In triangle ABC, cosa = 3 / 5, find the value of COS ^ (A / 2) - sin (B + C)

In triangle ABC, cosa = 3 / 5, find the value of COS ^ (A / 2) - sin (B + C)

Because: cosa = 2cos ^ 2 (A / 2) - 1,
cos^2(A/2)=(cosA+1)/2,
cosA=3/5,
sinA=√(1-cos^2A)=4/5.
cos^2(A/2)-sin(B+C)=(cosA+1)/2-sinA=(3/5+1)/2-4/5=0.

△ ABC, cosa = 1 / 3, find the value of sin Λ 2 (B + C) / 2 + cos Λ 2 (B + C)

[sin(B+C)/2]^2+[cos(B+C)]^2
[double angle formula]
=[1-cos(B+C)]/2+[cos(B+C)]^2
cos(B+C)=cos(π-A)=-cosA=-1/3
=(1+1/3)/2+1/9
=7/9

In triangle ABC, the angle a = 60 °, B = 1, and the area of the triangle is equal to root 3. How to find the length of the inscribed circle radius of triangle ABC?

Find the three sides first
According to area = (1 / 2) * b * c * Sina = √ 3
Find C = 4
Then, according to the cosine theorem, find a = √ (4 ^ 2 + 1 ^ 2-4 * 1 * cos60 °) = √ 15
Let the radius of the inscribed circle be r
Then triangle area = (1 / 2) a * r + (1 / 2) b * r + (1 / 2) c * r = 1 / 2 (a + B + C) * r = √ 3
Therefore, r = √ 3-3 √ 5 / 5

In triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively, and COS ^ 2 (A / 2) = B + C / 2C = 9 / 10, C = 5. Find the area of the inscribed circle of triangle ABC

From (B + C) / 2C = 9 / 10 and C = 5 = = = > b = 4cos ² (A / 2) = (1 + COSA) / 2 = 9 / 10 = = > cosa = 4 / 5 and cosa = 4 / 5 = (b) ²+ c ²- a ²)/ 2bc===>a=3,===>c ²= a ²+ b ² The triangle is a right triangle. If the radius of the inscribed circle is r, then R (a + B + C) / 2 = AB / 2 = =

In triangle ABC, if the angle B = 60 degrees, ab = 8, BC = 5, what is the area of the inscribed circle of triangle ABC RT

With the cosine theorem, AC = 7 can be obtained, and then the radius of the inscribed circle is set as R by the equal area theorem
Then 1 / 2 * 8 * 5 * sin60 = 1 / 2 * (8 + 5 + 7) * r
The solution is r = root 3
So the area of the inscribed circle is 3 π

In triangle ABC, the angle A60 degrees, the ratio of C to B is 8 to 5, and the area of the inscribed circle is 12 pie, then the radius of the circumscribed circle is?

Let C = 8x B = 5x cosa = (b ^ 2 + C ^ 2-A ^ 2) / 2BC get a = 7x
S = 0.5bc * Sina = 0.5 * 8x * 5x * root 3 * 0.5 = 10 root 3x ^ 2
Inscribed circle radius r = 2S / (a + B + C) = 20 root 3x ^ 2 / 20x = root 3x
Because π R ^ 2 = 12 π, r = 2 radical 3 = 3x x = 2 / 3 radical 3 a = 7x = 14 / 3 radical 3
Circumscribed circle radius 2R = A / Sina = 28 / 3, so r = 14 / 3
I don't know, do I?