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In the triangle ABC, the angle BAC is equal to 20 degrees, AB is equal to AC, D is a point on the edge of AB, and ad is equal to BC to find the degree of angle BDC
20 degrees
In the acute triangle ABC, AB is equal to AC greater than BC, point D is on edge AB, ad = BC, if angle BDC = 30 degrees, then angle A=————
Because ad = BC, the triangle ACD is isosceles triangle
So ∠ a = ∠ ACD
And ∠ a + ∠ ACD = ∠ BDC = 30 °
So ∠ a = 15 °
It is known as follows: in △ ABC, ∠ cab = 120 °, ab = 4, AC = 2, ad ⊥ BC, D are perpendicular feet. Find the length of AD
C is a graph with a high point,
Then ∠ CAE = 180 ° - 120 ° = 60 °,
In RT △ ace, ∠ CEA = 90 °,
∵sin∠CAE=CE
AC，cos∠CAE=AE
AC，
∴CE=AC•sin60°=2×
Three
2=
3，
AE=AC•cos60°=2×1
2=1
∴BE=AB+AE=5；
In RT △ CBE, from Pythagorean theorem, BC = 2
7，
∵AD⊥BC，
∴sin∠B=CE
BC＝AD
AB．
∴AD=AB•CE
BC＝2
Twenty-one
7．
It is known as follows: in △ ABC, ∠ cab = 120 °, ab = 4, AC = 2, ad ⊥ BC, D are perpendicular feet. Find the length of AD
C is a graph with a high point,
Then ∠ CAE = 180 ° - 120 ° = 60 °,
In RT △ ace, ∠ CEA = 90 °,
∵sin∠CAE=CE
AC，cos∠CAE=AE
AC，
∴CE=AC•sin60°=2×
Three
2=
3，
AE=AC•cos60°=2×1
2=1
∴BE=AB+AE=5；
In RT △ CBE, from Pythagorean theorem, BC = 2
7，
∵AD⊥BC，
∴sin∠B=CE
BC＝AD
AB．
∴AD=AB•CE
BC＝2
Twenty-one
7．
In the triangle ABC, AC = BC, In the triangle ABC, AC = BC,
The extension line of AC is f
In the triangle ABC, the angle c = 90, ad is the bisector of angle cab, intersect BC with D, BC = 4, CD = 1.5, and find the length of AC
Then the ACD of triangle is equal to AED of triangle, ed = CD = 1.5
In the right triangle EDB, EB = 2 (because ed = 1.5, BD = 2.5) can be obtained by Pythagorean theorem
Let AC = X. in the right triangle ABC, from the Pythagorean theorem, it can be obtained that:
X? 2 + 4? 2 = (2 + x) 2 16 = 4 + 4x, so x = 3
So AC = 3
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 in E, ∵ de ⊥ AB, AED = 90 °, ACB = ∠ AED = 90 ° and ? CAD = ≌△ AED, CD = ed, AC = AE, ∵ ACB = 90 °, AC = BC, ? B = 45 ° and EDB = 45 °, ed = EB, CD = E
As shown in the figure, in the triangle ABC, the angle c = 90 degrees, AC = BC, ad bisects the angle cab and intersects BC at point D. can a point e be determined on AB Make the perimeter of the triangle BDE equal to the length of ab. if you can, please make point E and give proof. If not, please explain the reason. The diagram is a right triangle cab (angle ACB is a right angle), and ad is the bisector of angle cab,
Because ad bisects angle cab, so CD = De, and because AC = ad, angle c = 90 degrees, so angle CAE = angle B = angle EDB = 45 degrees, so de = EB, so AB = de + EB + DB = BC + EB = AC + EB = AE + EB
In the triangle ABC, C is 90 ° and ad is the bisector of ∠ cab, where BC intersects with D, BC is 4 and CD is 1.5. Find the length of AC
Draw de ⊥ AB through D. the vertical foot e ∵ C is 90 ° and ad is the bisector of ∵ cab ᙽ CD = de = 1.5, AC = AE ∵ BC is 4 ᙽ BD = 4-1.5 = 2.5be = √ (2.5 ∵ 1.5)) = 2 ∵ ab ∵ ab ∵ ab ∵ AC ∵ AC ∵ AC ∵ AC ∵ AC = 1.5, AC = AE
In the triangle ABC, it is known that CAB=120 °, AB=4, AC=2, AD is vertical BC, and the perpendicular foot is D, so the length of BC is obtained
The area of the triangle ABC = 0.5 * 2 * 4 * sin120 = 2 times the root sign 3. According to the cosine theorem, CB = 2 times the root 7. Because the area of the triangle ABC = 0.5 * ad * CB, ad = (2 times the root 21) / 7
RELATED INFORMATIONS
- 1. In the triangle ABC, the angle cab120 degrees, ab = 4, AC = 2, ad is perpendicular to BC, and the perpendicular foot is d? Students are in the second year of junior high school, just learning Pythagorean theorem. We hope to solve it in other ways,
- 2. An inner angle of the quadrilateral ABCD is 120 ° connecting AC to obtain equilateral △ ABC and right triangle ACD. It is known that the side length of equilateral △ ABC is 2 (1) Find the height on the base BC of △ ABC (2) find the area of △ ACD
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- 8. As shown in the figure, rotate RT triangle ABC by 40 ° anticlockwise around point a to obtain RT triangle ab'c '. Point C' just falls on the hypotenuse AB and connects BB ', then ∠ bb'c' = °
- 9. Given vector a = (- 1 / 2, root 3 / 2), vector OA = vector a - vector B, vector ob = vector a + vector B, If △ AOB is an isosceles right triangle with o as its right vertex (1) coordinates of vector b (2) the angle between vector a and vector B
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- 16. In the triangle ABC, if edge B + C = radical 2 + 1, angle c = 45 degrees, angle B = 30 degrees, what are the values of edges B and C
- 17. In RT △ ABC, ∠ C = 90 ° s = 18 root sign 3, a
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- 19. In the RT triangle ABC, ∠ C = 90 °, ab = radical 10, AC: BC = 2:1, calculate the perimeter and area of RT triangle
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