It is known that in △ ABC, the side opposite the acute angle B = 7, the radius of the circumscribed circle r = 7 times the root number 3 / 3, and the triangle area s = 10 times the root number 3. Find the length of the other two sides of the triangle

It is known that in △ ABC, the side opposite the acute angle B = 7, the radius of the circumscribed circle r = 7 times the root number 3 / 3, and the triangle area s = 10 times the root number 3. Find the length of the other two sides of the triangle

Sine theorem: B / SINB = 2R, SINB = root 3 / 2
‡ B = 60 degrees
AC = 40 ① from S = 1 / 2Ac · sin60 °
Cosine theorem; a^2+c^2-2ac·cos60°=b^2
(a+c)^2-3ac=169
∴a+c=13②
It can be solved by ① ②: a = 5, C = 8 or a = 8, C = 5

In triangle ABC, B = 60 degrees, s triangle ABC area = 10 root sign 3. If the circumscribed circle radius of the triangle r = 7 / 3 * root sign 3, what is the circumference of the triangle?

From the title,
In triangle ABC, B = 60 °, s ⊿ ABC = 10 √ 3
Because s ⊿ ABC = 0.5acsinb
So, AC = 40
Radius of circumscribed circle of triangle r = 7 √ 3 / 3
So, B = 2rsinb = 7
Cosine theorem
b ²= a ²+ c ²- 2accosB
Namely a ²+ c ²= b ²+ 2accosB=7 ²+ 2*40*(1/2)=89
from
ac=40
a ²+ c ²= eighty-nine
Solution (assuming a > C)
a+c=13
a-c=3
Namely
a=8,c=5
So,
C⊿ABC=a+b+c=8+7+5=20
Hope to adopt~~~

In triangle ABC, we know a = 60, B = 1, s area = root 3 to find the circumscribed circle radius of the triangle

The height on B is 2 √ 3
AB=(2√3)/SIN60=4
The radius is 2

In triangle ABC, a = 60, B = root 3 + 1, C = 2, find the area of a and the circumscribed circle of triangle ABC

By cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2bccosa
---------->a^2=(√3+1)^2+2^2-2bccos60
---------->a^2=3+1+2√3+4-2(√3+1)
---------->a^2=8-2=6
---------->a==√6
Make a circumscribed circle. If the center of the circle is O, then ∠ BOC = 2 ∠ a ---- > ∠ BOC = 120 °
---->cos∠BOC=-1/2
According to the cosine theorem, if the radius of the circumscribed circle is r, then R ^ 2 + R ^ 2-A ^ 2 = 2R ^ 2cos ∠ BOC
---->2R^2-6=-R^2
---->3R^2=6
---->R^2=2
So the area of the circumscribed circle is s = π, R ^ 2 = 2 π

In the triangle ABC, given the angle a = 60 degrees, B = 1, and the area of the triangle ABC is root 3, what is the diameter of the circumscribed circle of the triangle

By the area of the triangle and B = 1, the angle a = 60 degrees
Calculate the value of A
A * bsin ∠ A / 2 = root 3
So a = 2
You can make a right triangle with an angle of 60 degrees,
It can be seen from the figure that the bevel is the diameter of the circle
Therefore, the execution of the circle is four-thirds root 3

In △ ABC, a = 60 degrees, B = 1, the area of this triangle is root 3, then what is the diameter of the circumscribed circle of △ ABC? Please help! Little brother, thank you first!

1 / 2BC times Sina = root 3
Find C
Cosa = (b square + C square - a square) / 2BC
a/sinA=2R
2R is the diameter