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As shown in the figure, it is known that ∠ AOB = 30 ° and the point P is inside ∠ AOB, Op = 6. If there is a moving point m on OA and a moving point n on ob, then the minimum perimeter of △ PMN is______ .
Make the symmetry points c and D of point P with respect to OA and ob, connect CD, respectively cross OA and ob at points m and N, and connect OP, OC, OD, PM and PN
As shown in the figure, given a point P in ∠ AOB and ∠ AOB, you can find a point Q and R on the edge of OA and ob, so that the circumference of the triangle composed of three points P, Q and R is minimized
As shown in the figure
It is known that: as shown in the figure, a point P in AOB, ∠ AOB = 60 ° and op = 6, make a point m, N on OA and OB to make the circumference of △ MPN the shortest, and find its value
As shown in the figure, we know that P is a point on the edge OA of ∠ AOB, and the two sides of ∠ MPN with P as the vertex respectively cross ray ob in proof: (2) in △ OPN and △ PMN, ∠ PON = ∠ MPN = 60 °, ONP = ∠ PNM, νΔ
As shown in the figure, ∠ AOB = 30 °, point P is a point in ∠ AOB, Op = 10, points m and N are respectively on OA and ob, and the minimum value of △ PMN perimeter is calculated
If we make the symmetry points P1 and P2 of point P with respect to OA and ob respectively, then the intersection of OA with m and the intersection of OB with N, then OP1 = OP = op2, ∠ p1oa = ∠ POA, ∠ POB = ∠ p2ob, MP = p1m, PN = p2n, then the minimum perimeter of △ PMN = p1p2 ﹤ p1op2 = 2 ﹣ AOB = 60 ° and ﹤ op1p2 is an equilateral triangle
In the triangle ABC, given a = 2, B = radical 2, C = radical 3 + 1, find a
In the triangle ABC, given a = 2, B = radical 2, C = radical 3 + 1, find a
Direct use of cosine theorem
cosA=(b²+c²-a²)2bc=2(1+√3)/2√2(1+-√3)=√2/2
A=45
In the triangle ABC, the angle c = 30 ° AC is equal to 4 roots 3, ad is the center line, and the values of ad ⊥ AC, 逑 AB and SINB are the values
Ad is the midline and ad ⊥ AC
The title is wrong
In the triangle ABC, if angle a = 75 degrees, SINB = root 3 / 2, then Tanc = 1, I don't know how to calculate it How is there no right angle in the triangle? How to find SINB
∵∠A=75° sinB=√3/2
∴0
In the triangle ABC, if the modulus of AB vector = radical 3, the modulus of BC vector = 1, Sina = SINB, then AC vector dot multiplies AB vector=
If Sina = SINB, then 2R Sina = 2R SINB, (2R is the circumscribed circle diameter of the triangle)
According to the sine theorem, a = B,
The modulus of BC vector is 1, so a = b = 1
The norm of AB vector = √ 3, that is, C = √ 3
According to cosine theorem, cosa = √ 3 / 2 is obtained
AC vector dot times AB vector = 1 ×√ 3 × cosa = 3 / 2
As shown in the figure, the square ABCD is connected to ⊙ o, e is the midpoint of DC, and the line be intersects ⊙ o at point F, if the radius of ⊙ o is 2, then the distance from o point to be OM=______ .
Make om ⊥ be to m, connect OE and BD,
∵∠DCB=90°，
/ / BD is the diameter,
∵OE=DE=1，
∴BE=
4+1=
5，
∵EF=DE•CE
BE=
Five
5，
∴BF=6
Five
5，
∴MF=3
Five
5，ME=2
Five
5，
∴OM=
1-4
5=
Five
5．
In △ ABC, BC = 1, ∠ B = 60 ° (1) if AC = radical 3, find AB (2) if cosa = (2 radical 7) / 7, find Tanc If you can't type the symbols, you can use words to describe them
In △ ABC, BC = 1, ∠ B = 60 °
(1) If AC = √ 3, find AB:
The sine theorem is applied
BC/sinA=AC/sinB
sinA=BC*sinB/AC
=BCsin60°/AC
=[1*√3/2]/√3
=1/2.
∴∠A=30°
Obviously, △ ABC is a right triangle, and ∠ C = 90 ° and ab is a hypotenuse
AB^2=AC^2+BC^2
=(√3)^2+1^2=4
∴AB=2.
(2) If cosa = 2 √ 7 / 7 = 0.7559, (∠ B = 60 °) [at this time △ ABC is not RT △!]
∠A=40.89°=40.9°
In △ ABC, ∠ C = 180 ° - 60 ° - 40.9 ° = 79.1 °
∴tanC=tan79.1°=5.929=5.93.
