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In △ ABC, a and B are acute angles, and the corresponding edges of angles a, B and C are a, B and C respectively, and cos2a = 3 / 5, SINB = 10 / 10 under the quadratic root sign If a + B = 2-1 under the square root sign, find the value of a, B, C [find detailed explanation]
In this paper, we make an analysis of the characteristics of the system, such as the size of the system, the size of the system, the size of the system, the size of the system, the size of the system, the size of the system, the size of the system, the number of the points, the number of the points, the number of the points, the number of the points, the number of the points, the number of the the points, the number of the the the the number of the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the the A / Sina = B / Si
Let the opposite sides of the inner angles a, B and C of the acute triangle ABC be a, B, C, a = 2bsina (I) find the size of B; (II) if a = 3 3, C = 5, find B
(2) BSA = 2,
According to the sine theorem, Sina = 2sinbsina, so SINB = 1
2，
If △ ABC is an acute triangle, B = π
6．
(II) according to the cosine theorem, B2 = A2 + c2-2accosb = 27 + 25-45 = 7
So, B =
7．
In the triangle ABC, the opposite sides of the inner angles a, B and C are a, B, C respectively, and the angle a is an acute angle, and 3b = 5asin B. (1) (2) If a = √ 2, the area of triangle ABC is 3 / 2. Find B, C
b/sinB=5a/3
a/sinA=b/sinB
a/sinA=5a/3
sinA=3/5 cosA=4/5
bcsinA=3,bc=5
a^2=b^2+c^2-2bc*cosA
b^2+c^2=10
b^2-2bc+c^2=10-2*5
(b-c)^2=0
b=c=√5
Let the opposite sides of the inner angles a, B and C of the acute triangle ABC be a, B, C, a = 2bsina (I) find the size of B; (II) if a = 3 3, C = 5, find B
(2) BSA = 2,
According to the sine theorem, Sina = 2sinbsina, so SINB = 1
2，
If △ ABC is an acute triangle, B = π
6．
(II) according to the cosine theorem, B2 = A2 + c2-2accosb = 27 + 25-45 = 7
So, B =
7．
Let the opposite sides of the inner angles a, B and C of the acute triangle ABC be a, B, C, a = 2bsina (I) find the size of B; (II) if a = 3 3, C = 5, find B
(2) BSA = 2,
According to the sine theorem, Sina = 2sinbsina, so SINB = 1
2，
If △ ABC is an acute triangle, B = π
6．
(II) according to the cosine theorem, B2 = A2 + c2-2accosb = 27 + 25-45 = 7
So, B =
7．
In △ ABC, if ∠ a, ∠ B satisfies the square of [cosa-1 / 2] + (sinb-2 root sign 2) = 0, then ∠ C =?
|cosA－1/2|＋(sinB－√2/2)²=0
Then:
cosA=1/2、sinB=√2/2
Results: a = 60 °, B = 45 ° or B = 135 ° [omitted]
Therefore, C = 180 ° - A-B = 75 °
In the triangle ABC, if the side is a, B, C, cosa = 2 / 3, SINB = root 5 times COSC, find Tanc? If a = radical 2, find the area of the triangle
From cosa = 2 / 3. Combining with 0 < a < 180, we can get: Sina = (√ 5) / 3. Cosa = 2 / 3. [[[[1]]]] combining SINB = (√ 5) COSC. COSC = cos [180 - (a + b)] = - cos (a + b) = - (cosacosb sinasinb) = sinasinb cosacosb = [(5 / 3) COSC]
(1 / 2) in the triangle ABC, the angles a, B and C are opposite to a, B, C respectively, and cosa = 2 / 5, root 5, SINB = 10 / 10, root 10 (1 / 2) in the triangle ABC, the angles a, B and C are opposite to a, B, C respectively, and cosa = 2 / 5, root sign 5, SINB = 10 / 10 root sign 10, find the angle c? If a
Cosa = 2 / 5, root 5
sinA=√1-cos^2A=√1-(4/5)=√5/5
In the same way
SINB = root of 10
cosB=3√10/10
cosC=-cos(A+B)=-cosAcosB+sinAsinB
=-2√5/5*3√10/10+√5/5*√10/10
=-√2/2
therefore
B = 135 degrees
In the triangle ABC, cosa = 2 times the 5 th root sign 5, SINB = 10 th root sign 10. (1) find the angle c (2) if A-B = root 2 minus 1, find C side
The first question can be solved by using the formula of double angle
Sina = 4 / 1-5 = 5 / 5
Similarly, CoSb is obtained
COS(A+B)=COSACOSB-SINASINB
Cos (π - (a + b)) = COSC = - cos (a + b) can solve the angle C in turn
The second question is opportunistic. The C side is 1
In RT △ ABC, ∠ C = 90, BC = 10, s △ ABC = 50 / 3, root sign 3, calculate the values of AC and AB and the degree of ∠ a
In RT △ ABC
Because: s △ ABC = 50 / 3
S △ ABC = 1 / 2 * 10ac = 50 / 3
AC = 10 * root 3 / 3
In RT △ ABC, according to Pythagorean theorem
AB^2=10*10+AC^2
=400/3
AB = 20 root 3 / 3
sinA=BC/AB
=10 * 3 / (20 pieces of 3)
=Radical 3 / 2
A = 60 degrees
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