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As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = BC, ∠ CAD = ∠ bad, AB = AC + CD
It is proved that the crossing point D is de ⊥ AB and E,
∵DE⊥AB，
∴∠AED=90°，
∴∠ACB=∠AED=90°，
And ? CAD = bad, ad = ad,
∴△ACD≌△AED，
∴CD=ED，AC=AE，
∵∠ACB=90°，AC=BC，
∴∠B=45°，
And ∠ AED = 90 °,
∴∠EDB=45°，
∴ED=EB，
∴CD=EB，
∴AB=AE+EB=AC+CD．
In the triangle, ∠ C = 90 °, ab = 2.6, BC = 2.2, find the sine value, cosine value and tangent value of ∠ a
It is known that ∠ C = 90 °, ab = 2.6, BC = 2.2
So AC = 0.2 * √ (169-121) = 0.8 * √ 3
So the sine value of ∠ a = BC / AB = 11 / 13
Cosine of ∠ a = AC / AB = 4 √ 3 / 13
Tangent value of ∠ a = BC / AC = 11 / (4 √ 3) = 11 √ 3 / 12
In a triangle, given C = 2, C = π / 3, if the area of the triangle ABC is equal to the arithmetic square root of 3, find the values of a and B
Area s = 1 / 2 * AB * sinc = √ 3 leads to ab = 4; cosine theorem COSC = 1 / 2 = (a * a + b * b * c * c) / 2Ab, substituting AB = 4, C = 2 to get a * a + b * b = 8; so a ^ 2 + B ^ 2-2ab = 0... (a-b) ^ 2 = 0; so a = B, and C = π / 3, so ABC is an equilateral triangle; so a = b = C = 2; I hope it can help you!
Given that in the triangle ABC, a = 1, B = root 3, C = 30 degrees, then a is equal to
Make a vertical line AC from B and cross AC to d
In RT △ BCD, ∠ C = 30, BC (i.e. a) = 1, then BD = 1 / 2, CD = √ 3 / 2
AC (i.e. b) = √ 3, so D is the midpoint of AC
BD is the center line and the height of AC. therefore, the triangle ABC is an isosceles triangle, BC = ab
So ∠ a = ∠ C = 30
Let's try to find the area of the triangle AB = 75 in ABC = 3B In the triangle ABC, ab = AC, angle B = 75 degrees, ab = 3cm. Try to find the area of triangle ABC
The area formula is s = AB * ac * sin (30 degrees) / 2 = 3 * 3 * 0.5 * 0.5 = 2.25
explain
Because AB = AC, the triangle is isosceles
So angle B = angle c = 75 degrees
So the angle a is 30 degrees
Given the triangle ABC, angle a = 75 degrees, angle c = 48 degrees, calculate AC ratio ab
∠B＝180°－75°－48°＝57°
According to the sine theorem, it is concluded that:
AC∶AB＝sinB∶sinC
＝sin57°∶sin48°
= 0.8387 ∶ 0.7431
The ratio is 1.13
In the triangle ABC, if AB = AC, CD is perpendicular to point D, CD = 3, and angle B = 75 degrees, then ab=
3 √ 3 + 3tan15 degrees
In the triangle ABC, BC = 1, B = π / 3, and the area of triangle ABC = root 3, then what is Tanc
Let the edges corresponding to the three angles a, B and C be a, B and C respectively, then
A=1
B=π/3
S = 1 / 2 acsinb = √ 3, so C = 4
According to the triangle interior angle sum theorem, the
A+C=π-B=2π/3
According to the sine theorem, the
a/sinA=c/sinC
asinC=csinA
sinC=4sin(2π/3-C)=4sin(C+π/3)=2sinC+2√3cosC
∴sinC=-2√3cosC
Obviously, COSC ≠ 0, divide both sides by COSC
tanC=-2√3
In the triangle ABC, the opposite sides of intersection a, B and C are a, B, C, Tanc = 3, radical 7 (1) Find cos C (2) If CB * CA = 5 / 2 and a + B = 9, find C (in the second problem, there are arrows on the top of CB and Ca to represent vectors) The process should be detailed and understandable
Tanc = sinc / COSC = 3 root number 7
So sincoc = 3
And sinc ^ 2 + COSC ^ 2 = 1
So COSC = ± 1 / 8
There is Tanc > 0
So COSC > 0
So COSC = 1 / 8
(2)
A * b = 5 / 2, and a + B = 9
So a ^ 2 + B ^ 2 = (a + b) ^ 2-2ab = 76
cosC=(a^2+b^2-c^2)/(2ab)=1/8
So C = (3 √ 134) / 4
In the triangle ABC, Tanc = 3, radical 7 In the triangle ABC, the opposite sides of intersection a, B and C are a, B, C, Tanc = 3, radical 7 (1) Find cos C (2) If CB * CA = 5 / 2 and a + B = 9, find C
I calculated C = 6
RELATED INFORMATIONS
- 1. In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and Tanc = 3 7． (I) find the value of COSC; (II) if CB• CA=5 2, and a + B = 9, find the length of C
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