Follow Us
মাধ্যম নির্বাচন করুন / Select Medium:
Eng (English) Beng (বাংলা) Hindi (हिन्दी)
পশ্চিমবঙ্গ মধ্যশিক্ষা পর্ষদ (WBBSE) • শ্রেণি X • Mathematics • অধ্যায় 25
আনুমানিক সময়: 110 minutes
অগ্রগতি: অধ্যয়নে সক্রিয়

Application of Trigonometric Ratios: Heights and Distances

Ratio and Proportion constitutes Chapter 5 of the WBBSE Class 10 Mathematics curriculum Ganit Prakash, providing essential algebraic techniques for comparative mathematics. A ratio is a mathematical comparison of two quantities of the same kind and in the same unit by division. In the ratio a:b, a is called the antecedent and b the consequent. Ratios are classified based on the relative sizes of terms into ratios of equality, greater inequality where antecedent a exceeds consequent b, and lesser inequality where antecedent a is less than consequent b. The chapter covers operations on ratios, including compound or mixed ratios formed by multiplying corresponding antecedents and consequents, as well as duplicate, sub-duplicate, triplicate, and sub-triplicate ratios. When two ratios are equal, they form a proportion a:b = c:d, where a and d are the extremes and b and c are the means, satisfying the fundamental equality ad = bc. In a continued proportion a:b = b:c, b is the mean proportional equal to the square root of ac, and c is the third proportional. The core strength of the chapter lies in proportion transformation properties: Invertendo, Alternendo, Componendo, Dividendo, Componendo and Dividendo ((a+b)/(a-b) = (c+d)/(c-d)), and the Theorem on Equal Ratios (Addendo). These properties, coupled with the powerful algebraic k-method where equal ratios are set to a non-zero parameter k, allow students to prove complex algebraic identities required in WBBSE Madhyamik examinations.

কখনও কি ভেবে দেখেছেন?

How did the ancient Greeks discover the Golden Ratio that sculpts the Parthenon, and why does every chemical reaction, map scale, financial partnership, and screen resolution hinge upon the laws of proportion? When two ratios are equated, powerful transformations like Componendo-Dividendo and the algebraic k-method unlock intricate proofs with astonishing mathematical elegance.

অধ্যায়টির গুরুত্ব

Ratio and Proportion carries heavy weightage in the Madhyamik algebra paper, regularly appearing as compulsory 3-mark identity proofs and short answer questions. Beyond the board curriculum, the laws of ratio and proportion form the foundation of geometry (similarity and Thales' theorem), trigonometry, physics (kinematics, Boyle's law, Ohm's law), chemistry (stoichiometric ratios), and commercial finance (partnership business and foreign exchange). Mastery of the k-method and Componendo-Dividendo equips students with rapid algebraic simplification skills vital for higher mathematics, IIT-JEE, and competitive aptitude tests.

পাঠের পূর্বে প্রয়োজনীয় ধারণা

  • Concept of fractions, lowest terms, and cross-multiplication (a/b = c/d => ad = bc).
  • Elementary algebraic identities like (a + b)^2, (a - b)^2, and a^3 - b^3.
  • Basic arithmetic operations on surds and radicals.
  • Simultaneous linear equations for finding proportional constants.

শিখন লক্ষ্যমাত্রা ও ফলাফল

  • Define and classify ratios: antecedent, consequent, inverse ratio, ratio of equality, and ratios of greater/lesser inequality.
  • Compute compound (mixed) ratios, duplicate (a^2:b^2), sub-duplicate (sqrt(a):sqrt(b)), triplicate, and sub-triplicate ratios.
  • Combine ratios to form continued ratios a:b:c by equating shared terms.
  • Define proportions a:b = c:d, verifying that the product of extremes equals the product of means (ad = bc), and find fourth proportionals.
  • Analyze continued proportions a:b = b:c, deriving mean proportional b = sqrt(ac) and third proportional c = b^2/a.
  • Apply classical transformation properties: Invertendo, Alternendo, Componendo, Dividendo, Componendo-Dividendo, and Addendo (Theorem on equal ratios).
  • Deploy the algebraic k-method to rigorously prove Madhyamik board algebraic identities and riders.

অধ্যায়ের বিষয়সূচি ও রূপরেখা

1 Module 1: Ratio Concepts, Classific...
2 Module 2: Proportion & Continued Pr...
3 Module 3: Transformation Properties...
4 Module 4: The Algebraic k-Method fo...

