![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
In the triangle ABC, ab = 14, BC = 14, s △ ABC = 84, find Tanc, Sina
The angle bisector passing through point B intersects at point D. because AB = BC = 14, it is an isosceles triangle, so BD is perpendicular to AC
BD is high and AC is the bottom of the triangle;
S△ABC=84
BD*AC/2=84
Isosceles triangle bisector bisector angle opposite side, so ad = DC = AC / 2
BD*AD=84 >> BD=84/AD
Right triangle theorem
BD^2+AD^2=AB^2
BD^2+AD^2=14^2
(84/AD)^2+AD^2=14^2
84^2/AD^2+AD^2=14^2
For the convenience of calculation, let x = ad ^ 2
84^2/X+X=14^2
84^2/X+X^2/X=14^2
(84^2+X^2)/X=14^2
84^2+X^2=14^2X
X^2-14^2X+84^2=0
Solve two roots of X, one of which is negative root invalid, ad ^ 2 is equal to this positive value
Then you can calculate all the side lengths. Tanc, Sina & nbsp; will come out naturally
In Δ ABC, (1) we know that sina = cosbcosc, and prove that Tanc + tanb = 1
tanC + tanB =sinC/cosC+sinB/cosB=(sinCcosB+sinBcosC)/cosBcosC=
sin(B+C)/sinA=sin(180-A)/sinA=1
In △ ABC, if Sina = cosbcosc, it is proved that tanb + Tanc is constant
A = 180-b-c (the sum of internal angles of a triangle is 180)
∴sin(180-(A+B))=cosBcosC
sinBcosC+sinCcosB=cosBCosC
∵ B, C ∈ (0180) (the number of C and B between 0 and 180) and Sina = cosbcosc ∵ B, C will not have one equal to 90
The equation is also multiplied by 1 / (sinbcosc)
1+（sinCcosB）/（sinBcosC）=（cosBCosC）/（sinBcosC）
1+tanC/tanB=1/tanB
Namely:
Tanb + Tanc = 1 (constant value)
In the triangle ABC, ab = AC = 5, BC = 6, find Sina
cosA=（c^2+b^2-a^2)/2cb
a=BC=6 b=c=AB=AC=5
Then soca = (C ^ 2 + B ^ 2-A ^ 2) / 2CB
=（25+25-36）/（2*5*5）
=0.56
Then (Sina) ^ 2 + (COSA) ^ 2 = 1
Then Sina ≈ 0.83
Finding Sina by triangle ABC, ab = BC = 2, AC = 1
Let's take the high BD on the edge of AC, ad = 0.5, ab = 2, and get the root 15 of BD = 2
So Sina = the root of 4 is 15
The bisectors of the three inner angles of a triangle intersect at the same point, which is called the center of the triangle. As shown in the figure, use a ruler and a compass to draw the center point P of the inner circle of △ ABC
Drawing with ruler and gauge:
According to the definition of the inner triangle, you just need to draw the intersection of the bisectors of the two inner angles, which is the inner P of the triangle
Given that the area of triangle ABC is 1, be = 2Ab, BC = CD, find the area of triangle BDE
Don't plagiarize, make a good formula
Because be = 2Ab, BC = CD,
So the BDE of triangle is 2 * (1 + 1) = 4 times of ABC of triangle
Then the area of triangle BDE is 1 * 4 = 4
A: (omitted)
Divide the two waists of the isosceles triangle ABC into three or four equal parts. It is known that the area of the shadow part is 25 square centimeters. How to find the area of the triangle ABC?
Another method to calculate the area of triangle is s = 1 / 2absin θ; θ is the angle between side lengths a and B
Set the shortest edge and use it to represent the rest. The area of the shadow should be the difference between the two triangles,
The area of two triangles can be expressed, the triangle ABC can also be expressed, and finally it can be solved
The area of triangle ABC is 180 square centimeters. Divide side length AB into two equal parts. Take the middle point as e and divide side length BC into three equal parts
According to the meaning of your title, G and F are the three equal points of BC
So the area of triangle EFB is 30, and the area of triangle GFE is 30
Make a median line through point E, which is easy to calculate
In triangle ABC, both sides of AB and AC are divided into 5 equal parts. The ratio of the area of the shadow part to the area of triangle ABC is 0______ ．
3: 5 & nbsp; as shown in the figure, let de = 1, then FG = 2, hi = 3, JK = 4, bc-5. The shadow part can be regarded as five triangles with de, FG, hi, JK and BC as the base respectively, while the non shadow part can be regarded as four triangles with JK, hi, FG and de as the base respectively, and the heights of these nine triangles are equal
RELATED INFORMATIONS
- 1. In the triangle ABC, D and E are close to the trisection point of a and B respectively. How many times is the area of triangle ABC?
