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Given the triangle ABC, if ∠ C = 90 °, the square of 4B + the square of 3C = four times the root sign 3bC, find the value of cosa
∵4b²+3c²=4√3bc
∴4b²+3c²-4√3bc=0
∴(2b)²-4√3bc+(√3c)²=0
∴(2b-√3c)²=0
∴2b=√3c
∴b/c=√3/2
That is, cosa = B / C = √ 3 / 2
In △ ABC, the opposite sides of the inner angles a, B and C are a, B, C respectively, if A2-B2= 3bc，sinC=2 3 SINB, then a = () A. 30° B. 60° C. 120° D. 150°
∵sinC=2
3sinB，∴c=2
3b，
∵a2-b2=
3bc，∴cosA=b2+c2−a2
2bc=2
3bc−
3bC
2bc=
Three
Two
∵ A is the inner angle of the triangle
∴A=30°
Therefore, a
In △ ABC, the opposite sides of the inner angles a, B and C are a, B, C respectively, if A2-B2= 3bc，sinC=2 3 SINB, then a = () A. 30° B. 60° C. 120° D. 150°
∵sinC=2
3sinB，∴c=2
3b，
∵a2-b2=
3bc，∴cosA=b2+c2−a2
2bc=2
3bc−
3bC
2bc=
Three
Two
∵ A is the inner angle of the triangle
∴A=30°
Therefore, a
In the triangle ABC, if a ^ 2 + B ^ 2
Using cosine theorem: C ^ 2 = a ^ 2 + B ^ 2-2 * a * b * COSC
∴a^2+b^2
In △ ABC, the opposite sides of the inner angles a, B and C are a, B, C respectively, if A2-B2= 3bc，sinC=2 3 SINB, then a = () A. 30° B. 60° C. 120° D. 150°
∵sinC=2
3sinB，∴c=2
3b，
∵a2-b2=
3bc，∴cosA=b2+c2−a2
2bc=2
3bc−
3bC
2bc=
Three
Two
∵ A is the inner angle of the triangle
∴A=30°
Therefore, a
In triangle ABC, sinA:sinB Sinc = 2 to 6 (root 3 + 1), minimum inner angle of triangle A 60 degrees B 45 degrees C30 degrees D and above are wrong
The answer is B
sinA:sinB;sinC=a : B: C = 2: (radical 6): (radical 3 + 1)
A is the smallest angle
Cosa = [6 + (radical 3 + 1) - 4] / [2 * radical 6 * (radical 3 + 1)] (cosine theorem)
=(radical 2) / 2
A = 45 degrees
In triangle ABC, sinA:sinB If sinc = 2: root 6: (root 3 + 1), then the minimum inner angle of the triangle is?
sinA:sinB;sinC=a : B: C = 2: (radical 6): (radical 3 + 1)
A is the smallest angle
Cosa = [6 + (radical 3 + 1) - 4] / [2 * radical 6 * (radical 3 + 1)] (cosine theorem)
=(radical 2) / 2
A = 45 degrees
In the right triangle ABC, ∠ C = 90 ° CD is the center line on the edge of AB, ∠ a = 30 ° AC = √ 3, and the circumference of △ ADC is obtained
Because ∠ C = 90 °, CD is the center line of AB edge, ∠ a = 30 ° AC = √ 3
So AB = 2, BC = 1, △ CDB is an equilateral triangle
So CD = ad = BD = 1
The circumference of △ ADC CD + AD + AC = 1 + 1 + √ 3 = 2 + √ 3
In the RT triangle ABC, ∠ C = 90 °, a = 30 ° and CD is the center line on AB side. If AC = 2, radical 3, BC = 2, then the circumference of the triangle ADC is equal to
The triangle ABC is a right triangle, ∠ a = 30 °
So the right angle of the right triangle ad = 2 is 2
CD is the center line of AB edge, CD = 2
So the circumference of the triangle ADC = AD + CD + AC = 2 + 2 + 2 √ 3 = 4 + 2 √ 3
In the RT triangle ABC, the angle c = 90, BC: AC = 1: radical 3, CD is perpendicular to AB and D, and s triangle is found CDB:S Triangle ABC
∵ BC: AC = 1: √ 3, each part of the ratio is t,
Then BC = t, AC = √ 3T
∵ angle c = 90, ᙽ AB = √ (AC ∵ BC ᙽ 2) = 2T
∵ CD is perpendicular to AB and D
∴ΔBDC∽ΔABC
∴BD:BC=BC:AB=√3:2
∵ the area ratio of the similar triangle is equal to the square of the similarity ratio
∴SΔ CDB:S ΔABC=3：4
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