Question

From the figure given, find the values of (i) tan x (ii) cos y (iii) sin y​

Answer 1

Answer:

wheres the figure

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

From right angled ΔACD,

By Pythagoras' theorem, we get

AC

2

=AD

2

+CD

2

AD

2

=AC

2

−CD

2

AD

2

=(13)

2

−(5)

2

AD

2

=169−25

AD

2

=144

AD

2

=12

2

AD=12

From right angled ΔABD,

By Pythagoras angled ΔABD

By Pythagoras' theorem, we get

AB

2

=AD

2

+BD

2

AB

2

=400

AB

2

=(20)

2

AB=20

(i) tan x = perpendicular/Base (in right-angled ΔACD)

=CD/AD

= 5/12

(ii) cos y = Base/Hypotenuse (in the right-angled ΔABD)

= BD/AB

= (20)/12 – (5/3)

Cot y = Base/Perpendicular (in right ΔABD)

=BD/AB

= 16/20 = 4/5

(iii) ) cos y = Hypotenuse/ perpendicular (in right-angled right-angled ΔABD)

BD/AB

= 20/12

= 5/3

Cot y = Base/Perpendicular (in right ΔABD)

AB/AD

= 16/12

= 4/3

Cosec

2

y−cot

2

y=(5/3)

2

−(4/3)

2

= (25/9) – (16/9)

= (25-16)/9

= 9/9

= 1

Hence, Cosec

2

y−cot

2

y=1

(iv) sin x = Perpendicular/Hypotenuse (in the right-angled  ΔACD)

= AD/AB

= 12/20

= 3/5

Cot y = Base/Perpendicular (in right-angled ΔABD)

= BD/AD

= 16/12From right angled ΔACD,

By Pythagoras' theorem, we get

AC

2

=AD

2

+CD

2

AD

2

=AC

2

−CD

2

AD

2

=(13)

2

−(5)

2

AD

2

=169−25

AD

2

=144

AD

2

=12

2

AD=12

From right angled ΔABD,

By Pythagoras angled ΔABD

By Pythagoras' theorem, we get

AB

2

=AD

2

+BD

2

AB

2

=400

AB

2

=(20)

2

AB=20

(i) tan x = perpendicular/Base (in the right-angled ΔACD)

=CD/AD

= 5/12

(ii) cos y = Base/Hypotenuse (in the right-angled ΔABD)

= BD/AB

= (20)/12 – (5/3)

Cot y = Base/Perpendicular (in right ΔABD)

=BD/AB

= 16/20 = 4/5

(iii) ) cos y = Hypotenuse/ perpendicular (in the right-angled ΔABD)

BD/AB

= 20/12

= 5/3

Cot y = Base/Perpendicular (in right ΔABD)

AB/AD

= 16/12

= 4/3

Cosec

2

y−cot

2

y=(5/3)

2

−(4/3)

2

= (25/9) – (16/9)

= (25-16)/9

= 9/9

= 1

Hence, Cosec

2

y−cot

2

y=1

(iv) sin x = Perpendicular/Hypotenuse (in the right-angled  ΔACD)

= AD/AB

= 12/20

= 3/5

Cot y = Base/Perpendicular (in the right-angled ΔABD)

= 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

= 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