## Maharashtra State Board Class 9 Maths Solutions Chapter 8 Trigonometry Practice Set 8.2

Question 1.

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

Solution:

i. cos θ = \(\frac { 35 }{ 37 }\) …(i) )[Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ BC = 35k and AC = 37k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

∴ (37k)^{2} = AB^{2}+ (35k)^{2}

1369k^{2} = AB^{2} + 1225k^{2}

AB2 = 1369k^{2} – 1225k^{2}

= 144k^{2}

AB = 144k^{2}

AB = \(\sqrt { 2ghK }\)^{2} … [Taking square root of both sides]

= 12k

ii. sin θ = \(\frac { 11 }{ 61 }\) …..(i) [Given]

In right angled ∆ABC, ∠C = θ.

Let the common multiple be k.

AB = 11k and AC = 61k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

∴ (61k)^{2} = (11k)^{2} + BC^{2}

∴ 3721k^{2} = 121k^{2} + BC^{2}

∴ BC^{2} = 3721k^{2} – 121k^{2} = 3600k^{2}

BC = \(\sqrt { 3600{ k }^{ 2 } }\) .. .[Taking square root of both sides]

= 60k

iii. tan θ = 1 = \(\frac { 1 }{ 1 }\) ..(i) [Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ AB = 1k and BC = 1k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

= K^{2} + K^{2}

= 2K^{2}

∴ AC = \(\sqrt { 2{ k }\)

iv. sin θ = \(\frac { 1 }{ 2 }\) ..(i) [Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ AB = 1k and BC = 2k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

∴ 2K^{2} = K^{2} + BC^{2}

∴ 4K^{2} = K^{2} + BC^{2}

∴ BC^{2} = 4K^{2} – K^{2} = 3K^{2}

∴ BC = \(\sqrt { 3{ k }^{ 2 } }\) .. .[Taking square root of both sides]

= \(\sqrt { 3{ k }\)

v. cos θ = \(\frac { 1 }{ \sqrt { 3 } } \) ..(i) [Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ AB = 1k and BC = √3k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

∴ (√3K)^{2} = AB^{2} + K^{2}

∴ 3K^{2} = 3K^{2} – K^{2} = 2K^{2}

∴ AB = \(\sqrt { 2{ k }^{ 2 } }\) .. .[Taking square root of both sides]

AB = √2K

vi. cos θ = \(\frac { 21 }{ \sqrt { 20 } } \) ..(i) [Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ AB = 21k and BC = 20k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

= (21)K^{2} + (20K)^{2}

= 441K^{2} – 400^{2}

= 841K^{2}

∴ AB = \(\sqrt { 841{ k }^{ 2 } }\) .. .[Taking square root of both sides]

= 29K

vii. tan θ = \(\frac { 8 }{ 15 } \) ..(i) [Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ AB = 8k and BC = 15k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

= (8)K^{2} + (15K)^{2}

= 64K^{2} – 225^{2}

= 289K^{2}

∴ AC = \(\sqrt { 289{ k }^{ 2 } }\) .. .[Taking square root of both sides]

= 17K

viii. sin θ = \(\frac { 3 }{ 5 } \) ..(i) [Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ AB = 3k and AC = 5k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

∴ (5)K^{2}= (3)K^{2} + BC^{2}

∴ 25K^{2} = 9K^{2} – 225^{2}

∴ BC^{2} = 25K^{2} – 9K^{2}

∴ BC = \(\sqrt { 16{ k }^{ 2 } }\) .. .[Taking square root of both sides]

= 4K

ix. tan θ = \(\frac { 1 }{ 2\sqrt { 2 } }\) ..(i) [Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ AB = 1k and AC = 2√2 k

Now, AC^{2} = AB^{2} + BC^{2} …[Pythagoras theorem]

= K^{2} + (2√2 k )^{2}

= K^{2} – 225^{2}

= 25K^{2} + 8K^{2}

= 9K^{2}

∴ AC = \(\sqrt { 9{ k }^{ 2 } }\) .. .[Taking square root of both sides]

= 3K

Question 2.

Find the values of:

i. 5 sin 30° + 3 tan 45°

ii. \(\frac { 4 }{ 5 }\)tan^{2} 60° + 3 sin^{2} 60°

iii. 2 sin 30° + cos 0° + 3 sin 90°

iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)

v. cos^{2} 45° + sin^{2} 30°

vi. cos 60° x cos 30° + sin 60° x sin 30°

Solution:

i. sin 30° = \(\frac { 1 }{ 2 }\) and tan 45° = 1

ii. \(\frac { 4 }{ 5 }\)tan^{2} 60° + 3 sin^{2} 60°

iii. 2 sin 30° + cos 0° + 3 sin 90°

2 sin 30° + cos0° + 3 sin 90° = 2 (\(\frac { 1 }{ 2 }\)) + 1 + 3(1)

= 1 + 1 + 3

∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

iv. \(\frac{\tan 60^{\circ}}{\sin 60^{\circ}+\cos 60^{\circ}}\)

v. cos^{2} 45° + sin^{2} 30°

vi. cos 60° x cos 30° + sin 60° x sin 30°

Question 3.

