**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 âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ BC = 35k and AC = 37k

Now, AC

^{2}= AB

^{2}+ BC

^{2}â€¦[Pythagoras theorem]

âˆ´ (37k)

^{2}= AB

^{2}+ (35k)

^{2}

1369k

^{2}= AB

^{2}+ 1225k

^{2}

AB2 = 1369k

^{2}â€“ 1225k

^{2}

= 144k

^{2}

AB = 144k

^{2}

= 12k

In right angled âˆ†ABC, âˆ C = Î¸.

Let the common multiple be k.

AB = 11k and AC = 61k

Now, AC^{2} = AB^{2} + BC^{2} â€¦[Pythagoras theorem]

âˆ´ (61k)^{2} = (11k)^{2} + BC^{2}

âˆ´ 3721k^{2} = 121k^{2} + BC^{2}

âˆ´ BC^{2} = 3721k^{2} â€“ 121k^{2} = 3600k^{2}

= 60k

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ AB = 1k and BC = 1k

Now, AC^{2} = AB^{2} + BC^{2} â€¦[Pythagoras theorem]

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

= 2K^{2}

âˆ´ AC = \(\sqrt { 2{ k }\)

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ AB = 1k and BC = 2k

Now, AC^{2} = AB^{2} + BC^{2} â€¦[Pythagoras theorem]

âˆ´ 2K^{2} = K^{2} + BC^{2}

âˆ´ 4K^{2} = K^{2} + BC^{2}

âˆ´ BC^{2} = 4K^{2} â€“ K^{2} = 3K^{2}

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

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ AB = 1k and BC = âˆš3k

Now, AC^{2} = AB^{2} + BC^{2} â€¦[Pythagoras theorem]

âˆ´ (âˆš3K)^{2} = AB^{2} + K^{2}

âˆ´ 3K^{2} = 3K^{2} â€“ K^{2} = 2K^{2}

AB = âˆš2K

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ AB = 21k and BC = 20k

Now, AC^{2} = AB^{2} + BC^{2} â€¦[Pythagoras theorem]

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

= 441K^{2} â€“ 400^{2}

= 841K^{2}

= 29K

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ AB = 8k and BC = 15k

Now, AC^{2} = AB^{2} + BC^{2} â€¦[Pythagoras theorem]

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

= 64K^{2} â€“ 225^{2}

= 289K^{2}

= 17K

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ AB = 3k and AC = 5k

Now, AC^{2} = AB^{2} + BC^{2} â€¦[Pythagoras theorem]

âˆ´ (5)K^{2}= (3)K^{2} + BC^{2}

âˆ´ 25K^{2} = 9K^{2} â€“ 225^{2}

âˆ´ BC^{2} = 25K^{2} â€“ 9K^{2}

= 4K

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ AB = 1k and AC = 2âˆš2 k

Now, AC^{2} = AB^{2} + BC^{2} â€¦[Pythagoras theorem]

= K^{2} + (2âˆš2 k )^{2}

= K^{2} â€“ 225^{2}

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

= 9K^{2}

= 3K

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

^{2}60Â° + 3 sin

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

^{2}45Â° + sin

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

i. sin 30Â° = 1/2 and tan 45Â° = 1

iii. 2 sin 30Â° + cos 0Â° + 3 sin 90Â°

= 1 + 1 + 3

âˆ´ 2 sin 30Â° + cos 0Â° + 3 sin 90Â° = 5

v. cos^{2} 45Â° + sin^{2} 30Â°

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

**Question 3.**

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ AB = 4k and AC = 5k

Now, AC

^{2}= AB

^{2}+ BC

^{2}â€¦ [Pythagoras theorem]

âˆ´ (5 k)

^{2}= (4k)

^{2}+ BC

^{2}

âˆ´ 25k

^{2}= 16k

^{2}+ BC

^{2}

âˆ´ BC

^{2}= 25k

^{2}â€“ 16k

^{2}= 9k

^{2}

= 3k

**Question 4.**

In right angled âˆ†ABC,

âˆ C = Î¸.

Let the common multiple be k.

âˆ´ BC = 15k and AC = 17k

Now, AC

^{2}= AB

^{2}+ BC

^{2}â€¦ [Pythagoras theorem]

âˆ´ (17 k)

^{2}= AB

^{2}+ (15K)

^{2}

âˆ´ 289k

^{2}= AB

^{2}+ 225

^{2}

âˆ´ AB

^{2}= 289k

^{2}â€“ 225k

^{2}

= 64k

^{2}

= 8k

**Intext Questions and Activities**

**Question 1.In right angled âˆ†PQR, âˆ Q = 900. Therefore âˆ P and âˆ R are complementary angles of each other. Verify the following ratios.i. sin Î¸ = cos (90 â€“ Î¸)ii. cos Î¸ = sin (90 â€“ Î¸)iii. sin 30Â° = cos (90Â° â€“ 30Â°) = cos 60Â°iv. cos 30Â° = sin (90Â° â€“ 30Â°) = sin 60Â° (Textbook pg. no. 107)Solution:**

In âˆ†PQR, âˆ Q = 90Â°, âˆ P = Î¸

âˆ´ âˆ R = 90 â€“ Î¸

i. sin Î¸ = cos (90 â€“ Î¸)

ii. cos Î¸ = sin (90 â€“ Î¸)

iii. Let âˆ P = Î¸ = 30Â°

âˆ´ âˆ R = 90Â° â€“ 30Â°

sin 30Â° = cos (90Â° â€“ 30Â°) â€¦ [From (i) and (ii)]

sin 30Â° = cos 60Â°

iv. cos 30Â° = sin (90Â° â€“ 30Â°) = sin 60Â°

âˆ´ cos 30Â° = sin (90Â° â€“ 30Â°) .,.[From (i) and (ii)]

âˆ´ cos 30Â° = sin 60Â°

**Question 2.**

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

In right angled âˆ†PQR, âˆ R = Î¸

Let the common multiple be k.

âˆ´ PQ = 5k and PR = 13k

Find QR by using Pythagoras theorem.

PR

^{2}= PQ

^{2}+ QR

^{2}â€¦ [Pythagoras theorem]

âˆ´ (13k)

^{2}= (5k)

^{2}+ QR

^{2}

âˆ´ 169k

^{2}= 25k

^{2}+ QR

^{2}

âˆ´ QR

^{2}= 169k

^{2}â€“ 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 â€˜sin ^{2} Î¸ + cos^{2} Î¸ = 1â€™ is true when Î¸ = 0Â° or Î¸ = 90Â°.(Textbook pg. no. 112)Solution:**
sin

^{2}Î¸ + cos

^{2}Î¸ = 1

i. lf Î¸ = 0Â°,

LH.S. = sin

^{2}Î¸ + cos

^{2}Î¸

= sin

^{2}0Â° + cos

^{2}0Â°

= 0 + 1 â€¦[âˆµ sin 0Â° = 0, cos 0Â° = 1]

= R.H.S.

âˆ´ sin

^{2}Î¸ + cos

^{2}Î¸ = 1

ii. If Î¸ = 90Â°,

L.H.S.= sin^{2} Î¸ +cos^{2} Î¸

= sin^{2} 90Â° + cos^{2} 90Â°

= 1 + 0 â€¦ [ âˆµ sin 90Â° = 1, cos 90Â° = 0]

= 1

= R.H.S.

âˆ´ sin^{2} Î¸ + cos^{2} Î¸ = 1