Height And Distance- Trigonometry Basics With Question & Answer 

9 mins read

Regarding height and distance in trigonometry, we often use trigonometric ratios to solve problems involving measuring height and distance that are difficult to measure directly. Trigonometry provides us with the tools to calculate heights and distances accurately.

Height And Distance Important Terms

Important terms to understand when working with height and distance include the following definitions:

Line of Sight –  The line drawn from the observer's eyes to the point being viewed on the object is known as the line of sight.

Angle of Elevation: The angle of elevation is the angle between the line of sight and the horizontal level when looking up at a point on an object.

The angle of Depression – The angle of depression is the angle formed by the line of sight with the horizontal level when an observer views a point on an object below them.

Right Angled Triangle In Trigonometry

Trigonometry deals with angles and sides of triangles, especially right triangles. The primary trigonometric functions are sine (sin), cosine (cos), and tangent (tan). Understanding these functions helps in calculating unknown angles or sides in a triangle.

In the right triangle ABC,

sin θ = Opposite/Hypotenuse = AB/AC

cos θ = Adjacent/Hypotenuse = BC/AC

tan θ = Opposite/Adjacent = AB/BC

Formula For Height & Distance

The most important formula for calculating height and distance is the tangent formula. This formula involves using the tangent of an angle in a right triangle to determine the height or distance of an object. The formula is:

Tangent (θ) = Opposite / Adjacent

By rearranging this formula, you can solve for the height or distance depending on the information given. This formula is essential for various applications, such as surveying, navigation, and trigonometry.

Trigonometric Ratios

Here is a table of the most commonly used trigonometric ratios and degrees for height and distance:

Solved Examples For Better Understanding

Example 1: A person standing on a cliff spots a boat at an angle of elevation of 30 degrees. If the cliff is 50 meters high, how far is the boat from the base of the cliff?

Solution: Using the tangent ratio: tan(30) = opposite/adjacent, we find that the boat is approximately 28.87 meters away from the base of the cliff.

Example 2: A kite flying at an angle of elevation of 60 degrees has its string anchored to the ground. Given that the string is 100 meters long, find the height of the kite.

Solution: Applying sine: sin(60) = opposite/hypotenuse, we determine that the height of the kite is around 86.60 meters above ground level.

Click here to enhance and upskill your quantitative aptitude in trigonometry related to height and distance right away!

Height And Distance Applications In Trigonometry

Trigonometry's height and distance application covers various concepts like:

• Measuring tall towers and huge mountains.

• Determining how far the shore is from the sea.

• Calculating the space between two heavenly bodies.

Tips For Height & Distance

Here are some tips for you to tackle problems related to height and distance:

• Draw diagrams to visualize the problem accurately.

• Label components clearly to avoid confusion while applying trigonometric functions.

• Practice regularly to enhance your understanding and speed in solving such problems.

Height And Distance Practice Questions & Answers

Provided below are some of the selected questions related to height and distance with solved answers for practice to enhance your quantitative aptitude prowess:

Question 1. One side of a staircase is to be closed in by rectangular planks from the floor to each step. The width of each plank is 9 inches, and their height is successively 6 inches, 12 inches, 18 inches and so on. There are 24 planks required in total. Find the area in square feet.

(a) 112.5 

(b) 107  

(c) 118.5 

(d) 105

Solution: a) 112.5 

Explanation: The series of plank sizes would be: 

0.75 × 0.5, 0.75 × 1, 0.75 × 1.5….0.75 × 12. 

The sum of this AP is 112.5.

Question 2. The average height of 40 girls out of a class of 50 is 150 cm, and that of the remaining is 155 cm. The average height of the whole class is

(a) 151 cm 

(b) 152 cm  

(c) 156 cm 

(d) 153 cm

Solution: a) 151 cm 

Explanation: Average height of the whole class = (40 x 150 + 10 x 155)/50 = 151

Question 3. Raunak generally wears his father’s coat. Unfortunately, his cousin Vikas told him one day that he was wearing a coat of length more than his height by 15%. If the length of Raunak’s father’s coat is 345 cm, then find the actual length (in cm) of his coat.

(a) 110 

(b) 345  

(c) 300 

(d) 105

Solution: c) 300 

Explanation: Let Raunak’s height be H. 

Then, H X 1.15 = 345, 

H = 345/1.15 = 300.

Question 4. Increasing the height of an equilateral triangle by 30% poses a question: what is the maximum percentage increase in the base's length to limit the area increase to a maximum of 90%?

(a) 33.33% 

(b) 20.67% 

(c) 46.15% 

(d) 25.34%

Solution: c) 46.15% 

Explanation: The area of a triangle depends on the product: base x height. 

Since the height increases by 30% and the area has to increase by 90% overall, the following PCG will give the answer. Let 100 be the original area.

