. Find the height of each pole and the distance of point P from the poles.  [CBSE 2015, ‘18C, ‘19]

    Solution:

    Let, AB and CD are two poles standing on either side of 

    Road AC.

    Given:

    AC= width of the road = 80 m.

    Angel of elevation of point B from point P = BPA = 600

    Also, angle of depression from D to point P = XDP = 300

     Let, AB = CD = height of each pole = h m (say)

    Let AP = distance between A and P = x m (say)

    Then distance PC = (80 – x) m

    From the figure it is clear that XDP = CPD = 300          [alternate angles]

    From right ΔPAB, we have tan 600 =AB/AP

    √3 = AB/AP

    √3 = h/x

    h = √3 x …………………………………………(i)

    Again, from right ΔPCD, we have tan 300 =DC/PC

    1/√3 = DC/PC

    1/√3 = h/ (80 – x) m

    h = (80 – x)/ √3.…………………………(ii)

    Evaluating h from (i) and (ii) we get,

    √3 x = (80 – x)/ √3

    √3 x* √3 = (80 – x)

    3x + x = 80

    4x = 80

    x = 20 m

    Therefore, height of each pole = h = √3 x = 20√3 m

    Some practice questions:

    Question 1:

    As observed from the top of 100m high light house from the sea level the angle of depression of two ships are 300 and 450. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships.

    [CBSE 2018]

    Question 2:

    A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angel of depression of the car changes from 300 to 450 in 12 minutes, find the time taken by the car now to reach the tower. [CBSE 2017]

    Question 3:

    From a point on the ground, the angle of elevation of the top of a tower is observed to be 600. From a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is 300. Find the height of the tower and its horizontal distance from the point of observation.                 [CBSE 2016]

    Question 4:

    A man standing at the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 600 and the angle of depression of the base of hill as 300. Find the distance of the hill from the ship and the height of the hill. [CBSE 2017]

    Question 5:

    Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, find the distance between the two ships.

    Question 6:

    A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 45º. What is the distance between the base of the tower and the point P?

    Question 7:

    If the angle of elevation of cloud from a point ‘h’ meter above a lake is and the angle of depression of its reflection in the lake is 𝛃, prove that the height of the cloud is 

    h (tan 𝛃 + tan )/ (tan 𝛃 – tan )

    Question 8:

    A vertical pole consists of two parts, the lower part being one-third of the whole. At a point in the horizontal plane through the base of the pole and distance 20 meters from it, the upper part of the pole subtends an angle whose tangent is 1/2. What are the possible heights of the pole?

    Question 9:

    Determine the height of the mountain if the elevation of it’s top at an unknown distance from the base is 30°and at a distance10km further off from the mountain along the same line the angle of elevation is 15°(use tan 15°=0.27)

    We hope this article on heights and distances has provided significant value to your knowledge. If you have any queries or suggestions, feel free to write them down in the comment section below. We will love to hear from you.

    In the coming series of this article, we shall cover other aspects of the question pattern of maths and time management for solving them. Best of luck for good practice and better results.