Introduction to Trigonometry - Exercise 8.1

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Overview

Step-by-step NCERT solutions and practice problems for CBSE Class 10 Introduction to Trigonometry (Exercise 8.1), covering trigonometric ratios ($sin, cos, an, cot, sec, ext{cosec}$), Pythagoras theorem applications, and conceptual validations.

Master Trigonometric Ratios & Pythagoras Theorem

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Basic Ratios Q1: Right Triangle Ratios Q2: tan P - cot R Q3: sin A = 5/13 Q4: 8 cot A = 15 Q5: sec θ = 5/3 Q6: cos A = cos B Q7: cot θ = 3/4 Q8: 3 cot A = 4 Identity Q9: tan A = 1 Q10: Right Triangle PQR Q11: True or False
Trigonometric Ratios in a Right Triangle:
$$sin heta = rac{ ext{Opposite Side (P)}}{ ext{Hypotenuse (H)}}, quad cos heta = rac{ ext{Adjacent Side (B)}}{ ext{Hypotenuse (H)}}, quad an heta = rac{ ext{Opposite Side (P)}}{ ext{Adjacent Side (B)}}$$ $$ ext{cosec } heta = rac{1}{sin heta} = rac{H}{P}, quad sec heta = rac{1}{cos heta} = rac{H}{B}, quad cot heta = rac{1}{ an heta} = rac{B}{P}$$
Question 1 NCERT Ex 8.1
In $Delta ABC$, right-angled at $B$, $AB = 5 ext{ cm}$, $BC = 12 ext{ cm}$. Determine:

(i) $sin A, cos A$
(ii) $sin C, cos C$
A B C 5 cm 12 cm 13 cm
Step 1: Find Hypotenuse ($AC$) Using Pythagoras Theorem
In right $Delta ABC$: $$AC^2 = AB^2 + BC^2 = 5^2 + 12^2 = 25 + 144 = 169$$ $$AC = sqrt{169} = 13 ext{ cm}$$
Step 2: Ratios for Angle $A$
For $angle A$: Opposite side $= BC = 12 ext{ cm}$, Adjacent side $= AB = 5 ext{ cm}$, Hypotenuse $= AC = 13 ext{ cm}$. $$sin A = rac{BC}{AC} = rac{12}{13}$$ $$cos A = rac{AB}{AC} = rac{5}{13}$$
Step 3: Ratios for Angle $C$
For $angle C$: Opposite side $= AB = 5 ext{ cm}$, Adjacent side $= BC = 12 ext{ cm}$, Hypotenuse $= AC = 13 ext{ cm}$. $$sin C = rac{AB}{AC} = rac{5}{13}$$ $$cos C = rac{BC}{AC} = rac{12}{13}$$
(i) $sin A = rac{12}{13}, cos A = rac{5}{13}$; (ii) $sin C = rac{5}{13}, cos C = rac{12}{13}$
Question 2 NCERT Ex 8.1
In the figure below, find $ an P - cot R$. (Given $PQ = 8 ext{ cm}$, $PR = 17 ext{ cm}$)
P Q R 8 cm 15 cm 17 cm
Step 1: Find Side $QR$ Using Pythagoras Theorem
In right $Delta PQR$: $$QR = sqrt{PR^2 - PQ^2} = sqrt{17^2 - 8^2} = sqrt{289 - 64} = sqrt{225} = 15 ext{ cm}$$
Step 2: Compute $ an P$ and $cot R$
For $angle P$: Opposite side $= QR = 15$, Adjacent side $= PQ = 8$: $$ an P = rac{QR}{PQ} = rac{15}{8}$$ For $angle R$: Adjacent side $= QR = 15$, Opposite side $= PQ = 8$: $$cot R = rac{QR}{PQ} = rac{15}{8}$$
Step 3: Evaluate $ an P - cot R$
$$ an P - cot R = rac{15}{8} - rac{15}{8} = 0$$
$ an P - cot R = 0$
Question 3 NCERT Ex 8.1
If $sin A = rac{5}{13}$, calculate $cos A$ and $ an A$.
Step 1: Assign Triangle Sides
Given $sin A = rac{ ext{Perpendicular}}{ ext{Hypotenuse}} = rac{5}{13}$.
Let $P = 5k$ and $H = 13k$ (where $k > 0$).
Step 2: Calculate Base ($B$)
$$B = sqrt{H^2 - P^2} = sqrt{(13k)^2 - (5k)^2} = sqrt{169k^2 - 25k^2} = sqrt{144k^2} = 12k$$
Step 3: Compute Required Ratios
$$cos A = rac{ ext{Base}}{ ext{Hypotenuse}} = rac{12k}{13k} = rac{12}{13}$$ $$ an A = rac{ ext{Perpendicular}}{ ext{Base}} = rac{5k}{12k} = rac{5}{12}$$
$cos A = rac{12}{13}$ and $ an A = rac{5}{12}$.
Question 4 NCERT Ex 8.1
Given $8 cot A = 15$, find $sin A$ and $sec A$.
Step 1: Express $cot A$
$$8 cot A = 15 implies cot A = rac{15}{8} = rac{ ext{Base}}{ ext{Perpendicular}}$$ Let $ ext{Base} = 15k$ and $ ext{Perpendicular} = 8k$.
Step 2: Compute Hypotenuse
$$H = sqrt{P^2 + B^2} = sqrt{(8k)^2 + (15k)^2} = sqrt{64k^2 + 225k^2} = sqrt{289k^2} = 17k$$
Step 3: Compute Ratios
$$sin A = rac{P}{H} = rac{8k}{17k} = rac{8}{17}$$ $$sec A = rac{H}{B} = rac{17k}{15k} = rac{17}{15}$$
$sin A = rac{8}{17}$ and $sec A = rac{17}{15}$.
Question 5 NCERT Ex 8.1
Given $sec heta = rac{5}{3}$, calculate all other trigonometric ratios.
Step 1: Sides of Right Triangle
$$sec heta = rac{H}{B} = rac{5}{3} implies H = 5k, ; B = 3k$$ $$P = sqrt{H^2 - B^2} = sqrt{(5k)^2 - (3k)^2} = sqrt{25k^2 - 9k^2} = sqrt{16k^2} = 4k$$
Step 2: Compute All 5 Remaining Ratios
$$sin heta = rac{P}{H} = rac{4k}{5k} = rac{4}{5}$$ $$cos heta = rac{B}{H} = rac{3k}{5k} = rac{3}{5}$$ $$ an heta = rac{P}{B} = rac{4k}{3k} = rac{4}{3}$$ $$cot heta = rac{B}{P} = rac{3k}{4k} = rac{3}{4}$$ $$ ext{cosec } heta = rac{H}{P} = rac{5k}{4k} = rac{5}{4}$$
$sin heta = rac{4}{5}$, $cos heta = rac{3}{5}$, $ an heta = rac{4}{3}$, $cot heta = rac{3}{4}$, $ ext{cosec } heta = rac{5}{4}$.
Question 6 NCERT Ex 8.1
If $angle A$ and $angle B$ are acute angles such that $cos A = cos B$, then show that $angle A = angle B$.
Step 1: Geometric Setup
Consider a right-angled triangle $Delta ABC$ where $angle C = 90^circ$, so $angle A$ and $angle B$ are acute angles.
For $angle A$: Adjacent side is $AC$, Hypotenuse is $AB$: $$cos A = rac{AC}{AB}$$ For $angle B$: Adjacent side is $BC$, Hypotenuse is $AB$: $$cos B = rac{BC}{AB}$$
Step 2: Equate and Conclude
Given $cos A = cos B$: $$ rac{AC}{AB} = rac{BC}{AB} implies AC = BC$$ In $Delta ABC$, since opposite sides $AC$ and $BC$ are equal, their opposite angles must also be equal: $$angle B = angle A implies angle A = angle B$$
Hence proved that $angle A = angle B$.
Question 7 NCERT Ex 8.1
If $cot heta = rac{3}{4}$, evaluate:

(i) $ rac{(1 + sin heta)(1 - sin heta)}{(1 + cos heta)(1 - cos heta)}$
(ii) $cot^2 heta$
Step 1: Evaluate Expression (i) Algebraically
Numerator: $(1 + sin heta)(1 - sin heta) = 1 - sin^2 heta = cos^2 heta$
Denominator: $(1 + cos heta)(1 - cos heta) = 1 - cos^2 heta = sin^2 heta$
Therefore: $$ rac{(1 + sin heta)(1 - sin heta)}{(1 + cos heta)(1 - cos heta)} = rac{cos^2 heta}{sin^2 heta} = left( rac{cos heta}{sin heta} ight)^2 = cot^2 heta$$
Step 2: Substitute Given Value $cot heta = rac{3}{4}$
$$cot^2 heta = left( rac{3}{4} ight)^2 = rac{9}{16}$$
Step 3: Part (ii)
$$cot^2 heta = left( rac{3}{4} ight)^2 = rac{9}{16}$$
(i) $ rac{9}{16}$; (ii) $ rac{9}{16}$.
Question 8 NCERT Ex 8.1
If $3 cot A = 4$, check whether $ rac{1 - an^2 A}{1 + an^2 A} = cos^2 A - sin^2 A$ or not.
Step 1: Find Triangle Sides
$$3 cot A = 4 implies cot A = rac{4}{3} implies an A = rac{3}{4}$$ Let Base $= 4k$, Perpendicular $= 3k$.
Hypotenuse $H = sqrt{(4k)^2 + (3k)^2} = sqrt{16k^2 + 9k^2} = sqrt{25k^2} = 5k$.
$$cos A = rac{4}{5}, quad sin A = rac{3}{5}$$
Step 2: Evaluate Left Hand Side (LHS)
$$ ext{LHS} = rac{1 - an^2 A}{1 + an^2 A} = rac{1 - left( rac{3}{4} ight)^2}{1 + left( rac{3}{4} ight)^2} = rac{1 - rac{9}{16}}{1 + rac{9}{16}} = rac{ rac{7}{16}}{ rac{25}{16}} = rac{7}{25}$$
Step 3: Evaluate Right Hand Side (RHS)
$$ ext{RHS} = cos^2 A - sin^2 A = left( rac{4}{5} ight)^2 - left( rac{3}{5} ight)^2 = rac{16}{25} - rac{9}{25} = rac{7}{25}$$
Yes, $ ext{LHS} = ext{RHS} = rac{7}{25}$.
Question 9 NCERT Ex 8.1
In $Delta ABC$, right-angled at $B$, if $ an A = 1$, find the value of:

(i) $sin A cos C + cos A sin C$
(ii) $cos A cos C - sin A sin C$
Step 1: Find Ratios for Angles $A$ and $C$
Given $ an A = 1 = rac{BC}{AB} implies BC = AB = k$.
Hypotenuse $AC = sqrt{AB^2 + BC^2} = sqrt{k^2 + k^2} = ksqrt{2}$.
Therefore: $$sin A = rac{BC}{AC} = rac{1}{sqrt{2}}, quad cos A = rac{AB}{AC} = rac{1}{sqrt{2}}$$ $$sin C = rac{AB}{AC} = rac{1}{sqrt{2}}, quad cos C = rac{BC}{AC} = rac{1}{sqrt{2}}$$
Step 2: Evaluate (i)
$$sin A cos C + cos A sin C = left( rac{1}{sqrt{2}} ight)left( rac{1}{sqrt{2}} ight) + left( rac{1}{sqrt{2}} ight)left( rac{1}{sqrt{2}} ight) = rac{1}{2} + rac{1}{2} = 1$$
Step 3: Evaluate (ii)
$$cos A cos C - sin A sin C = left( rac{1}{sqrt{2}} ight)left( rac{1}{sqrt{2}} ight) - left( rac{1}{sqrt{2}} ight)left( rac{1}{sqrt{2}} ight) = rac{1}{2} - rac{1}{2} = 0$$
(i) $1$; (ii) $0$.
Question 10 NCERT Ex 8.1
In $Delta PQR$, right-angled at $Q$, $PR + QR = 25 ext{ cm}$ and $PQ = 5 ext{ cm}$. Determine the values of $sin P, cos P$ and $ an P$.
Step 1: Determine Side Lengths using Pythagoras Theorem
Let $QR = x ext{ cm}$. Then $PR = (25 - x) ext{ cm}$.
In right $Delta PQR$: $$PR^2 = PQ^2 + QR^2$$ $$(25 - x)^2 = 5^2 + x^2$$ $$625 - 50x + x^2 = 25 + x^2$$ $$625 - 25 = 50x implies 600 = 50x implies x = rac{600}{50} = 12 ext{ cm}$$ Hence, $QR = 12 ext{ cm}$ and $PR = 25 - 12 = 13 ext{ cm}$.
Step 2: Compute Trigonometric Ratios for Angle $P$
$$sin P = rac{QR}{PR} = rac{12}{13}$$ $$cos P = rac{PQ}{PR} = rac{5}{13}$$ $$ an P = rac{QR}{PQ} = rac{12}{5}$$
$sin P = rac{12}{13}$, $cos P = rac{5}{13}$, $ an P = rac{12}{5}$.
Question 11 NCERT Ex 8.1
State whether the following are true or false. Justify your answer:

(i) The value of $ an A$ is always less than 1.
(ii) $sec A = rac{12}{5}$ for some value of angle $A$.
(iii) $cos A$ is the abbreviation used for the cosecant of angle $A$.
(iv) $cot A$ is the product of $cot$ and $A$.
(v) $sin heta = rac{4}{3}$ for some angle $ heta$.
Part (i)
False. Since $ an A = rac{ ext{Perpendicular}}{ ext{Base}}$, the perpendicular can be greater than the base in a right triangle. For instance, $ an 60^circ = sqrt{3} approx 1.732 > 1$.
Part (ii)
True. $sec A = rac{ ext{Hypotenuse}}{ ext{Adjacent}}$. Since hypotenuse is always greater than or equal to adjacent side, $sec A geq 1$. Here $ rac{12}{5} = 2.4 > 1$, which is completely valid.
Part (iii)
False. $cos A$ is the abbreviation used for the cosine of angle $A$. The abbreviation for cosecant is $ ext{cosec } A$.
Part (iv)
False. $cot A$ denotes the cotangent ratio of angle $A$. It is a single unified mathematical symbol; '$cot$' separated from $A$ has no meaning.
Part (v)
False. $sin heta = rac{ ext{Perpendicular}}{ ext{Hypotenuse}}$. Since hypotenuse is the longest side in a right-angled triangle, $sin heta leq 1$. Here $ rac{4}{3} approx 1.33 > 1$, which is impossible.
(i) False, (ii) True, (iii) False, (iv) False, (v) False.
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