সম্পূর্ণ তত্ত্ব ও ধারণাগত আলোচনা

Module 1: Ratio Concepts, Classification & Derived Ratios

1.1 The Concept of Ratio (অনুপাতের ধারণা)

A ratio is a comparative division of two quantities of the same kind measured in the same physical unit. The ratio of quantity $a$ to quantity $b$ ($b eq 0$) is written as $a : b = rac{a}{b}$.

  • Antecedent (পূর্বপদ): The first term $a$.
  • Consequent (উত্তরপদ): The second term $b$.
  • Dimensionless Property: A ratio is an abstract pure real number and carries no physical units.
1.2 Classification of Ratios
Ratio Type Condition Fraction Value Example
Ratio of Equality (সমানুপাত / সাম্যানুপাত) $a = b$ $ rac{a}{b} = 1$ $5 : 5 = 1 : 1$
Ratio of Greater Inequality (গুরু অনুপাত) $a > b$ $ rac{a}{b} > 1$ $7 : 4$ (Antecedent exceeds consequent)
Ratio of Lesser Inequality (লঘু অনুপাত) $a < b$ $ rac{a}{b} < 1$ $3 : 8$ (Antecedent is less than consequent)
Inverse Ratio (ব্যস্ত বা বিপরীত অনুপাত) Interchanging terms $b : a = rac{b}{a}$ Inverse of $4 : 9$ is $9 : 4$
1.3 Derived Ratios & Continued Ratios
  • Compound / Mixed Ratio (মৌলিক বা মিশ্র অনুপাত): Formed by taking the product of antecedents to the product of consequents. For $a:b$ and $c:d$, the compound ratio is $(a \cdot c) : (b \cdot d)$.
  • Duplicate Ratio (দ্বিগুণানুপাত): $a^2 : b^2$.
  • Sub-duplicate Ratio (দ্বিভাজিত অনুপাত): $\sqrt{a} : \sqrt{b}$.
  • Triplicate Ratio (ত্রিগুণানুপাত): $a^3 : b^3$.
  • Sub-triplicate Ratio (ত্রিভাজিত অনুপাত): $\sqrt[3]{a} : \sqrt[3]{b}$.
  • Continued Ratio (ধারাবাহিক অনুপাত $a:b:c$): If $a:b = 2:3$ and $b:c = 4:5$, equate the common term $b$ by taking $ ext{LCM}(3, 4) = 12$: $a:b = 8:12$ and $b:c = 12:15 \implies a:b:c = 8:12:15$.

Module 2: Proportion & Continued Proportion

2.1 Proportion (সমানুপাত)

An equality of two ratios constitutes a proportion. If $a:b = c:d$, we write $a : b :: c : d$, which algebraically signifies:

$$ rac{a}{b} = rac{c}{d} \iff a \cdot d = b \cdot c$$
$$ extbf{Product of Extremes} = extbf{Product of Means}$$

Here, $a$ and $d$ are called the extremes (প্রান্তীয় পদ), while $b$ and $c$ are called the means (মধ্যপদ). The term $d$ is called the fourth proportional to $a, b, c$, given by $d = rac{bc}{a}$.

2.2 Continued Proportion (ক্রমিক সমানুপাত)

Three non-zero quantities $a, b, c$ are said to be in continued proportion if the ratio of the first to the second equals the ratio of the second to the third:

$$ rac{a}{b} = rac{b}{c} \iff b^2 = ac \iff b = \pm \sqrt{ac}$$
  • $b$ is the Mean Proportional (মধ্য সমানুপাতী) between $a$ and $c$: $b = \sqrt{ac}$ (taking positive sign for positive quantities).
  • $c$ is the Third Proportional (তৃতীয় সমানুপাতী) to $a$ and $b$: $c = rac{b^2}{a}$.

