**Chapter 8 Trigonometry Practice Set 8.2**

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:

Solution:

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}

= 12k

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}

= 60k

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 = \(\sqrt { 2{ k }\)

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}

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

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 = √2K

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}

= 29K

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}

= 17K

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}

= 4K

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}

= 3K

**Question 2.Find the values of:i. 5 sin 30° + 3 tan 45°ii. 45tan**

^{2}60° + 3 sin

^{2}60°iii. 2 sin 30° + cos 0° + 3 sin 90°v. cos

^{2}45° + sin

^{2}30°vi. cos 60° x cos 30° + sin 60° x sin 30°Solution:

i. sin 30° = 1/2 and tan 45° = 1

iii. 2 sin 30° + cos 0° + 3 sin 90°

= 1 + 1 + 3

∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

v. cos^{2} 45° + sin^{2} 30°

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

**Question 3.**

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}

= 3k

**Question 4.**

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}

= 8k

**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.**

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}

= 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:**

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