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ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios
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(a) From the figure (1) given below, find the values of:
(i) sin B
(i) cos C
(iii) sin B + sin C
(iv) sin B cos C + sin C cos B
(b) From the figure (2) given below, find the values of……..
(1) sin B
(ii) cos C
From the right-angled triangle ABC,
By Pythagoras theorem, we get
\begin{array}{l} \mathrm{BC}^{2}=\mathrm{AC}^{2}+\mathrm{AB}^{2} \\ \mathrm{AC}^{2}=\mathrm{BC}^{2}-\mathrm{AB}^{2} \\ \mathrm{AC}^{2}=(10)^{2}-(6)^{2} \\ \mathrm{AC}^{2}=100-36 \\ \mathrm{AC}^{2}=64 \\ \mathrm{AC}^{2}=8^{2} \\ \mathrm{AC}=8 \end{array}
(i) sin B = perpendicular/ hypotenuse
= AC/BC
= 8/10
= 4/5
(ii) cos C = Base/hypotenuse
= AC/BC
= 8/10
= 4/5
(iii) sin B = Perpendicular/hypotenuse
= AC/BC
= 8/10
= 4/5
Sin C = perpendicular/hypotenuse
= AB/BC
= 6/10
= 3/5
Now,
Sin B + sin C = (4/5) + (3/5)
= (4 + 3)/5
= 7/5
(iv) sin B = 4/5
Cos C = 4/5
Sin C = perpendicular/ hypotenuse
= AB/BC
= 6/10
= 3/5
Cos B = Base/Hypotenuse
= AB/BC
= 6/10
= 3/5
sin B cos C + sin C cos B
= (4/5) × (4/5) × (3/5) × (3/5)
= (26/25) × (9/25)
= (16+9)/25
= 25/25
= 1
From Figure
AC = 13, CD = 5, BC =21,
BD = BC – CD
= 21 – 5
= 16
From right angled ∆ACD,
By Pythagoras theorem we get
\begin{array}{l} A C=A D^{2}+C D^{2} \\ A D^{2}=A C^{2}-C D^{2} \end{array}
\begin{array}{l} A D^{2}=(13)^{2}-(5)^{2} \\ A D^{2}=169-25 \\ \left.A D^{2}=144\right) \\ A D^{2}=(12)^{2} \\ A D=12 \end{array}
From right angled ∆ABD,
By Pythagoras angled ∆ABD
By Pythagoras theorem we get
\begin{array}{l} A B^{2}=A D^{2}+B D^{2} \\ A B^{2}=400 \\ A B^{2}=(20)^{2} \\ A B=20 \end{array}
(i) tan x = perpendicular/Base (in right angled ∆ACD)
=CD/AD
= 5/12
(ii) cos y = Base/Hypotenuse (in right angled ∆ABD)
= BD/AB
= (20)/12 – (5/3)
Cot y = Base/Perpendicular (in right ∆ABD)
=BD/AB
= 16/20 = 4/5
(iii) ) cos y = Hypotenuse/ perpendicular (in right angled ∆ABD)
BD/AB
= 20/12
= 5/3
Cot y = Base/Perpendicular (in right ∆ABD)
AB/AD
= 16/12
= 4/3
\begin{array}{l} \operatorname{cosec}^{2} y-\cot ^{2} y=(5 / 3)^{2}-(4 / 3)^{2} \\ =(25 / 9)-(16 / 9) \\ =(25-16) / 9 \\ =9 / 9 \\ =1 \end{array}
\text { Hence, } \operatorname{cosec}^{2} y-\cot ^{2} y=1
(iv) sin x = Perpendicular/Hypotenuse (in right angled ∆ACD)
= AD/AB
= 12/20
= 3/5
Cot y = Base/Perpendicular (in right angled ∆ABD)
= BD/AD
= 16/12
= 4/3
(5/sin x) + (3/sin y) – 3cot y
= 5/(5/13) + 3/(3/5) – 3 × 4/3
= 5 × 13/5 + 3 × 5/3 – 3 × 4/3
= 1 × 13/1 + 1 × 5/1 – 1 × 4/1
= 13 + 5 – 4 = 18 – 4
= 14
Hence , 5/sin x + 3/sin y – 3cot y = 14
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