Question

In triangle ABC, AB=c, BC=a, CA=b
if b+c : c+a : a+b = 4 : 5 : 6, try to find
sin A : sin b : sin C.
​

Answer 1

Answer:

Step-by-step explanation:

Correct option is

B

AE  

2

=  

9

73

​  

 

C

tanθ=  

8

3

​  

 

Hence

BD=DE=EC=  

3

5

​  

 and △ABC is a right angled triangle.

Now  

cosC=  

2.AC.EC

AC  

2

+EC  

2

−AE  

2

 

​  

 

Or  

AE  

2

=AC  

2

+EC  

2

−2.AC.EC.cosC

Or

=4  

2

+(  

3

5

​  

)  

2

−2.8.  

3

5

​  

.cosC

=16+  

9

25

​  

−  

3

2×4×5

​  

.  

5

4

​  

 

Or

=  

9

169

​  

−  

3

32

​  

 

=  

9

169−96

​  

 

=  

9

73

​  

.

Hence

AE  

2

=  

9

73

​  

.

cosθ=  

2.AE.4

AE  

2

+4  

2

−  

3

5

​  

 

2

 

​  

 

=  

73

​  

 

8

​  

.

Hence

secθ=  

8

73

​  

 

​  

 

Now  

tanθ=  

sec  

2

θ−1

​  

 

=  

64

73

​  

−1

​  

 

=  

8

3

​  

.

solution

Expand-image

Correct option is

B

AE  

2

=  

9

73

​  

 

C

tanθ=  

8

3

​  

 

Hence

BD=DE=EC=  

3

5

​  

 and △ABC is a right angled triangle.

Now  

cosC=  

2.AC.EC

AC  

2

+EC  

2

−AE  

2

 

​  

 

Or  

AE  

2

=AC  

2

+EC  

2

−2.AC.EC.cosC

Or

=4  

2

+(  

3

5

​  

)  

2

−2.8.  

3

5

​  

.cosC

=16+  

9

25

​  

−  

3

2×4×5

​  

.  

5

4

​  

 

Or

=  

9

169

​  

−  

3

32

​  

 

=  

9

169−96

​  

 

=  

9

73

​  

.

Hence

AE  

2

=  

9

73

​  

.

cosθ=  

2.AE.4

AE  

2

+4  

2

−  

3

5

​  

 

2

 

​  

 

=  

73

​  

 

8

​  

.

Hence

secθ=  

8

73

​  

 

​  

 

Now  

tanθ=  

sec  

2

θ−1

​  

 

=  

64

73

​  

−1

​  

 

=  

8

3

​  

 

solution

Expand-image