Question

In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

sin θ		`11/61`		`1/2`				`3/5`
cos θ	`35/37`				`1/sqrt3`
tan θ			`1`			`21/20`	`8/15`		`1/(2sqrt2)`

Answer 1

sin θ	`bb(12/37)`	`11/61`	`bb(1/sqrt2)`	`1/2`	`bb(sqrt2/sqrt3)`	`bb(21/29)`	`bb(8/17)`	`3/5`	`bb(1/3)`
cos θ	`35/37`	`bb(60/61)`	`bb(1/sqrt2)`	`bb(sqrt3/2)`	`1/sqrt3`	`bb(20/29)`	`bb(15/17)`	`bb(4/5)`	`bb((2sqrt2)/3)`
tan θ	`bb(12/35)`	`bb(11/60)`	1	`bb(1/sqrt3)`	`bb(sqrt2)`	`21/20`	`8/15`	`bb(3/4)`	`1/(2sqrt2)`

Explanation:

(i) cos θ = `35/37`            ...(i)[Given]

In right angled ΔABC,

∠C = θ

cos θ = `"Adjacent side of θ"/"Hypotenuse"`

∴ cos θ = `"BC"/"AC"`         ...(ii)

∴ `"BC"/"AC" = 35/37`        ...[From (i) and (ii)]

Let the common multiple be k.

∴ BC = 35k and AC = 37k

Now, AC² = AB² + BC²    ...[Pythagoras theorem]

∴ (37k)² = AB² + (35k)²

1369k² = AB² + 1225k²

AB² = 1369k² – 1225k²

AB² = 144k²

AB² = (12k)²

Taking square root of both sides,

AB = 12k

∴ sin θ = `"Opposite side of θ"/"Hypotenuse" = "AB"/"AC" = "12k"/"37k" = 12/37`.

tan θ = `"Opposite side of θ"/"Adjacent side of θ" =  "AB"/"BC" = "12k"/"35k" = 12/35`.

(ii) sin θ = `11/61`       ...(i)[Given]

In right angled ΔABC, ∠C = θ

sin θ = `"Opposite side of θ"/"Hypotenuse"`

∴ sin θ = `"AB"/"AC"`        ...(ii)

∴ `"AB"/"AC" = 11/61`       ...[From (i) and (ii)]

Let the common multiple be k.

AB = 11k and AC = 61k

Now, AC² = AB² + BC²         ...[Pythagoras theorem]

∴ (61k)² = (11k)² + BC²

∴ 3721k² = 121k² + BC²

∴ BC² = 3721k² – 121k²

∴ BC² = 3600k²

∴ BC² = (60k)²

Taking square root of both sides,

BC = 60k

∴ cos θ = `"Adjacent side of θ"/"Hypotenuse" = "BC"/"AC" = "60k"/"61k" = 60/61`

tan θ = `"Opposite side of θ"/"Adjacent side of θ" =  "AB"/"BC" = "11k"/"60k" = 11/60`.

(iii) tan θ = 1    ...(i) [Given]

In right angled ∆ABC, ∠C = θ

tan θ = `"Opposite side of θ"/"Adjacent side of θ"`

∴ tan θ = `"AB"/"BC"`    ...(ii)

∴ `"AB"/"BC" = 1/1`      ...[From (i) and (ii)]

Let the common multiple be k.

∴ AB = 1k and BC = 1k

Now, AC² = AB² + BC²           ...[Pythagoras theorem]

∴ AC² = k² + k²  

∴ AC² = 2k²

Taking square root of both sides,

∴ AC = `sqrt2k`

∴ sin θ = `"Opposite side of θ"/"Hypotenuse" = "AB"/"AC" = "1k"/(sqrt2"k") = 1/sqrt2`

∴ cos θ = `"Adjacent side of θ"/"Hypotenuse" = "BC"/"AC" = "1k"/(sqrt2"k") = 1/sqrt2`

(iv) sin θ = `1/2`      ...(i)[Given]

In right angled ∆ABC, ∠C = θ

sin θ = `"Opposite side of θ"/"Hypotenuse"`

∴ sin θ = `"AB"/"AC"`      ...(ii)

∴ `"AB"/"AC" = 1/2`        ...[From (i) and (ii)]

Let the common multiple be k.

∴ AB = 1k and AC = 2k

Now, AC² = AB² + BC²         ...[Pythagoras theorem]

∴ (2k)² = k² + BC²

∴ 4k² = k² + BC²

∴ BC² = 4k² – k²

∴ BC² = 3k²

Taking square root of both sides,

∴ BC = `sqrt3"k"`

∴ cos θ = `"Adjacent side of θ"/"Hypotenuse" = "BC"/"AC" = (sqrt3"k")/(2"k") = (sqrt3)/2`

tan θ = `"Opposite side of θ"/"Adjacent side of θ" = "AB"/"BC" = (1"k")/(sqrt3"k") = 1/sqrt3`

(v) cos θ = `1/sqrt3`      ...(i)[Given]

In right angled ∆ABC, ∠C = θ

∴ cos θ = `"Adjacent side of θ"/"Hypotenuse"`

∴ cos θ = `"BC"/"AC"`       ...(ii)

∴ `"BC"/"AC" = 1/sqrt3`     ...[From (i) and (ii)]

 Let the common multiple be k.

