In △ ABC, points D and E are AB and AC respectively. Let CD and be intersect at point O. if ∠ a = 60 °, DCB, EBC = half a, find an angle equal to a

In △ ABC, points D and E are AB and AC respectively. Let CD and be intersect at point O. if ∠ a = 60 °, DCB, EBC = half a, find an angle equal to a

∠EOC=∠DOB=∠A

In △ ABC, the points o and E are on AB and AC respectively, ∠ DCB = ∠ EBC = 1,2 ∠ a, be and CD intersect at point O to prove BD = CE

∵∠DCB＝∠EBC=1\2∠A
∴DC:BA=CB:AC=DB:BC=1:2
EB:BA=BC:AC=EC:BC=1:2
∵DB:BC=1:2
EC:BC=1:2
∴DB=EC

In the triangle ABC, a; B; C = 7; 8; 13, the cosine value of the largest angle in this △ is

According to the big side to the big angle, angle c is the largest, so COSC = - 1 / 2

In △ ABC, if a = 7, B = 8, COSC = 1314, then the cosine of the largest angle is ()
A. −15B. −16C. −17D. −18

∵ in △ ABC, a = 7, B = 8, COSC = 1314, ∵ C2 = A2 + b2-2abcosc = 49 + 64-2 × 7 × 8 × 1314 = 9, C = 3 ∵ b > a > C, ∵ the largest edge is B, and B is the largest angle. Therefore, CoSb = 49 + 9 − 642 × 7 × 3 = − 17, that is, the cosine value of the largest angle is − 17

In △ ABC, if a = 7, B = 8, COSC = 1314, then the cosine of the largest angle is ()
A. −15B. −16C. −17D. −18

∵ in △ ABC, a = 7, B = 8, COSC = 1314, ∵ C2 = A2 + b2-2abcosc = 49 + 64-2 × 7 × 8 × 1314 = 9, C = 3 ∵ b > a > C, ∵ the largest edge is B, and B is the largest angle. Therefore, CoSb = 49 + 9 − 642 × 7 × 3 = − 17, that is, the cosine value of the largest angle is − 17

In the triangle ABC, a = 7, B = 8, COSC = 13 / 14, find the cosine of C and the largest angle

C & sup2; = A & sup2; + B & sup2; - 2abcosc = 49 + 64-2 * 7 * 8 * 13 / 14 = 9, so C = 3;
b> A > C, so the maximum angle is B, CoSb = (A & sup2; + C & sup2; - B & sup2;) / 2Ac = (49 + 9-64) / 2 * 7 * 3 = - 1 / 7

In △ ABC, if a = 7, B = 8, COSC = 1314, then the cosine of the largest angle is ()
A. −15B. −16C. −17D. −18

∵ in △ ABC, a = 7, B = 8, COSC = 1314, ∵ C2 = A2 + b2-2abcosc = 49 + 64-2 × 7 × 8 × 1314 = 9, C = 3 ∵ b > a > C, ∵ the largest edge is B, and B is the largest angle. Therefore, CoSb = 49 + 9 − 642 × 7 × 3 = − 17, that is, the cosine value of the largest angle is − 17

In the triangle ABC, if (c + b): (a + C): (a + b) = 4:5:6, then the minimum cosine of the internal angle of the triangle ABC is

（c+b):(a+c):(a+b)=4:5:6
set up
c+b=4k
a+c=5k
a+b=6k
The solution is as follows
a=7k/2
b=5k/2
c=3k/2
So: A: B: C = 7:5:3
Then: the minimum internal angle is C
According to the cosine theorem, we get the following results
cosC=(a²+b²-c²)/2ab
=(49+25-9)/(2×7×5)
=65/70
=13/14

In the triangle ABC, the opposite sides of the angle ABC are ABC. We know that sinc / Sina = A2 + c2-b2 / A2 to find the angle B (2) let t = sin square a + sin square C to find the range of T

sinC/sina=c/

It is known that the opposite sides of the three inner angles a, B and C of △ ABC are a, B and C respectively, and B2 + C2 = A2 + BC (1) find the value of sina; (2) if a = 2, find the maximum value of B + C

(1) ∵ B2 + C2 = A2 + BC, ∵ A2 = B2 + C2 BC, combined with cosine theorem, we know that cosa = 12, ∵ a = π 3, ∵ Sina = 32 (6 points) (2) from a = 2, combined with sine theorem, B + C = 433 SINB + 433 sinc (8 points) = 433sinb + 433sin (2 π 3-B) (9 points) = 23sinb + 2cosb (10 points) = 4sin (B + π 6) (11 points) and B ∈ (0, 2 π 3), so B + π 6 ∈ (π 6, 5 π 6), so when B + π 6 = π 2, that is, B = π 3, the maximum value of B + C is 4 (13 points)