Find the values of: sin B

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:

(i) tan x

(ii) cos y

(iii) cosec² y – cot² y

(iv) 5/sin x + 3/sin y – 3 cot y.

Answer :

(a) From right angled triangle ABC,

By Pythagoras theorem, we get

BC² = AC² + AB²

⇒ AC² = BC² – AB²

⇒ AC² = (10)² – (6)²

⇒ AC² = 100 – 36

⇒ AC² = 64

⇒ AC² = 8²

⇒ AC = 8

(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

(b) From Figure

AC = 13, CD = 5, BC =21,

BD = BC – CD

= 21 – 5

= 16

From right angled ∆ACD,

AC = AD² + CD²

⇒ AD² = AC² – CD²

⇒ AD² = (13)² – (5)²

⇒ AD² = 169 – 25

⇒ AD² = 144)

⇒ AD² = (12)²

⇒ AD = 12

From right angled ∆ABD,

AB² = AD² + BD²

⇒ AB² = 400

⇒ AB² = (20)²

⇒ AB = 20

(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

cosec² y – cot² y = (5/3)² – (4/3)²

= (25/9) – (16/9)

= (25-16)/9

= 9/9

= 1

Therefore,

cosec² y – cot² 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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