This MCQ module is based on: Trigonometric Ratios
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Trigonometric Ratios
🎓 Class 10
Mathematics
CBSE
Theory
Ch 8 — Introduction to Trigonometry
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Trigonometric Ratios
Targeting Class 10 level in Trigonometry, with Intermediate difficulty.
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8.1 Introduction
Suppose a student standing in front of the Qutub Minar wishes to find its height. She cannot climb the monument. Or suppose a girl on the balcony of her house observes a temple across a river — how far is the river? Such problems are solved using trigonometryi, the study of relationships between the sides and angles of triangles.
Did you know?
The word trigonometry comes from the Greek tri (three), gon (sides) and metron (measure). The Indian mathematician Aryabhata (476 CE) gave the first table of sines, calling the function jya. The Arabs adopted it as jiba, and later translators rendered it in Latin as sinus — a fold or bay — from which the modern sine is derived.
8.2 Trigonometric Ratios
Consider a right triangle ABC, right-angled at B, and let angle A be the acute angle under study. With respect to angle A:
- Opposite side (perpendicular) = BC
- Adjacent side (base) = AB
- Hypotenuse = AC (always opposite the right angle)
Fig. 8.4 — Opposite, Adjacent and Hypotenuse with respect to ∠A
Six Trigonometric Ratios
\[\sin A=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{AC},\quad \cos A=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AB}{AC},\quad \tan A=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AB}\]
\[\csc A=\frac{1}{\sin A}=\frac{AC}{BC},\quad \sec A=\frac{1}{\cos A}=\frac{AC}{AB},\quad \cot A=\frac{1}{\tan A}=\frac{AB}{BC}\]
Also \(\tan A=\dfrac{\sin A}{\cos A}\) and \(\cot A=\dfrac{\cos A}{\sin A}\).
Note. The ratios depend only on the angle A, not on the size of the triangle — because all right triangles with the same acute angle are similar, so their corresponding side ratios are equal.
Why ratios are fixed for a given angle
Take two right triangles \(\triangle ABC\) and \(\triangle APQ\) both right-angled at B and Q respectively, sharing angle A (see Fig. 8.6). By the AA criterion, \(\triangle ABC\sim\triangle APQ\), so \(\dfrac{AB}{AP}=\dfrac{BC}{PQ}=\dfrac{AC}{AQ}\). Therefore \(\dfrac{BC}{AC}=\dfrac{PQ}{AQ}\) — the sine of angle A is the same in both triangles.
Fig. 8.6 — Similar right triangles give the same trig ratios for angle A
Activity — Trig Ratios with a Protractor
Materials: protractor, ruler, graph paper
Predict: Will sin 40° be the same in a small right triangle as in a larger one?
- Draw two right triangles, both with one acute angle = 40°, but of different sizes.
- Measure the opposite side and hypotenuse in each.
- Compute opposite/hypotenuse for each. Compare.
Both ratios come out ≈ 0.643. Sin 40° depends only on the angle — similar triangles, same ratio. This confirms why trigonometric ratios are functions of the angle alone.
Example 1
Given \(\tan A=\dfrac{4}{3}\), find the other trigonometric ratios of angle A.
Solution. In right \(\triangle ABC\), right-angled at B, with \(\tan A=BC/AB=4/3\). Let \(BC=4k,\;AB=3k\). By Pythagoras, \(AC^2=(3k)^2+(4k)^2=25k^2\Rightarrow AC=5k\). Then
\[\sin A=\frac{4k}{5k}=\frac{4}{5},\;\cos A=\frac{3}{5},\;\cot A=\frac{3}{4},\;\csc A=\frac{5}{4},\;\sec A=\frac{5}{3}.\]Example 2
If \(\angle B\) and \(\angle Q\) are acute angles such that \(\sin B=\sin Q\), prove \(\angle B=\angle Q\).
Solution. In right \(\triangle ABC\) (right at C) and \(\triangle PQR\) (right at R), \(\sin B=AC/AB\) and \(\sin Q=PR/PQ\). Given these are equal, \(AC/AB=PR/PQ\), hence by similar triangles \(\triangle ACB\sim\triangle PRQ\), so \(\angle B=\angle Q\).
Example 3
Consider \(\triangle ACB\), right-angled at C, in which \(AB=29\), \(BC=21\) and \(\angle ABC=\theta\). Find \(\cos^2\theta+\sin^2\theta\) and \(\cos^2\theta-\sin^2\theta\).