Module 3: Transformation Properties of Proportion

3.1 The Six Master Transformation Properties

Let $ rac{a}{b} = rac{c}{d}$. The following transformations are universally valid:

Transformation Bengali Term Transformed Form Derivation Principle
1. Invertendo ব্যস্ত প্রক্রিয়া $ rac{b}{a} = rac{d}{c}$ Inverting both ratios.
2. Alternendo একান্তর প্রক্রিয়া $ rac{a}{c} = rac{b}{d}$ Interchanging the means $b$ and $c$.
3. Componendo যোগ প্রক্রিয়া $ rac{a + b}{b} = rac{c + d}{d}$ Adding $1$ to both sides: $ rac{a}{b} + 1 = rac{c}{d} + 1$.
4. Dividendo ভাগ প্রক্রিয়া $ rac{a - b}{b} = rac{c - d}{d}$ Subtracting $1$ from both sides: $ rac{a}{b} - 1 = rac{c}{d} - 1$.
5. Componendo & Dividendo যোগ-ভাগ প্রক্রিয়া $ rac{a + b}{a - b} = rac{c + d}{c - d}$ Dividing Componendo by Dividendo equation.
6. Addendo (Theorem on Equal Ratios) সংযোজন প্রক্রিয়া $ ext{Each ratio} = rac{a + c}{b + d}$ If $ rac{a}{b} = rac{c}{d} = k$, then $ rac{a+c}{b+d} = rac{bk+dk}{b+d} = k$.

Module 4: The Algebraic k-Method for WBBSE Board Proofs

4.1 The k-Method Architecture

The k-method is the gold-standard technique for proving symmetrical algebraic identities in WBBSE examinations:

Standard Proportion k-Method:

If $ rac{a}{b} = rac{c}{d}$, let $ rac{a}{b} = rac{c}{d} = k$ (where $k eq 0$ is a constant).
Then express the antecedents in terms of consequents: $\mathbf{a = bk}$ and $\mathbf{c = dk}$.
Substitute these values into both Left Hand Side (L.H.S.) and Right Hand Side (R.H.S.) to show $ ext{L.H.S.} = ext{R.H.S.}$

Continued Proportion k-Method:

If $a, b, c$ are in continued proportion: $ rac{a}{b} = rac{b}{c} = k$ ($k eq 0$).
Then $b = ck$, and $a = bk = (ck)k = \mathbf{ck^2}$.
If four quantities $a, b, c, d$ are in continued proportion: $ rac{a}{b} = rac{b}{c} = rac{c}{d} = k$.
Then $c = dk$, $b = ck = \mathbf{dk^2}$, and $a = bk = \mathbf{dk^3}$.

গুরুত্বপূর্ণ গাণিতিক সূত্র, অভেদ ও উপপাদ্য

Ratio Antecedent and Consequent
a : b = a / b
Compound Ratio
(a * c) : (b * d)
Duplicate and Sub-duplicate Ratio
$$a^2 : b^2 and sqrt(a) : sqrt(b)$$
Proportion Fundamental Law
ad = bc
Fourth Proportional
x = (b * c) / a
Mean Proportional
b = sqrt(a * c)
Third Proportional
$$c = b^2 / a$$
Componendo and Dividendo
(a + b) / (a - b) = (c + d) / (c - d)
Theorem on Equal Ratios (Addendo)
$$Each ratio = (a_1 + a_2 + ...) / (b_1 + b_2 + ...)$$
Continued Proportion k-Form
$$b = ck, a = ck^2$$

সমাধানকৃত উদাহরণ ও প্রয়োগ (Solved Examples)

