Question

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

Sr. No.	i.	ii.
sin θ	3/5
cos θ
tan θ		1/2√2

Answer 1

i. sin θ = 3/5 ...(i) [Given] 

In right angled ∆ABC, ∠C = θ.

Let the common multiple be k.

∴ AB = 3k and AC = 5k 

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

∴ (5)K² = (3)K² + BC² 

∴ 25K² = 9K² – 225² 

∴ BC² = 25K² – 9K² 

∴ BC² = 25k² - 9k²

∴ BC = √(16k²) .. .[Taking square root of both sides] 

= 4K

ii. tan θ = 1/2√2 ...(i) [Given] 

In right angled ∆ABC, ∠C = θ.

Let the common multiple be k. 

∴ AB = 1k and AC = 2√2 k 

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

= K² + (2√2k)² = K² – 225² = 25K² + 8K² = 9K²

∴ AC = √(9k²) .. .[Taking square root of both sides]

= 3K

Sr. No.	i.	ii.
sin θ	3/5	1/3
cos θ	4/5	2√2/3
tan θ	3/4	1/2√2