A high school mathematics problem about solving triangle In the known △ ABC, a = 12, B = 10 √ 2, a = 45 °, find B and C I have the answer to this question, but I don't understand it. SINB = 5 / 6, and ∵ b > A, B is an obtuse angle or an acute angle. That is, B ≈ 56 ° 27 'or B ≈ 123 ° 33'. I want to know, how can I know the two values of B through the value of SINB? The explanation is detailed, I add scores. Please

A high school mathematics problem about solving triangle In the known △ ABC, a = 12, B = 10 √ 2, a = 45 °, find B and C I have the answer to this question, but I don't understand it. SINB = 5 / 6, and ∵ b > A, B is an obtuse angle or an acute angle. That is, B ≈ 56 ° 27 'or B ≈ 123 ° 33'. I want to know, how can I know the two values of B through the value of SINB? The explanation is detailed, I add scores. Please

In △ ABC, D is the midpoint of the edge BC, ab = 2, AC = 1, ∠ bad = 30 °, then ad=______ ．

Extend ad to e, make de = ad, connect be, then ∵ BD = CD, ∵ ADC = ∵ EDB ≌ BDE ≌ CDA ≌ be = AC = 1. In △ Abe, ab = 2, be = 1, ∵ bad = 30 °, according to sine theorem, ∵ AEB = 90 °, so AE = 3, ≌ ad = 32

High school mathematics solution triangle problem!
To the process of ah! Hard everyone! Thank you!
1. In △ ABC, the opposite sides of angles a, B, and C are a, B, and C respectively. If B-C = 2acos (c + one-third), find angle A
Thank you for your hard work!

According to the sine theorem, SINB sinc = 2sinacos (c + π / 3), SINB = sin (a + C) = sin acosc + cos asinc, cos (c + π / 3) = cos c * 1 / 2-sinc * radical number 3 / 2, which is replaced by 3 / 2 * cos a + 1 / 2 * sin a = 1 / 2, that is sin (a + π / 6) = 1 / 2, so a + π / 6 = π / 6 or 5 π / 6, so a = 5 π / 6

1. In triangle ABC, it is known that lgtanb is the arithmetic mean of lgtana and lgtanc?
2. Given Sina + SINB + sinc = 0 and cosa + CoSb + COSC = 0, what is the value of COS (a-b)
3. In triangle ABC, Sina + cosa = 2 / 3, then the shape of triangle ABC is?

1. Lgtanb is the arithmetic mean of lgtana and lgtanc, so: lgtana + lgtanc = 2lgtanb = LG (tanb) ^ 2 = LG [(Tana) · (Tanc)], so, (tanb) ^ 2 = (Tana) (Tanc), because B = π - (a + b), so: - tanb = Tan (a + C) = [Tana + Tanc] / [1 - (ta

It is proved that: (1) sin3a = 3sina-4sin ^ 3a
(2)cos3a=4cos^3a-3cosa

sin(3a) =sin(2a+a) =sin(2a)cosa+cos(2a)sina =(2sinacosa)cosa+(1-2sin^a)sina =2sina(1-sin^a)+sina-2sin^3 a) =3sina-4sin^3 a cos(3a) =cos(2a+a) =cos(2a)cosa-sin(2a)sina =(2cos^a-1)cosa-2(sinacosa)sina =...

Y = (SiNx + 1) (cosx + 1), X belongs to [- 30 °, 90 °]

y=(sinxcosx+sinx+cosx+1)
Then let SiNx + cosx = M;
Then (SiNx + cosx) ^ 2 = 1 + 2sinxcosx;
Sinxcosx = (m ^ 2-1) / 2; if you bring it in, you can get
y=(m+1)^2/2;
Root 2 * sin15

Why learn to count?

Because it is very important, many of them are used in reality. For example, the numerical control in the hardware industry is inseparable from the function

(2006 · Huai'an) both positive and negative scale function images pass through points (1, 4). In the first quadrant, the value range of the independent variable x above the negative scale function image of the positive scale function image is ()
A. x＞1B. 0＜x＜1C. x＞4D. 0＜x＜4

It can be seen from the image that the value range of the independent variable X of the positive scale function image above the inverse scale function image in the first quadrant is x > 1

A basic problem of inverse implication number in senior one
Arc Tan (Tan 5 / 4)
How to calculate?

The range of arctan is (- π / 2, π / 2), so the original formula = arctan [Tan (π + π / 4)] = arctan (Tan π / 4) = π / 4

One time implication

The number of primary inclusions: y = KX + B (K ≠ 0)