∴ `"AB" = 1"k" and "BC" = sqrt3"k"`

Now, AC² = AB² + BC²     ...[Pythagoras theorem]

`∴ (sqrt3"k")^2 = "AB"^2 + "k"^2`

∴ 3k² = AB² – k²

∴ AB² = 3k² – k²

∴ AB² = 2k²

Taking square root of both sides,

AB = `sqrt2`k 

∴ sin θ = `"Opposite side of θ"/"Hypotenuse" = "AB"/"AC" = (sqrt2"k")/(sqrt3"k") = sqrt2/sqrt3` 

tan θ = `"Opposite side of θ"/"Adjacent side of θ" = "AB"/"BC" = (sqrt2"k")/(1"k") = sqrt2`

(vi) tan θ = `21/20`             ...(i)[Given]

In right angled ∆ABC, ∠C = θ

tan θ = `"Opposite side of θ"/"Adjacent side of θ"`

∴ tan θ = `"AB"/"BC"`      ...(ii)

∴ `"AB"/"BC" = 21/20`      ...[From (i) and (ii)]

Let the common multiple be k.

∴ AB = 21k and BC = 20k

Now, AC² = AB² + BC²           ...[Pythagoras theorem]

AC² = (21k)² + (20k)²

AC² = 441k² + 400k²

AC² = (841k)²

Taking square root of both sides,

AC = 29k

∴ sin θ = `"Opposite side of θ"/"Hypotenuse" = "AB"/"AC" = (21"k")/(29"k") = 21/29`

cos θ = `"Adjacent side of θ"/"Hypotenuse" = "BC"/"AC" = (20"k")/(29"k") = 20/29` 

(vii) tan θ = `8/15`      ...(i)[Given]

In right angled ∆ABC, ∠C = θ.

tan θ = `"Opposite side of θ"/"Adjacent side of θ"`

∴ tan θ = `"AB"/"BC"`      ...(ii)

∴ `"AB"/"BC" = 8/15`      ...[From (i) and (ii)]

Let the common multiple be k.

∴ AB = 8k and BC = 15k

Now, AC² = AB² + BC²         ...[Pythagoras theorem]

∴ AC² = (8k)² + (15k)² 

∴ AC² = 64k² + 225k²

∴ AC² = (289k)²

Taking square root of both sides,

∴ AC = 17k

∴ sin θ = `"Opposite side of θ"/"Hypotenuse" = "AB"/"AC" = (8"k")/(17"k") = 8/17`

cos θ = `"Adjacent side of θ"/"Hypotenuse" = "BC"/"AC" = (15"k")/(17"k") = 15/17` 

(viii) sin θ = `3/5`       ...(i)[Given]

In right angled ΔABC, ∠C = θ

sin θ = `"Opposite side of θ"/"Hypotenuse"`

∴ sin θ = `"AB"/"AC"`      ...(ii)

∴ `"AB"/"AC" = 3/5`        ...[From (i) and (ii)]

Let the common multiple be k.

∴ AB = 3k and AC = 5k

Now, AC² = AB² + BC²         ...[Pythagoras theorem]

∴ (5k)² = (3k)² + BC²

∴ 25k² = 9k² + BC²

∴ BC² = 25k² – 9k²

∴ BC² = 16k²

Taking square root of both sides,

∴ BC = 4k

∴ cos θ = `"Adjacent side of θ"/"Hypotenuse" = "BC"/"AC" = (4"k")/(5"k") = 4/5`

tan θ = `"Opposite side of θ"/"Adjacent side of θ" = "AB"/"BC" = (3"k")/(4"k") = 3/4`

(ix) tan θ = `1/(2sqrt2)`      ...(i)[Given]

In right angled ∆ABC, ∠C = θ

tan θ = `"Opposite side of θ"/"Adjacent side of θ"`

∴ tan θ = `"AB"/"BC"`      ...(ii)

∴ `"AB"/"BC" = 1/(2sqrt2)`      ...[From (i) and (ii)]

Let the common multiple be k.

∴ AB = 1k and BC = `2sqrt2"k"`

Now, AC² = AB² + BC²         ...[Pythagoras theorem]

∴ `"AC"^2 = (1"k")^2 + (2sqrt2"k")^2 `

∴ AC² = 1k² + 8k²

∴ AC² = (9k)²

Taking square root of both sides,

∴ AC = 3k

∴ sin θ = `"Opposite side of θ"/"Hypotenuse" = "AB"/"AC" = (1"k")/(3"k") = 1/3`

cos θ = `"Adjacent side of θ"/"Hypotenuse" = "BC"/"AC" = (2sqrt2"k")/(3"k") = (2sqrt2)/3`