RELATED INFORMATIONS
- 1. In the triangle ABC, if AC = radical 2, BC = radical 7, ab = 3, then cosa =?
- 2. Known: as shown in the figure, in the right triangle ABC, the angle c = 90 degrees, the point O is the center of the circle, the length of OA is the radius of the circle, and AC, AB intersect with point D.E, and the angle CBD = angle a.bd is the tangent line of circle O. find: if ad: Ao = 8:5, BC = 2, find the length of BD
- 3. Point O is a point in △ ABC, ∠ ABO = 25 degrees, ∠ ACO = 35 degrees, and the degree of ∠ BOC is 5 degrees less than 2 times of ∠ a In △ ABC, the top angle is ∠ a, the left is ∠ ABC, and the right is ∠ ACB
- 4. In △ ABC, if the bisectors of ∠ a = 50 °, B and C intersect at point O, then the degree of ∠ BOC is () A. 65° B. 100° C. 115° D. 130°
- 5. As shown in the figure, in the RT triangle ABC, the angle B = 90 degrees. If the distance between the point O and the triangle three times is equal, what is the angle AOC? For detailed explanation
- 6. It is known that: as shown in the figure, in △ ABC, the bisector of ∠ B and the bisector of external angle of △ ABC intersect at point D, ∠ a = 90 °
- 7. It is known that the distance between point O and AB and AC is equal, and ob = OC (1) As shown in Figure 1, if point O is on edge BC, it is proved that ab = AC; (2) As shown in Figure 2, if point O is inside △ ABC, it is proved that ab = AC; (3) If point O is outside △ ABC, is ab = AC true? Please draw a picture to show it
- 8. In the triangle ABC, BC = 2, Sina = 2 / 3, root 2, then what is the maximum product of AB vector and AC vector
- 9. In △ ABC, Tanc = 3, radical 7 (1) find COSC (2) if vector CB · vector CA = 5 / 2 and a + B = 9 find C Process
- 10. In the triangle ABC, a = 60 degrees, AC = 16, area s = 220, root number 3, then BC = what?
- 11. As shown in the figure, the side of the cone is cut along OA, and the fan-shaped OAB is expanded into a plane figure (1) What is the relationship between the length of the sector arc AB and the length of the circle at the bottom of the cone? What is the position relationship between point a and point B on the side of the cone? (2) If the angle ∠ AOB = 90 °, what is the relationship between the radius r of the cone bottom circle and the radius r of fan-shaped OAB (i.e. OA or OB)? (3) If point a moves on the side of the cone and then returns to its original position, how should the shortest distance of point a be designed? If R 2 = 0.5, ∠ AOB = 90 °, find the shortest distance of point a
- 12. The right angle vertex P of the triangle ruler slides on the ray OM, and the two right angle sides intersect OA and ob at point CD respectively Conjecture the quantitative relationship between PC and PD
- 13. As shown in the figure, OA, OB and OC are the radii of circle O. if ∠ AOB = 2 ∠ BOC, please judge whether ∠ ACB = 2 ∠ BAC is tenable. Why?
- 14. As shown in the figure, the ray OA, ob, OC, OD have common endpoint o, and ∠ AOB = 90 °, COD = 90 °, AOD = 5 4 ﹤ AOC. calculate the degree of ﹤ BOC
- 15. In the plane rectangular coordinate system, the image of function y = - 3 / 4x + 6 intersects X axis and Y axis at point a and B respectively, line BC and X axis intersect at C, and point C is the midpoint of line OA Find out whether there is a point m on the straight line AB so that △ BCM is an isosceles triangle. If there is, find out the coordinates of point M
- 16. Known: as shown in the figure, find a point P on the straight line Mn, so that the distance between the point P and the two sides of ∠ AOB is equal (the method is required to be written, the drawing trace shall be retained, and the conclusion shall be written)
- 17. As shown in the figure, ∠ AOB = 30 ° OC bisection ∠ AOB, P is a point on OC, PD ‖ OA intersects ob at point D, PE ⊥ OA at e, OD = 4cm, then PE ⊥ OA is at point E=______ .
- 18. The bottom surface of four pyramids P-ABCD is square, and the bottom surface of PB ⊥ ABCD proves that no matter how the height of four pyramids PB changes, the surface PAD and the surface PCD cannot be vertical
- 19. It is known that: as shown in the figure, P is the interior point of the square ABCD, ∠ pad = ∠ PDA = 15 °. Verification: △ PBC is an equilateral triangle
- 20. Ray OA, ob, OC, OD have a common endpoint o, and OA is perpendicular to ob, OC is perpendicular to OD, ∠ AOD = 5 / 4 ∠ AOC, calculate the degree of ∠ BOC ∠ AOD = 5 / 4 ∠ AOC! No, ∠ AOD = 5 / 4 ∠ BOC