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Trigonometric Ratios | Trigonometric Ratios Exercise 17

Question 2

(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

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  • Solution

  • Transcript

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

"hey guys welcome to lido q a video and we neet your lido tutor bringing you this question on your screen question is from the figure given below find the values of sine b cos c sine b plus sine c sine b cos c plus sine c cos b right now we have two angles here so we need the opposite and hypotenuse with respect to both the angles opposite adjacent and hypotenuse so let us write it down right this is angle b and this is angle c so with respect to angle c opposite is a b right adjacent is ac and hypotenuse is bc with respect to both the angles that is now we are the sine and cos so just let let us just write it down sign b is equal to sohcahtoa remember sohcahtoa always so sine is opposite by hypotenuse so a b by b c right and sine c sorry cos b is adjacent by hypotenuse right so we have ac by bc that's all for angle b now with respect sorry this was angle c so let me fix that real quick so this is angle c now with respect to angle b so this is the opposite ac is the opposite and a b is the adjacent right and bc is the hypotenuse therefore sine b will be equal to opposite by hypotenuse and sine okay cos b will be equal to a b by bc okay now let us find the values now in this figure this is 6 this is 10 so by pythagoras theorem ac square is equal to a b square minus sorry b c square minus a b square this is equal to 10 square minus 6 square this is equal to 64. therefore ac is equal to 8. now we know all the sides and we know this this is one two three and four right so let us see number one sine b is equal to sine b is equal to ac by bc from 1 so this is equal to 8 by 10 or 4 by 5 again cos c is equal to so cos c from 2 is equal to ac by bc from 2 so this is equal to 4 by 5 again right number 3 sine b plus sine c is equal to ac by bc plus sine c is equal to a b by bc so this is equal to 6 by 10 plus 4 by 10 sorry 8 by 10 so this is equal to 14 by 10 is equal to 7 by 5. and the last one sine a cos c plus sine c cos b so sine a is sorry sine b is 4 by 5 into 4 by 5 plus sine c is 6 by 10 that is 3 by 5 into cos b is also 3 by 5 so this is equal to 16 plus 9 by 25 this is equal to 1 isn't that easy guys if you still have a doubt please leave a comment below do like the video and subscribe to our channel right now let us look at the part 2 don't go away yet okay part 2 says from the figure given below find the values of sine v and cos c so sine b is this so from the figure if we are taking angle b right so opposite is 80 opposite is 80 so let us do number one sine b is equal to a d by b d right now this is 21 and this is 5 so 21 minus 5 will be 16 80 is 16 80 is 16 now let us uh sorry b d is 16 let us find e d right by pythagoras theorem v d sorry a d square is equal to a c square minus d c square this implies a d square is equal to 169 minus 25 okay so 169 minus 25 is 144 so a d is equal to root over 144 that is equal to 2n therefore sine b will be equal to 12 by 16 so this is equal to 3 by 4. now let us look at number 2 cos c cos c this is angle c so opposite is 80 but cos is car so adjacent adjacent is 5 by hypotenuse is 13. okay yeah okay sign b is not uh yeah sign b is not a d by b d but a d by is a b sorry yeah a d by a b because hypotenuse so this is a d by a b right so a d is 16. and 80 is 12 right so 16 16 square plus 12 square is equal to 400 so this will be 20. right so let me fix that real quick so this is 20 so this will be equal to 3 by 5 right so i'll see you in our next video until then bye guys keep practicing keep flourishing "

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Question 2

(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

  • Solution

  • Transcript

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

"hey guys welcome to lido q a video and we neet your lido tutor bringing you this question on your screen question is from the figure given below find the values of sine b cos c sine b plus sine c sine b cos c plus sine c cos b right now we have two angles here so we need the opposite and hypotenuse with respect to both the angles opposite adjacent and hypotenuse so let us write it down right this is angle b and this is angle c so with respect to angle c opposite is a b right adjacent is ac and hypotenuse is bc with respect to both the angles that is now we are the sine and cos so just let let us just write it down sign b is equal to sohcahtoa remember sohcahtoa always so sine is opposite by hypotenuse so a b by b c right and sine c sorry cos b is adjacent by hypotenuse right so we have ac by bc that's all for angle b now with respect sorry this was angle c so let me fix that real quick so this is angle c now with respect to angle b so this is the opposite ac is the opposite and a b is the adjacent right and bc is the hypotenuse therefore sine b will be equal to opposite by hypotenuse and sine okay cos b will be equal to a b by bc okay now let us find the values now in this figure this is 6 this is 10 so by pythagoras theorem ac square is equal to a b square minus sorry b c square minus a b square this is equal to 10 square minus 6 square this is equal to 64. therefore ac is equal to 8. now we know all the sides and we know this this is one two three and four right so let us see number one sine b is equal to sine b is equal to ac by bc from 1 so this is equal to 8 by 10 or 4 by 5 again cos c is equal to so cos c from 2 is equal to ac by bc from 2 so this is equal to 4 by 5 again right number 3 sine b plus sine c is equal to ac by bc plus sine c is equal to a b by bc so this is equal to 6 by 10 plus 4 by 10 sorry 8 by 10 so this is equal to 14 by 10 is equal to 7 by 5. and the last one sine a cos c plus sine c cos b so sine a is sorry sine b is 4 by 5 into 4 by 5 plus sine c is 6 by 10 that is 3 by 5 into cos b is also 3 by 5 so this is equal to 16 plus 9 by 25 this is equal to 1 isn't that easy guys if you still have a doubt please leave a comment below do like the video and subscribe to our channel right now let us look at the part 2 don't go away yet okay part 2 says from the figure given below find the values of sine v and cos c so sine b is this so from the figure if we are taking angle b right so opposite is 80 opposite is 80 so let us do number one sine b is equal to a d by b d right now this is 21 and this is 5 so 21 minus 5 will be 16 80 is 16 80 is 16 now let us uh sorry b d is 16 let us find e d right by pythagoras theorem v d sorry a d square is equal to a c square minus d c square this implies a d square is equal to 169 minus 25 okay so 169 minus 25 is 144 so a d is equal to root over 144 that is equal to 2n therefore sine b will be equal to 12 by 16 so this is equal to 3 by 4. now let us look at number 2 cos c cos c this is angle c so opposite is 80 but cos is car so adjacent adjacent is 5 by hypotenuse is 13. okay yeah okay sign b is not uh yeah sign b is not a d by b d but a d by is a b sorry yeah a d by a b because hypotenuse so this is a d by a b right so a d is 16. and 80 is 12 right so 16 16 square plus 12 square is equal to 400 so this will be 20. right so let me fix that real quick so this is 20 so this will be equal to 3 by 5 right so i'll see you in our next video until then bye guys keep practicing keep flourishing "

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