In a triangle, two of the angles are 20◦ and 55◦ and the included side has length (1 +√3). Its circumradius is?
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use sine property ,
sinA/a = sinB/b = sinC/c = 1/2R
for ∆ ,
A + B + C = 180°
20° + 55° + C = 180°
C = 180 - 75° = 105°
so, sinC/c = 1/2R
sin105°/(1 + √3) = 1/2R
2R = (1 + √3)/sin105°
sin(105°) = sin(60+45°) = (1+√3)/2√2
put this above ,
2R = (1+ √3)/( 1+√3)/2√2
2R = 2√2
R = √2
sinA/a = sinB/b = sinC/c = 1/2R
for ∆ ,
A + B + C = 180°
20° + 55° + C = 180°
C = 180 - 75° = 105°
so, sinC/c = 1/2R
sin105°/(1 + √3) = 1/2R
2R = (1 + √3)/sin105°
sin(105°) = sin(60+45°) = (1+√3)/2√2
put this above ,
2R = (1+ √3)/( 1+√3)/2√2
2R = 2√2
R = √2
vishagh:
thnq abhi :)
my pl
is this correct na ???
yes its perfect
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Answered by
0
using sines law
a/sina = b/sinb = c/sinc = 2r
a,b,c are sides of triangles
and r is the radius
a=> 20° b=>55°
we know that a+b+c = 180°
so 20+55+c=180°
c= 105°
we know side
applying sine rule
(1+√3)/sin(105) = c/sinc
now
value of sin (105) = (1+√3)/2√2
so we get
c/sinc= (1+√3)(2√2)/1+√3)=2√2
c/sinc= 2r
so we can equate our equation
2r = 2√2
r= √2
so dude its circumradius is √2
mark it as the best and click on the read heart button <3
a/sina = b/sinb = c/sinc = 2r
a,b,c are sides of triangles
and r is the radius
a=> 20° b=>55°
we know that a+b+c = 180°
so 20+55+c=180°
c= 105°
we know side
applying sine rule
(1+√3)/sin(105) = c/sinc
now
value of sin (105) = (1+√3)/2√2
so we get
c/sinc= (1+√3)(2√2)/1+√3)=2√2
c/sinc= 2r
so we can equate our equation
2r = 2√2
r= √2
so dude its circumradius is √2
mark it as the best and click on the read heart button <3
thnq :)
u need to learn the value of sin(105) without solving it u can easily do in exam
i have already learned its value so i just direcltly solve this if u learn values u will save time in exam
ok thnq gohan :)
or u can use formula sin(a+b) = sinacosb+cosasinb for better understanding
to calculate sin(60+45)
thanks man ! love u <3
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