100(30% increase, + 30 effect of increase in base)= 130

130( effect of increase in height)=190 therefore , 60/130 X 100 = 46.15%

Question 5. When the height of an equilateral triangle measures 10 cm, what will its area be?

  (a) 100√3 cm²   

(b) 100/3√3 cm² 

(c) 100/3 cm²  

(d) 200√3/3 cm²

Solution: b) 100/3√3 cm² 

Explanation: h = 10 cm

H = a√3/2⇒ a= 10 X 2/√3 = 20/√3cm

Area = 1/2 X 20/√3 X 10 = 100/√3cm² or 100√3/3cm²

Question 6. The volume of a right circular cone is 100p cm3, and its height is 12 cm. Find its curved surface area. 

(a) 130 π cm2 

(b) 65 π cm2 

(c) 204 π cm2 

(d) 65 π cm2 

Solution: b) 65 π cm2

Explanation: Volume of a cone = πr²h/3

Then 100 π= πr²12/3

⇒r=5 cm

Curved surface area = πrl

l = √h² + √r²

⇒12² + 5² = 13

πrl = π x 13 x 5 = 65 π cm².

Question 7. The diameters of the two cones are equal. If their slant height is in the ratio 5: 7, find the ratio of their curved surface areas. 

(a) 25:7 

(b) 25:49 

(c) 5:49 

(d) 5:7

Solution: d) 5:7

Explanation: Let the radius of the two cones be = x cm 

Let the slant height of 1st cone = 5 cm and 

The slant height of 2nd cone = 7 cm 

Then the ratio of covered surface area = πX 5/ πX 7 = 5 : 7

Question 8. The ratio of radii of a cylinder to that of a cone is 1 : 2. If their heights are equal, find the ratio of their volumes. 

(a) 3:1 

(b) 2:3 

(c) 3:4 

(d) 2: 1

Solution: c) 3:4

Here's a step-by-step explanation:

Let the radius of cylinder = 1(r) 

Then the radius of cone be = 2(R) 

Then, as per question = πr²h/πR²h/3

⇒3πr²h/πR²h

⇒3r²/3R²

⇒3:4

Question 9. The ratio of the areas of two isosceles triangles with equal angles is 16:25. What is the ratio of their corresponding heights?

(a) 4/5  

(b) 4/6  

(c) 5/4  

(d) 5/7 

Solution: a) 4/5

Explanation: (Ratio of corresponding sides)²= Ratio of area of similar triangles 

The ratio of corresponding sides in this question= √16/√25 = 4/5.

Question 10. What is the height of a tower that casts a shadow of 60 meters on the ground, given that a vertical stick 30 meters long casts a shadow of 20 meters simultaneously?

(a) 105 m 

(b) 120 m 

(c) 25 m 

(d) 200 m

Solution: b) 120m

Explanation: When the length of the stick = 30 m, 

then the length of shadow = 20 m 

Here, the length formula is = 2 x shadow. 

Assuming the same angle of inclination of the sun, the tower's length that casts a shadow of 60 m 2 x 60m = 120m, i.e. the height of the tower = 120 m.

Conclusion

The exploration of trigonometry's application in determining heights and distances has shed light on the fundamental principles governing these calculations. By delving into trigonometric formulas and ratios, solving practical examples, and understanding the relationships between various elements, a comprehensive understanding of height and distance problems has been achieved.

Identifying different types of triangles and their relevance to real-world scenarios further solidifies the practicality of these mathematical concepts.

Time For A Small Quiz

Frequently Asked Questions (FAQs)

1. What is the significance of understanding heights and distances in trigonometry?

Knowing about heights and distances in trigonometry is super important for things like building stuff, finding your way, and measuring land. You can calculate how tall or far away something is using trigonometric ratios.

2. How are Trigonometry Formulas explained for calculating heights and distances?

Trigonometry formulas play a vital role in calculating height and distance by providing relationships between angles and side lengths of triangles. These formulas, such as sine, cosine, and tangent, help determine unknown measurements accurately.

3. Can you explain the relationship between height and distance in trigonometry?

The relationship between height and distance in trigonometry involves using angles and side lengths of triangles to determine the height or distance of an object from a specific point. This relationship is based on trigonometric functions like sine, cosine, and tangent.

4. Why are solved examples important for better understanding heights and distances?

Solved examples provide practical applications of trigonometric concepts related to heights and distance. By analyzing these examples step by step, learners can grasp how to apply trigonometry principles effectively to solve height and distance-related problems.

5. How do types of triangles play a role in understanding height and distance concepts?

Different types of triangles (such as right-angled triangles) are fundamental in understanding height and distance concepts in trigonometry. These triangles' properties help determine angles and side lengths and apply trigonometric ratios to calculate heights or distances accurately.

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