- 2. In the acute triangle ABC, AB equals 14, BC equals 14, the area of triangle ABC is 84, find the value of Tanc and sinc I have worked it out, but the result is too complicated. If you want to verify it, please give the result
- 3. In the acute triangle ABC, AB > AC, CD and be are the heights on the sides of AB and AC respectively. The extension line of De and BC is called t, and the vertical line of BC crossed by D, BC crossed by F, It is proved that f, G and T are collinear
- 4. In △ ABC, a, B and C are the opposite sides of angles a, B and C respectively, a = 60 ° and (b-2c) / ACOS (60 ° + C)=
- 5. As shown in the figure, △ ABC, ab = AC, points D, e and F are on edges BC, AB and AC respectively, and BD = CF, ∠ EDF = ∠ B. is there a triangle congruent with △ BDE in the figure? And explain the reason
- 6. Triangle ABC, angle bisector of angle a intersects with D, BD = 3, DC = 4, find the inscribed circle radius of triangle ABC
- 7. It is known that in △ ABC, ab = AC = 13, BC = 10, then the radius of the inscribed circle of △ ABC is () A. 103B. 125C. 2D. 3
- 8. In RT triangle ABC, angle ACB is equal to 90 degrees, angle a = 30 degrees, BD is the bisector of triangle ABC, De is perpendicular to point E 1, In RT triangle ABC, angle ACB is equal to 90 degrees, angle a = 30 degrees, BD is the bisector of triangle ABC, De is perpendicular to point E Q: connect EC and prove that the triangle EBC is an equilateral triangle
- 9. As shown in the figure, in △ ABC, the angle a = 60 ° and the circle O with BC as diameter intersects with AB and AC at points D and e respectively. Try to judge the shape of △ deo and prove it
- 10. As shown in the figure, in △ ABC, ab = AC, D is a point on the edge of BC, with AB and BD as adjacent sides, make parallelogram ABDE and connect AD and EC
- 11. Triangle ABC cosa = 1 / 2 tanb= ∠c=90°
- 12. A necessary and sufficient condition for the equation AX2 + BX + C = 0 of X to have a root of - 1 is__________ . Better have analysis! ***********It's "one root is - 1", not "only one root is - 1"************
- 13. Let the equation AX ^ 2 + BX + C = 0 and the coefficients a, B and C are odd numbers
- 14. In △ ABC, ab = AC, De is the vertical bisector of AB, the perpendicular foot is D, intersecting AC with E, connecting be. (1) if ∠ Abe = 40 °, calculate the degree of ∠ C; (2) if the perimeter of △ ABC is 82cm, one side is 30cm, calculate the perimeter of △ BCE
- 15. As shown in the figure, in ABC, three bisectors ad, be, CF intersect at O, og ⊥ BC at g, and prove: BOD = GOC!
- 16. As shown in the figure, in the isosceles trapezoid ABCD, ab ∥ CD, DC = ad = BC, and the diagonal AC is perpendicular to the waist BC, calculate the degree of each internal angle of the trapezoid
- 17. In the triangle ABC, angle c equals 90 degrees. If angle a equals 45 degrees and C equals 2, find a and B
- 18. If △ ABC ∽ def, ab = 2, BC = 2 times radical 3, de = radical 2, then EF is emergency
- 19. D point on AB, ad: ab = 1:3, e point on BC, be: BC = 1:4, f point on Ca, CF: CA = 1:5, s △ def = 19, find s △ ABC?
- 20. Two identical triangles can be combined into one () A. Parallelogram B. trapezoid C. rectangle