for ∆ABC,sin(B+C)/2=..............?
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Answered by
0
the answer is cos A/2
yash842004:
please mark as brainliest...
plzz giv me full solution
please brainliest
for ∆A+B+C=180 sin(B+C/2). SO ..SIN (180-A/2). SO..SIN(90-A/2). SO...COSA/2
dhruvsh ka dekh kar kiya he tune
nahi. maine khud kiya hai
i am also in std 10
Answered by
1
For a ∆ABC,
A+B+C = π
So,
A+B+C/2 = π/2
So,
(B+C)/2 = π/2 - A/2
So,
Sin(B+C)/2 = sin(π/2 - A/2) = cos A/2
Now,
cos A/2 = √ {s(s-a)/bc}
where, s is the semiperimeter of the triangle, and a, b ,c are the lengths of the sides opposite to vertices A,B & C respectively.
A+B+C = π
So,
A+B+C/2 = π/2
So,
(B+C)/2 = π/2 - A/2
So,
Sin(B+C)/2 = sin(π/2 - A/2) = cos A/2
Now,
cos A/2 = √ {s(s-a)/bc}
where, s is the semiperimeter of the triangle, and a, b ,c are the lengths of the sides opposite to vertices A,B & C respectively.
may i know your standard, because maybe it's high level
tx bro
10th
yep no prob
ok cool
all the very best for your exam
u r from
std?
plzz tell me
I'm in 12th
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