Solution. \(AC=\sqrt{29^2-21^2}=\sqrt{841-441}=\sqrt{400}=20\).
\[\sin\theta=\frac{AC}{AB}=\frac{20}{29},\;\cos\theta=\frac{BC}{AB}=\frac{21}{29}.\] \[\cos^2\theta+\sin^2\theta=\frac{400+441}{841}=1,\;\;\cos^2\theta-\sin^2\theta=\frac{441-400}{841}=\frac{41}{841}.\]Example 4
In a right \(\triangle ABC\), right-angled at B, if \(\tan A=1\), verify that \(2\sin A\cos A=1\).
Solution. \(\tan A=BC/AB=1\Rightarrow BC=AB=k\). Then \(AC=k\sqrt2\), giving \(\sin A=1/\sqrt2\) and \(\cos A=1/\sqrt2\). Therefore \(2\sin A\cos A=2\cdot\dfrac{1}{\sqrt2}\cdot\dfrac{1}{\sqrt2}=1\). Verified.
Example 5
In \(\triangle OPQ\) right-angled at P, \(OP=7\) cm and \(OQ-PQ=1\) cm. Determine \(\sin Q\) and \(\cos Q\).
Solution. Let \(PQ=x\), so \(OQ=x+1\). Pythagoras: \((x+1)^2=7^2+x^2\Rightarrow 2x+1=49\Rightarrow x=24\). So \(PQ=24,\;OQ=25\). Then \(\sin Q=OP/OQ=7/25\), \(\cos Q=PQ/OQ=24/25\).
Competency-Based Questions
Sunlight strikes the ground at an acute angle θ. A vertical pole of height 3 m casts a horizontal shadow of length 4 m.
Q1. Determine all six trigonometric ratios of θ.
L3 ApplyOpposite = 3, adjacent = 4, hypotenuse = 5. sin θ = 3/5, cos θ = 4/5, tan θ = 3/4, cosec θ = 5/3, sec θ = 5/4, cot θ = 4/3.
Q2. Compare how sin θ changes if the pole is raised to 4 m while the shadow remains 4 m.
L4 AnalyseNew hypotenuse = √(16+16) = 4√2. New sin θ = 4/(4√2) = 1/√2 ≈ 0.707, up from 0.6. A taller pole gives a larger sin θ — because the angle of elevation increases.
Q3. A student claims, "If tan θ = 1, then sin θ = cos θ = 1". Evaluate the claim.
L5 EvaluateIncorrect. tan θ = 1 means opposite = adjacent, hence sin θ = cos θ = 1/√2, not 1. Sine and cosine of an acute angle cannot exceed 1.
Q4. Design a right triangle whose sin A = 5/13 and compute all other ratios of A.
L6 CreateOpposite = 5k, hypotenuse = 13k, so adjacent = √(169 − 25)·k = 12k. cos A = 12/13, tan A = 5/12, cosec A = 13/5, sec A = 13/12, cot A = 12/5.
Assertion & Reason
A: The value of sin A lies between 0 and 1 for any acute angle A.
R: In a right triangle, the opposite side is always shorter than the hypotenuse.
R: In a right triangle, the opposite side is always shorter than the hypotenuse.
Answer: A. The hypotenuse is the longest side of a right triangle; therefore opposite/hypotenuse < 1, and it is positive for an acute angle — so 0 < sin A < 1. R explains A.
A: sec A × cos A = 1 for any acute angle A.
R: sec A is defined as the reciprocal of cos A.
R: sec A is defined as the reciprocal of cos A.
Answer: A. By definition sec A = 1/cos A, hence their product is 1.
A: tan A = sin A / cos A.
R: sin A = opposite/hypotenuse and cos A = adjacent/hypotenuse.
R: sin A = opposite/hypotenuse and cos A = adjacent/hypotenuse.
Answer: A. sin A / cos A = (opp/hyp) / (adj/hyp) = opp/adj = tan A.
Frequently Asked Questions
What is the definition of sine of an angle?
In a right triangle, the sine of an acute angle A is the ratio of the length of the side opposite to A to the length of the hypotenuse.
What is the relationship between sine and cosecant?
Cosecant is the reciprocal of sine: cosec A = 1/sin A. Similarly, secant is the reciprocal of cosine and cotangent is the reciprocal of tangent.
Why don't trigonometric ratios depend on triangle size?
Because all right triangles with the same acute angle are similar, their corresponding sides are in the same ratio. So the trigonometric ratios depend only on the angle, not the size.
What is tan A in terms of sin and cos?
tan A = sin A / cos A, obtained by dividing the opposite over adjacent as (opposite/hypotenuse) divided by (adjacent/hypotenuse).
Why is the hypotenuse always the largest side?
The hypotenuse is opposite the right angle (90 degrees), which is the largest angle in a right triangle. By the triangle-angle-side relationship, the side opposite the largest angle is the longest.
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