In the triangle ABC, the angle c = 90 ° and CE is the middle line of the triangle ABC. If AC = 4 and BC = 3, what is the area of the triangle AEC?

In the triangle ABC, the angle c = 90 ° and CE is the middle line of the triangle ABC. If AC = 4 and BC = 3, what is the area of the triangle AEC?

Because CE is the middle line of triangle ABC, so AE = be, according to the triangle area formula s = 0.5 bottom * height, take AE and be as the bottom respectively, and the height is equal (you can make a high auxiliary line), the area of ACE and BCE is equal, which is half of triangle ABC

In known triangle ABC, angle a = 90 °, angle B = 60 ° and a = 8, then B=_____ ,c=______ .

B equals four times the root sign three, C equals four

In the triangle ABC, the angle c = 90 ° is known, C = 8 root sign 3, ∠ a = 60 ° and B, a, B are calculated

B = 90-60 = 30 B = C / 2 = 4 radical 3 a = C * sin30 = 12

In the triangle ABC sin (a-b) / sin (a + b) = (C-B) / C, then the triangle must contain A.30 ° internal angle b.45 ° internal angle c.60 ° internal angle d.90 ° internal angle

The opposite sides of the inner angles a, B and C of △ ABC are a, B and C respectively. Given a-c = 90 ° and a + C = 2B, find C

From a-c = 90 °, a = C + 90 °, B = π - (a + C) = 90 ° - 2C (in fact, 0 ° < C < 45 °), from a + C = 2B, according to the sine theorem, there are: Sina + sinc = 2sinb, ｜ sin (c + 90 °) + sinc = 2Sin (90 ° - 2C), that is, COSC + sinc = 2coc2c = 2 (cos2c − sin2c) = 2 (COSC + sinc)

The opposite sides of the inner angle a B C of △ ABC are a B C, a-c = 90 ° a + C = B × √ 2 to find the angle C

A-c = 90 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\, C ≠ 0, We can get the c-corner formula, COSC / sinc + 1 = (COSC / sinc + 1) (COSC / sinc + 1 = (cos2c) \c) \\\\\\\35\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\c + 2coscsinc), 1 / 2 = 1 + 2coscsin, C = 1 + sin2c, sin2c = 1 / 2, ∧ 2C = 30 °, that is, C = 15 °. The solution is finished

In △ ABC, ∠ C = 90 ° is known, and the right triangle is solved according to the following conditions: (1) a = 30 ° and B = 3 (the result retains the root sign)
(2)a=3.245,c=4.876

1、BC/AC=tan30°,
BC=（√3/3）*3=√3,
BC/AB=sin30°,
AB=2BC=2√3,
〈CBA=90°-30°=60°.
2. According to Pythagorean theorem, B = √ (C ^ 2-A ^ 2) = √ (23.7754-10.53) = 3.6394,
sinA=a/c=0.6655,
∴A=41°44‘.
∴B=90°-41°44’=48°16‘.

In △ ABC, we know that C = 2, radical 6, a = 45 ° and a = 4, and solve right triangle

sinA:a=sinC : C √ 2 / 2:4 = sinc: 2 √ 6, so sinc = √ 3 / 2 C = 60 ° or 120 ° corresponds to B = 75 ° or 15 °

In the triangle ABC, it is known that the opposite sides of angles a, B and C are a, B and C respectively, and (a + B + C) * (B + C-A) = 3bC. If B = 105 degrees and C = 4, then B =?

(a+b+c)*(b+c-a)=3bc
(b+c+a)(b+c-a)=3bc
(b+c)^2-a^2=3bc
b^2+2bc+c^2-a^2=3bc
b^2+c^2-a^2=bc
cosA=(b^2+c^2-a^2)/2bc
=bc/2bc
=1/2
∵A

In △ ABC, a, B and C are the sides opposite to the inner angles a, B and C. if (a + B + C) (B + C-A) = 3bC, then the value of angle a is ()
A. 60°B. 90°C. 120°D. 150°

∵ (a + B + C) (B + C-A) = 3bC ∵ [(B + C) + a] [(B + C) - A] = 3bC ∵ (B + C) 2-A2 = 3bC ∵ B2 + 2BC + c2-a2 = 3bC ∵ BC = B2 + c2-a2. According to the cosine theorem, cosa = B2 + C2 − a22bc ∵ cosa = 12 ∵ angle a is the inner angle of △ ABC ∵ a = 60 ° so a is selected