Question

In right angled △PMO, ∠PMO = 90° and ∠POM = θ. Find the following ratio.

(i) sinθ (ii) cosθ (iii) tanθ​

Answer 1

Step-by-step explanation:

sin =pm/po

cos=mo/po

tan=pm/mo

Answer 2

Answer:

Question

QuestionBookmark

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ=

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ= 25

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ= 2524

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ= 2524

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ= 2524 , find sinθandtanθ Similarly , find (sin

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ= 2524 , find sinθandtanθ Similarly , find (sin 2

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ= 2524 , find sinθandtanθ Similarly , find (sin 2 θ) and (cos

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ= 2524 , find sinθandtanθ Similarly , find (sin 2 θ) and (cos 2

QuestionBookmarkIn right angles ΔLMN If ∠N=θ,∠M=90 o ,cosθ= 2524 , find sinθandtanθ Similarly , find (sin 2 θ) and (cos 2 θ)

Step-by-step explanation:

Solution

In right angles ΔLMN

∠N=θ,

cosθ=

25

24

⇒

LN

MN

=

25

24

Let MN=24K and LN=25K .

Using Pythagoras theorem , we have

LN

2

=LM

2

+MN

2

⇒(25k)

2

=LM

2

+(24k

2

)

⇒LM

2

=625K

2

−576k

2

=49k

2

⇒LM

2

=(7k)

2

⇒LM=7k

Now,

sinθ=

LN

LM

=

25k

7k

=

25

7

tanθ=

NM

LM

=

24k

7k

=

24

7

Also ,

sin

2

θ=(

25

7

)

2

=

625

49

cos

2

θ=(

25

24

)

2

=

625

576