As shown in the figure, in triangle ABC, ab = AC = 20, angle B is equal to 15 degrees, then the area of triangle ABC

As shown in the figure, in triangle ABC, ab = AC = 20, angle B is equal to 15 degrees, then the area of triangle ABC

Because: ab = AC = 20, so: ∠ B = ∠ C = 15 °
Therefore: ∠ BAC = 120 °
Ad = 1 / 2Ab = 10 when passing through point B as BD and perpendicular to point D
Therefore, according to Pythagorean theorem, BD = 10 times root sign 3
Area of triangle ABC = 1 / 2 * 20 * 10 root number 3 = 100 times root number 3

A = 2. C = root 19 and angle c = 120, then the area of triangle ABC

From the cosine theorem: C ²= a ²+ b ²- 2ab*cosC
Given that a = 2. C = root 19 and angle c = 120, then:
19=4+b ²- 2*2*b*cos120°
Namely: B ²+ 2b-15=0
(b+5)(b-3)=0
Since b > 0, the solution is: B = 3
Then area: s △ ABC = (1 / 2) * AB * sinc = (1 / 2) * 2 * 3 * sin120 ° = 3 (root sign 3) / 2

B = 120 degrees, B = root sign 13, a + C = 4, find the area of triangle ABC

B=120°,b=Sqrt[13],a+c=4
a^2 + c^2 - 2 a c Cos[120 °] == b^2
a + c ==4,
The solution is {{a = 1, C = 3}
S=0.5 a c Sin[120°]=3 Sqrt[3]/4

In triangle ABC, the angles of three sides a, B and C are a, B and C respectively. Given that a = 2, root number 3 and B = 2, the area of triangle ABC s = root number 3, then C =? The answer is only 30? Or 30 and 150?

S=1/2ab (sinC)
Sinc = 1 / 2
So, C = 30 or 150 degrees

In triangle ABC, the opposite sides of angle ABC are ABC, a = π / 3, SINB = root 3 / 3 respectively. Find CoSb (1) and find the value of CoSb; (2) If 2C = B + 2, find In triangle ABC, the opposite sides of angle ABC are ABC, a = π / 3, SINB = root 3 / 3 respectively. Find CoSb (1) and find the value of CoSb; (2) If 2C = B + 2, find the side length B

Make the vertical line of AB through point C, and the vertical foot is point D. if ≓∠ a = π / 3 = 60 °∠ ACD = 30 °, set ad = x, then AC = b = 2x, from the Pythagorean theorem, CD = √ 3x, in the right angle △ CDB, from SINB = √ 3 / 3, CB = a = 3x ‰ from the Pythagorean theorem, DB = √ 6x CoSb = √ 6x / (3x) = √ 6 / 3, ab = C = x + √ 6x ‰ from 2C

In △ ABC, cosa = five 5，cosB＝ ten 10． (I) angle c; (II) let AB = 2. Find the area of △ ABC

(I) from cosa = 55 and CoSb = 1010, a and B ∈ (0, π 2), so Sina = 25 and SINB = 310. (3 points) because COSC = cos [π - (a + b)] = - cos (a + b) = - cosacosb + sinasinb = 22, (6 points) and 0 < C < π, so C = π 4. (7 points) (II) according to the sine theorem, absinc = a