RS Aggarwal Class 10 Chapter 14 Solutions (Heights and Distances)
RS Aggarwal Solutions for Class 10 Maths Chapter 14 is about heights and distances which is very important from CBSE Class 10 board exams point of view. It is advised that you solve most of the questions in this chapter of RS Aggarwal Class 10 Maths Solutions so as to get a good command on this topic of heights and distances which is actually an application of trigonometric ratios. The questions follow the CBSE guidelines for Class 10 Maths and have an adequate level of difficulty so that you can prepare well for the exams.
In this chapter, there are 2 sets of exercises with a total of 58 questions. The questions are a mix of short and long answer types along with some objective type questions. The problems make you learn the fundamentals of heights and distances, calculate the angle of elevation, depression along with some word problems on heights and distances. Problems usually involve at most two rightangle triangles and angle of elevation/depression are 30, 45 or 60 degrees in the problems.
Our unique explanation and the simple stepwise RS Aggarwal Solutions for Class 10 Maths for Chapter 14 will help you understand the concepts behind each of the problems. You will be able to build a strong knowledge of the basics which is fundamental to solving problems of any difficulty. Our academic team follows CBSE guidelines to solve the problems hence going through these solutions will make you aware of what is to be expected in your exams.
Important Topics for RS Aggarwal Solutions for Class 10 Chapter 14: Heights and Distances
Trigonometric ratios and Pythagoras theorems learned in the earlier chapter are used here to find the distance between two or more than two points or to find the height of an object or the angle that an object subtends at a point. Here we will take a look at some of the important definitions and formulas which are used to solve these problems.
 Line of Sight – If a line is drawn from the eye of the observer to the point on the object which is being viewed by the observer, this line forms the line of sight.
 Horizontal Line – It is the horizontal line drawn from the observer to the object.
 Angle of Elevation – Angle of elevation is relevant and can be measured if the object is above the horizontal line. If O is the observer’s eye, OA is the horizontal line drawn from the observer’s eye to the object and P is the position of the object (which is above the horizontal line) then the angle ∠AOP is the angle of elevation. It is called an angle of elevation since the observer needs to raise (or elevate) his eyes from the horizontal level to the line of sight to look at the object.
 Angle of Depression – The angle of depression is relevant and can be measured if the object is below the horizontal line. If O is the observer’s eye, OA is the horizontal line drawn from the observer’s eye to the object and P is the position of the object (which is below the horizontal line) then the angle ∠AOP is the angle of depression. It is called an angle of depression since the observer needs to lower (or depress) his eyes from the horizontal level to the line of sight to look at the object.
 Measuring Heights and Distances – The directions of the objects are described by angles of elevation and angle of depression. Trigonometric ratios (Sin , Cos , Tan , Cosec , Sec , Cot) are used to measure heights and distances. An instrument called Theodolite is used to measure the angle of elevation and depression. This instrument is based on the principles used in trigonometry.
 Hypotenuse – The side opposite to 90 degrees in a rightangles triangle is called the hypotenuse
 Heights and Distance Formula
Here, is the reference angle (a reference angle is any of the other 2 angles which are not 90 degrees). “Opposite” refers to the side of the triangle opposite to this reference angle and “adjacent” refers to the side of the triangle which is adjacent to the reference angle.

 Angle Bisector Theorem – In any triangle ABC, if we bisect angle A such that the bisecting line falls on opposite side BC at point D, then
BD/DC = AB/BC. An angle bisector is a line that divides angle A into 2 equal halves.
 Pythagoras theorem
 Cos^{2}θ+ Sin^{2}θ = 1
 Cot^{2}θ +1 = Cosec^{2}θ
 Trigonometric Quotient formula
 Tan θ= Sin θ/Cos θ
 Cot θ= Cos θ/ Sin θ
 Trigonometric Reciprocal formula
 Cosec θ = 1/Sin θ
 Sec θ=1/ Cos θ
 Cot θ=1/ Tan θ
 Some Important Values:

 2π radians = 360 degrees
 √2 = .1414
 √3 = 1.732
 √5 = 2.236

 Trigonometric Ratios of a few specific angles :
T Ratio  0º  30º  45º  60º  90º 
Sin  0  ½  1/√2  √3/2  1 
Cos  1  √3/2  1/√2  ½  0 
Tan  0  1/√3  1  √3  U 
Cosec  U  2  √2  2/√3  1 
Sec  1  2/√3  √2  2  U 
Cot  U  √3  1  1/√3  0 
 Few important tips and points to remember while solving heights and distances problems:

 If the height of the observer is unknown then the observer is represented as a point
 An object is either represented as a line segment or a point if the height Is not known
 The angle of elevation and depression are always acute angles.
 The angle of elevation increases as the observer moves towards the object and decreases as s/he moves away from the object.
 The length of the shadow of the object is dependent on the angle of elevation of Sun if that angle decreases then the shadow length increases and vice versa.
Exercise Discussions of RS Aggarwal Solutions for Class 10 Chapter 14: Heights and Distances
Exercise 14
The exercise has 33 questions that require short or long answers in finding the height of objects like a tower or a kite, given the angle of elevation. In some problems, you need to find the distance between objects given their angles of elevation from a common point. There are problems where the angle of depression of an observer from a tower is given and one needs to find the distance between objects on the ground.
MultipleChoice Questions
After the exercise, there are 25 multiplechoice questions. These mostly require small steps to derive at the correct answer. They are based on the concepts of the length of the shadow of an object, angle of elevation and depression and applying trigonometric ratios.
Benefits of RS Aggarwal Solutions for Class 10 Chapter 14: Heights and Distances
 The RS Aggarwal Solutions for Chapter 14 include accurate, stepwise and easy to understand solutions for the chapter.
 These solutions will act as a guide and will help you manage your time much better during the stressful Class 10 boards exams.
 The diagrammatic approach towards problems, wherever required, will give a clear picture to you to understand the solution better.
 The solutions are free of cost so that you never have a lack of quality resources at hand for your revision or exam preparation.