If sin θ = \(\frac { 4 }{ 5 }\) , then find cos θ.

Solution:

sin θ = \(\frac { 4 }{ 5 }\) .. .(i)[Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ AB = 4k and AC = 5k

Now, AC^{2} = AB^{2} + BC^{2} … [Pythagoras theorem]

∴ (5 k)^{2} = (4k)^{2} + BC^{2}

∴ 25k^{2} = 16k^{2} + BC^{2}

∴ BC^{2} = 25k^{2} – 16k^{2} = 9k^{2}

∴ BC = \(\sqrt { 9{ k }^{ 2 } }\) . .[Taking square root of both sides]

= 3k

Question 4.

If cos θ = \(\frac { 15 }{ 17 }\) , then find sin θ.

Solution:

cos θ = \(\frac { 15 }{ 17 }\) .. .(i)[Given]

In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.

∴ BC = 15k and AC = 17k

Now, AC^{2} = AB^{2} + BC^{2} … [Pythagoras theorem]

∴ (17 k)^{2} = AB^{2} + (15K)^{2}

∴ 289k^{2} = AB^{2} + 225^{2}

∴ AB^{2} = 289k^{2} – 225k^{2}

= 64k^{2}

∴ AB = \(\sqrt { 64{ k }^{ 2 } }\) . .[Taking square root of both sides]

= 8k

**Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.2 Intext Questions and Activities**

Question 1.

In right angled ∆PQR, ∠Q = 900. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.

i. sin θ = cos (90 – θ)

ii. cos θ = sin (90 – θ)

iii. sin 30° = cos (90° – 30°) = cos 60°

iv. cos 30° = sin (90° – 30°) = sin 60° (Textbook pg. no. 107)

Solution:

In ∆PQR, ∠Q = 90°, ∠P = θ

∴ ∠R = 90 – θ

i. sin θ = cos (90 – θ)

ii. cos θ = sin (90 – θ)

iii. Let ∠P = θ = 30°

∴ ∠R = 90° – 30°

sin 30° = cos (90° – 30°) … [From (i) and (ii)]

sin 30° = cos 60°

iv. cos 30° = sin (90° – 30°) = sin 60°

∴ cos 30° = sin (90° – 30°) .,.[From (i) and (ii)]

∴ cos 30° = sin 60°

Question 2.

In right angled ∆PQR, ∠Q = 90°, ∠R = θ and if sin θ = \(\frac { 5 }{ 13 }\), then find cos θ and tan θ. (Textbook pg. no. 110)

Solution:

i. Take the given trigonometric ratio as 13k equation (i).

sin θ = \(\frac { 5 }{ 13 }\) .. .(i)[Given]

By using the definition write the trigonometric ratio of sin O and take it as equation (ii).

In right angled ∆PQR, ∠R = θ

Let the common multiple be k.

∴ PQ = 5k and PR = 13k

Find QR by using Pythagoras theorem.

PR^{2} = PQ^{2} + QR^{2} … [Pythagoras theorem]

∴ (13k)^{2} = (5k)^{2} + QR^{2}

∴ 169k^{2} = 25k^{2} + QR^{2}

∴ QR^{2} = 169k^{2} – 25k^{2}

= 144k^{2}

∴ QR = \(\sqrt { 144{ k }^{ 2 } }\) . . . [Taking square root of both sides]

= 12k

Question 3.

While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)

Solution:

\(\frac { PQ }{ PR }\) = \(\frac { 5 }{ 13 }\) … [Given]

Here, the ratio of the lengths of sides PQ and PR is 5 : 13.

The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving.

Question 4.

While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Tcxtbook pg. no. 111)

Solution:

Yes, we can take lengths of PQ and PR as 5 and 13.

In that case, we will have to take k = 1 and solve the problem accordingly.

Question 5.

Verify that the equation ‘sin^{2} θ + cos^{2} θ = 1’ is true when θ = 0° or θ = 90°.

(Textbook pg. no. 112)

Solution:

sin^{2} θ + cos^{2} θ = 1

i. lf θ = 0°,

LH.S. = sin^{2} θ + cos^{2} θ

= sin^{2} 0° + cos^{2} 0°

= 0 + 1 …[∵ sin 0° = 0, cos 0° = 1]

= R.H.S.

∴ sin^{2} θ + cos^{2} θ = 1

ii. If θ = 90°,

L.H.S.= sin^{2} θ +cos^{2} θ

= sin^{2} 90° + cos^{2} 90°

= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0]

= 1

= R.H.S.

∴ sin^{2} θ + cos^{2} θ = 1