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In the triangle ABC, the angle B is equal to 30 degrees, the angle c is equal to 45 degrees, AB is equal to two centimeters, then what is BC equal to
AB = 2
BD = AD = √2
AC = 2 AD = 2√2
For Pythagorean theorem, CD = √ {AC & # 178; - Ad & # 178;} = √ 6
BC = BD + CD = √2 + √6
In the triangle ABC, a = 80, B = 100, a = 45, the solution of this triangle is
Angle a is not the angle between a and B, so you can not determine a triangle. You can draw a line AC with a length of 5, make a 45 ° angle with point a, ray AP, and then take point B as the center of the circle with a radius of 4, (100:80 = 5:4). You will find that there are two focal points in the circle and AP, one is an obtuse triangle, and the other is an acute triangle
In the triangle ABC, B = 2, a = 60, C = 45, find the minimum side length of the triangle
A=60,C=45
So B = 75
So C is the smallest
sin75=sin(45+30)=sin45cos30+cos45sin30=(√6+√2)/4
So C / sinc = B / SINB
c=bsinC/sinB=2*(√2/2)/[(√6+√2)/4]=2√3-2
Known: as shown in the figure, in △ ABC, ∠ B = 45 °, C = 60 ° and ab = 6, find the length of BC. (results the root sign is reserved)
As shown in the figure, through point a, ad ⊥ BC is made at point D, in RT △ abd, ∠ B = 45 ° and 〈 ad = BD, let ad = x, and ∵ AB = 6, in 〈 RT △ abd, X2 + x2 = 62, and the solution is x = 32, that is, ad = BD = 32, in RT △ ACD, ∠ ACD = 60 ° and ∵ CAD = 30 ° and Tan 30 ° = CDAD, that is, 33 = CD32, ∵ CD = 6, ∵ BC = BD + DC = 32 + 6. (7 points)
In △ ABC, the length of AB is obtained when ∠ a = ∠ B = 45 ° and BC = 3 are known
∵∠A=∠B=45°，∴AC=BC=3，∠C=90°，∴AB=AC2+BC2=32．
In ABC, if tanatanb = tanatanc + tanctanctanb, then C squared (a squared plus B squared) =?
stay
ABC
Medium, if
tan
A
tan
B
=tan
A
tan
C
+tanctan
B
, then
two
two
two
c
b
A is C squared (a squared plus B squared) =?, write down the detailed steps!
A ^ 2 + B ^ 2 = C ^ 2 + 2abcosc, the original formula is 1 + 2abcosc / C ^ 2. Because tanatanb = tanatanc + tanctanb, 1 / Tanc = 1 / tanb + 1 / Tana is simplified to sinasinbcosc / sinc = sinc, that is, sinasinbcosc = sin ^ 2C and abcosc = C ^ 2, the original formula is 1 + 2abcosc / C ^ 2 = 1 + 2 = 3
In triangle ABC, if tanatanb = tanatanc + tanctanb and C = 2, the maximum area of triangle ABC is obtained
Tanatanb = tanatanc + tanctanb, 1 / Tanc = 1 / tanb + 1 / Tana, let the area of the triangle ABC be s, the height of the side of AB be h, ab = C = 2, then s = h, 1 / Tanc = 1 / tanb + 1 / Tana = C / h = 2 / s, s = 2tanc; tanb = tanctana / (Tana Tanc), in the triangle. Tanatanbtanc = Tana + tanb + Tanc
If Tan & # 178; a (Tan & # 178; C-1) - tanctana + Tan & # 178; C = 0, Δ = Tan & # 178; c-4tan & # 178; C (Tan & # 178; C-1) ≥ 0, then 0 ≤ s ≤√ 5 and the maximum area of triangle ABC √ 5
In △ ABC, if tanatanb = tanatanc + tanctanb, then & nbsp; A2 + b2c2=______ ．
The known equation is & nbsp; sinasinbcosb = sinasinccosacosc + sinbsinccosbcosc, i.e. sinasinbsincc = sin (a + b) COSC, i.e. sinasinbcoscsin2c = 1, i.e. abcoscc2 = 1. Therefore, A2 + B2 − c22c2 = 1, so A2 + b2c2 = 3
In triangle ABC, the degree of angle a is five times that of angle c, and the degree of angle B is three times that of angle C. what are the degrees of angle a, angle B and angle c?
It's urgent! Come on!
Degree of angle c = 180 ÷ (5 + 3 + 1) = 20 (degree)
Degree of angle a = 20 × 5 = 100 (degree)
Degree of angle B = 20 × 3 = 60 (degree)
In angle ABC, angle a, angle B, angle c = 1:3:5, what are the degrees of these three triangles
The inner angle of a triangle is 180 degrees, so let the angle ax degrees, then: x + 3x + 5x = 180 degrees, the solution is x = 20 degrees, then the angle B = 60 degrees, and the angle c = 100 degrees
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