উদাহরণ 1
If a : b = 3 : 4 and b : c = 8 : 9, find a : c and the continued ratio a : b : c.
ধাপে ধাপে সমাধান / উত্তর:
Step 1: Write given ratios as fractions: a/b = 3/4 and b/c = 8/9. Step 2: Find a:c by multiplying the fractions: a/c = (a/b) * (b/c) = (3/4) * (8/9) = (3 * 8) / (4 * 9) = 24 / 36 = 2/3. Hence, a : c = 2 : 3. Step 3: To find a : b : c, equalize the common term b: The value of b is 4 in the first ratio and 8 in the second. LCM of 4 and 8 is 8. Multiply terms of the first ratio by 2: a : b = (3 * 2) : (4 * 2) = 6 : 8. Second ratio is already: b : c = 8 : 9. Step 4: Combine into a single continued ratio: a : b : c = 6 : 8 : 9. Hence, a : c = 2 : 3 and a : b : c = 6 : 8 : 9.
উদাহরণ 2
Find: (i) the fourth proportional to 4, 9, 12; (ii) the third proportional to 16 and 24; (iii) the mean proportional between 9 and 25.
ধাপে ধাপে সমাধান / উত্তর:
Part (i): Let the fourth proportional be x. Then 4 : 9 :: 12 : x => 4/9 = 12/x 4x = 9 * 12 = 108 => x = 108 / 4 = 27. Hence, fourth proportional = 27. Part (ii): Let the third proportional to 16 and 24 be y. Then 16, 24, y are in continued proportion: 16 : 24 :: 24 : y 16/24 = 24/y => 16y = 24 * 24 = 576 y = 576 / 16 = 36. Hence, third proportional = 36. Part (iii): Let the mean proportional between 9 and 25 be m. Then m = sqrt(9 * 25) = sqrt(225) = 15. Hence, mean proportional = 15.
উদাহরণ 3
If [sqrt(1 + x) + sqrt(1 - x)] / [sqrt(1 + x) - sqrt(1 - x)] = a, prove using Componendo and Dividendo that x = 2a / (a^2 + 1).
ধাপে ধাপে সমাধান / উত্তর:
Step 1: Write the given equation with denominator 1 on RHS: [sqrt(1 + x) + sqrt(1 - x)] / [sqrt(1 + x) - sqrt(1 - x)] = a / 1 Step 2: Apply Componendo and Dividendo [(Numerator + Denominator) / (Numerator - Denominator)]: Numerator + Denominator = [sqrt(1+x) + sqrt(1-x)] + [sqrt(1+x) - sqrt(1-x)] = 2*sqrt(1 + x) Numerator - Denominator = [sqrt(1+x) + sqrt(1-x)] - [sqrt(1+x) - sqrt(1-x)] = 2*sqrt(1 - x) So: [2*sqrt(1 + x)] / [2*sqrt(1 - x)] = (a + 1) / (a - 1) sqrt(1 + x) / sqrt(1 - x) = (a + 1) / (a - 1) Step 3: Square both sides to eliminate square roots: (1 + x) / (1 - x) = (a + 1)^2 / (a - 1)^2 = (a^2 + 2a + 1) / (a^2 - 2a + 1) Step 4: Apply Componendo and Dividendo once more: [(1 + x) + (1 - x)] / [(1 + x) - (1 - x)] = [(a^2 + 2a + 1) + (a^2 - 2a + 1)] / [(a^2 + 2a + 1) - (a^2 - 2a + 1)] 2 / (2x) = (2a^2 + 2) / (4a) 1 / x = 2(a^2 + 1) / (4a) = (a^2 + 1) / (2a) Step 5: Invert both sides: x = 2a / (a^2 + 1). Hence, proved.
উদাহরণ 4
If a, b, c are in continued proportion, prove using the k-method that (a^2 + b^2) / (b^2 + c^2) = a / c.
ধাপে ধাপে সমাধান / উত্তর:
Step 1: Set up the continued proportion k-method: Since a, b, c are in continued proportion: a/b = b/c = k (where k != 0). Then b = ck and a = bk = (ck)k = ck^2. Step 2: Evaluate the Left Hand Side (L.H.S.): L.H.S. = (a^2 + b^2) / (b^2 + c^2) Substitute a = ck^2 and b = ck: Numerator = (ck^2)^2 + (ck)^2 = c^2 k^4 + c^2 k^2 = c^2 k^2 (k^2 + 1) Denominator = (ck)^2 + c^2 = c^2 k^2 + c^2 = c^2 (k^2 + 1) L.H.S. = [c^2 k^2 (k^2 + 1)] / [c^2 (k^2 + 1)] = k^2. Step 3: Evaluate the Right Hand Side (R.H.S.): R.H.S. = a / c = (ck^2) / c = k^2. Step 4: Compare: Since L.H.S. = k^2 and R.H.S. = k^2, we have L.H.S. = R.H.S. Hence, (a^2 + b^2)/(b^2 + c^2) = a/c. (Proved)
উদাহরণ 5
If x / (b + c - a) = y / (c + a - b) = z / (a + b - c), prove that (b - c)x + (c - a)y + (a - b)z = 0.
ধাপে ধাপে সমাধান / উত্তর:
Step 1: Set each equal ratio to k (where k != 0): x / (b + c - a) = y / (c + a - b) = z / (a + b - c) = k Then: x = k(b + c - a) y = k(c + a - b) z = k(a + b - c) Step 2: Substitute x, y, z into the expression L.H.S.: L.H.S. = (b - c)x + (c - a)y + (a - b)z = (b - c)[k(b + c - a)] + (c - a)[k(c + a - b)] + (a - b)[k(a + b - c)] = k[(b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)] Step 3: Expand the products inside brackets using difference of squares: (b - c)(b + c - a) = (b - c)(b + c) - a(b - c) = (b^2 - c^2) - ab + ac (c - a)(c + a - b) = (c - a)(c + a) - b(c - a) = (c^2 - a^2) - bc + ab (a - b)(a + b - c) = (a - b)(a + b) - c(a - b) = (a^2 - b^2) - ca + bc Step 4: Sum all terms: (b^2 - c^2 + c^2 - a^2 + a^2 - b^2) + (-ab + ac - bc + ab - ca + bc) = 0 + 0 = 0. Step 5: Multiply by k: L.H.S. = k * 0 = 0 = R.H.S. Hence, proved.

