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Trigonometric Ratios | Trigonometric Ratios Exercise 17

Question 1

(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

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(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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Question 1

(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

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

(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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