For ΔABC, express in terms of s, a, b, and c.
Answers
Answered by
2
Solution :
************************************
We know that ,
sinC/2 = √[(s-a)(s-b)]/ab
SinA/2 = √[(s-b)(s-c)]/2
***************************************
Now ,
asin²C/2 + cSin²A/2
=a × [(s-a)(s-b)]/ab + c × [(s-b)(s-c)]/bc
= [(s-a)(s-b)]/b + [(s-b)(s-c)]/b
= [(s-b)/b ] ( s - a + s - c )
= [(s-b)/b ] ( 2s - a - c )
= [ (s-b)/b ] ( a + b + c - a - c )
[ Since , 2s = a + b + c ]
= [ ( s - b )/b ] × b
= s - b
************************************
We know that ,
sinC/2 = √[(s-a)(s-b)]/ab
SinA/2 = √[(s-b)(s-c)]/2
***************************************
Now ,
asin²C/2 + cSin²A/2
=a × [(s-a)(s-b)]/ab + c × [(s-b)(s-c)]/bc
= [(s-a)(s-b)]/b + [(s-b)(s-c)]/b
= [(s-b)/b ] ( s - a + s - c )
= [(s-b)/b ] ( 2s - a - c )
= [ (s-b)/b ] ( a + b + c - a - c )
[ Since , 2s = a + b + c ]
= [ ( s - b )/b ] × b
= s - b
Answered by
0
HELLO DEAR,
Answer:
Step-by-step explanation:
We know that ,
sinC/2 = √[(s-a)(s-b)]/ab
SinA/2 = √[(s-b)(s-c)]/2
Now ,
asin²C/2 + cSin²A/2
=> a * [(s-a)(s-b)]/ab + c × [(s-b)(s-c)]/bc
=> [(s-a)(s-b)]/b + [(s-b)(s-c)]/b
=> [(s-b)/b ] ( s - a + s - c )
=> [(s-b)/b ] ( 2s - a - c )
=> [ (s-b)/b ] ( a + b + c - a - c )
[ we know, 2s = a + b + c ]
=> [ ( s - b )/b ] * b
=> s - b
I HOPE IT'S HELP YOU DEAR,
THANKS
Similar questions