- Rational and Irrational Numbers
- Compound Interest
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- Quadratic Equations
- Indices
- Logarithms
- Triangles
- Mid Point Theorem
- Pythagoras Theorem
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- Mensuration
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- Trigonometric Ratios and Standard Angles
- Coordinate Geometry
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ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios
(a) From the figure (1) given below, find the values of: (i) sin θ
(ii) cos θ
(iii) tan θ
(iv) cot θ
(v) sec θ
(vi) cosec θ
(b) From the figure (2) given below, find the values of:
(i) sin θ
(ii) cos θ
(iii) tan θ
(iv) cot θ
(v) sec θ
(vi) cosec θ
(a) From right-angled triangle OMP,
By Pythagoras theorem, we get
\begin{array}{l} O P^{2}=O M^{2}+M P^{2} \\ M P^{2}=O P^{2}+O M^{2} \\ M P^{2}=(15)^{2}-(12)^{2} \\ M P^{2}=225-144 \\ M P^{2}=81 \\ M P^{2}=9^{2} \end{array}
MP = 9
(i) sin θ = MP/OP
= 9/15
= 3/5
(ii) cos θ = OM/OP
= 12/15
= 4/5
(iii) tan θ = MP/OP
= 9/12
= ¾
(iv) cot θ = OM/MP
= 12/9
= 4/3
(v) sec θ = OP/OM
= 15/12
= 5/4
(vi) cosec θ = OP/MP
= 15/9
= /3
(b) From the right-angled triangle ABC,
By Pythagoras theorem, we get
\begin{array}{l} A B^{2}=A C^{2}+B C^{2} \\ A B^{2}=(12)^{2}+(5)^{2} \\ A B^{2}=144+25 \\ A B^{2}=169 \\ A B^{2}=13^{2} \\ A B=13 \end{array}
(i) sin A = BC/AB
= 5/13
(ii) cos A = AC/AB
= 12/13
\begin{aligned} &\text { (iii) } \sin ^{2} A+\cos ^{2} A=(B C / A B)^{2}+(A C / A B)^{2}\\ &=(5 / 13)^{2}+(12 / 13)^{2}\\ &=(25 / 169)+(144 / 169)\\ &=(25+144) / 169\\ &=169 / 169\\ &=1\\ &\sin ^{2} A+\cos ^{2} A=1 \end{aligned}
\begin{array}{l} \text { (iv) } \operatorname{Sec}^{2} A-\tan ^{2} A=(A B / A C)^{2}-(B C / A C)^{2} \\ =(13 / 12)^{2}-(5 / 12)^{2} \\ =(169 / 144)-(25 / 144) \\ =(169-25) / 144 \\ =144 / 144 \\ =1 \\ \operatorname{Sec}^{2} A-\tan ^{2} A=1 \end{array}
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