সাধারণ ভুলত্রুটি ও সতর্কতা (Common Traps)

সাধারণ ভুল ধারণা

Adding units to a ratio (e.g. writing 3 kg : 4 kg = 3/4 kg).

সঠিক পদ্ধতি ও সমাধান

A ratio is a pure real number without physical units: 3 : 4.

সাধারণ ভুল ধারণা

Comparing quantities in different units directly (e.g. 50 paise : ₹2 = 50 : 2).

সঠিক পদ্ধতি ও সমাধান

Convert all quantities to the same unit first: 50 paise : 200 paise = 50 : 200 = 1 : 4.

সাধারণ ভুল ধারণা

Confusing Sub-duplicate ratio with Duplicate ratio.

সঠিক পদ্ধতি ও সমাধান

Duplicate ratio is squared (a^2 : b^2); Sub-duplicate is square root (sqrt(a) : sqrt(b)).

সাধারণ ভুল ধারণা

Assuming a, b, c in continued proportion means a = bk, b = ck without a common base.

সঠিক পদ্ধতি ও সমাধান

Express all variables in terms of the LAST variable c: b = ck, a = ck^2.

সাধারণ ভুল ধারণা

Applying Componendo-Dividendo to only one side of an equation.

সঠিক পদ্ধতি ও সমাধান

Componendo-Dividendo must be applied simultaneously to both LHS and RHS.

Concept Map: Ratio and Proportion (WBBSE Class 10 Ganit Prakash)

Ratio and Proportion (অনুপাত ও সমানুপাত) WBBSE Class 10 Mathematics • Chapter 5 • Properties, Transformations & k-Method 1. Ratio Concepts & Types • Ratio a:b = a/b (Antecedent: a, Consequent: b)• Inverse ratio: b:a; Compound: (ac):(bd)• Duplicate: a^2:b^2; Sub-duplicate: sqrt(a):sqrt(b)• Triplicate: a^3:b^3; Sub-triplicate: cbrt(a):cbrt(b) 2. Proportion & Mean Proportional • Proportion: a:b = c:d <=> ad = bc• Extremes: a, d; Means: b, c; 4th prop: x = bc/a• Continued Proportion: a/b = b/c <=> b^2 = ac• Mean proportional: b = sqrt(ac); 3rd prop: c = b^2/a 3. Proportion Transformations • Invertendo: b/a = d/c; Alternendo: a/c = b/d• Componendo: (a+b)/b = (c+d)/d• Dividendo: (a-b)/b = (c-d)/d• Componendo-Dividendo: (a+b)/(a-b) = (c+d)/(c-d) 4. Addendo & The k-Method • Addendo: Each ratio = (a1 + a2 + ...)/(b1 + b2 + ...)• k-method: a/b = c/d = k => a = bk, c = dk• Continued prop k-method: b = ck, a = ck^2• Board proofs: LHS & RHS evaluation via k

অধ্যায় সারসংক্ষেপ ও গুরুত্বপূর্ণ বিষয়

মূল বিষয় 1
A ratio a:b compares two quantities of the same kind by division and has no physical units.
মূল বিষয় 2
In a:b, a is the antecedent and b is the consequent; inverse ratio is b:a; compound ratio is (ac):(bd).
মূল বিষয় 3
Duplicate ratio = a^2:b^2; sub-duplicate = sqrt(a):sqrt(b); triplicate = a^3:b^3; sub-triplicate = cbrt(a):cbrt(b).
মূল বিষয় 4
A proportion a:b = c:d equates two ratios, satisfying: Product of Extremes = Product of Means (ad = bc).
মূল বিষয় 5
In continued proportion a:b = b:c, b is the mean proportional (b = sqrt(ac)) and c is the third proportional (c = b^2/a).
মূল বিষয় 6
Proportion transformations include Invertendo, Alternendo, Componendo, Dividendo, and Componendo-Dividendo.
মূল বিষয় 7
The Theorem on Equal Ratios (Addendo) states that each ratio equals the sum of antecedents divided by the sum of consequents.
মূল বিষয় 8
The k-method sets equal ratios to k, converting proportional variables to simple multiples for rigorous proofs.

স্ব-মূল্যায়ন অনুশীলন (Check Your Understanding)

মূল ধারণাগত স্পষ্টতা যাচাই করার জন্য অনুশীলন প্রশ্ন। উত্তর দেখার আগে নিজে সমাধান করার চেষ্টা করো।

1
What is the compound ratio of 2 : 3, 5 : 7, and 9 : 10?
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: Product of antecedents = 2 * 5 * 9 = 90. Product of consequents = 3 * 7 * 10 = 210. Compound ratio = 90 : 210 = 3 : 7.
2
Find the mean proportional between 4 and 16.
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: Mean proportional b = sqrt(4 * 16) = sqrt(64) = 8.
3
If 3, x, 12 are in continued proportion, find the positive value of x.
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: Since 3/x = x/12 => x^2 = 36 => x = 6.
4
State the transform property that changes a/b = c/d to a/c = b/d.
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: This property is called Alternendo (একান্তর প্রক্রিয়া).
5
If x : y = 3 : 4, find the value of (2x + 3y) : (3x + 4y).
উত্তর ও ব্যাখ্যা দেখুন
উত্তর: Let x = 3k, y = 4k. Then (2x + 3y)/(3x + 4y) = (6k + 12k)/(9k + 16k) = 18k / 25k = 18/25. Ratio is 18 : 25.
অধ্যায় পড়া শেষ হয়েছে?
অনুশীলন শুরু করো

অনলাইন মক টেস্ট দিয়ে প্রস্তুতি যাচাই করো

পশ্চিমবঙ্গ মধ্যশিক্ষা পর্ষদ (WBBSE) পাঠ্যক্রম অনুযায়ী বহু বিকল্পীয় প্রশ্ন (MCQ) সমাধান করো। তাৎক্ষণিক ফলাফল, সঠিক ব্যাখ্যা এবং নিজের স্কোর জেনে নাও।

শ্রেণি 10 Mathematics — সকল অধ্যায়

অধ্যায় 1: Quadratic Equations in One Variable অধ্যায় 2: Simple Interest অধ্যায় 3: Theorems related to Circle অধ্যায় 4: Rectangular Parallelopiped or Cuboid অধ্যায় 5: Ratio and Proportion অধ্যায় 6: Compound Interest and Uniform Rate of Increase or Decrease অধ্যায় 7: Theorems Related to Angles in a Circle অধ্যায় 8: Right Circular Cylinder অধ্যায় 9: Quadratic Surd অধ্যায় 10: Theorems Related to Cyclic Quadrilateral অধ্যায় 11: Construction of Circumcircle and Incircle of a Triangle অধ্যায় 12: Sphere অধ্যায় 13: Variation অধ্যায় 14: Partnership Business অধ্যায় 15: Theorems Related to Tangent to a Circle অধ্যায় 16: Right Circular Cone অধ্যায় 17: Construction of Tangent to a Circle অধ্যায় 18: Similarity অধ্যায় 19: Problems Related to Different Solid Objects অধ্যায় 20: Trigonometry: Concept of Measurement of Angle অধ্যায় 21: Construction: Determination of Mean Proportional অধ্যায় 22: Pythagoras Theorem অধ্যায় 23: Trigonometric Ratios and Trigonometric Identities অধ্যায় 24: Trigonometric Ratios of Complementary Angle অধ্যায় 25: Application of Trigonometric Ratios: Heights and Distances অধ্যায় 26: Statistics: Mean, Median, Ogive, Mode

AI সহায়ক

তাত্ক্ষণিক সমাধান

Application of Trigonometric Ratios: Heights and Distances অধ্যায়ে কোনো প্রশ্ন বা সন্দেহ আছে? আমাদের AI শিক্ষক থেকে সহজ সমাধান ও ব্যাখ্যা নিন।