(1) find the angle c;
(2) Find the maximum value of △ABC area.
Classroom exercises:
1. In △ABC, it is known that A = 5, C = 10 and A = 30, then ∠B= ().
(a)105 (b) 60 (c) 15 (d)105 or15.
2 in △ABC, if A = 2, B = 2 and C =+
30 (B) 45 (C) 60 (D) 75
3. In △ABC, it is known that three sides A, B and C satisfy (A+B+C) (A+B-C) = 3ab, then ∠C= ().
15 30 45 60
4. The sum of the maximum and minimum angles of triangles with side lengths of 5, 7 and 8 is ()
(A)90(B) 120(C) 135(D) 150
5. In △ABC, ∠ A = 60, A =, B = 4, then the conditional △ABC () is satisfied.
(a) There is one solution (b) There are two solutions (c) There is no solution (d) and it cannot be determined.
6. In parallelogram ABCD, AC=BD, then the maximum value of acute angle A is ().
30 (B) 45 (C) 60 (D) 75
7. In △ABC, if = =, the shape of △ABC is ().
(a) isosceles triangle (b) equilateral triangle (c) right triangle (d) isosceles right triangle
8. If all three sides of a right triangle are increased by the same length, the shape of the new triangle is ().
(a) acute triangle (b) right triangle (c) obtuse triangle (d) is determined by the increased length.
9. In △ABC, if a=50, B= 25 and a = 45, then B=.
10. If the lengths of two adjacent sides of a parallelogram are 4cm and 4cm respectively, and their included angle is 45, then the lengths of the two diagonals of this parallelogram are respectively.
1 1. In the isosceles triangle ABC, it is known that sinA∶sinB= 1∶2 and BC= 10, then the circumference of △ABC is.
12. In △ABC, if ∠ B = 30, AB = 2 and AC = 2, the area of △ABC is.
13. In an acute triangle, sides A and B are two sides of the equation x2-2x+2 = 0, and angles A and B satisfy 2sin (a+b)-= 0. Find the degree of angle c, the length of side c and the area of △ABC.
14. In △ABC, it is known that side C = 10 and = =. Find the radius of the inscribed circle of a, b and △ABC.
15. In the known quadrilateral ABCD, BC = a, DC=2a, and the ratio of degrees of four corners, A, B, C and D, is 3: 7: 4: 10. Find the length of AB.
16. In △ABC, it is known that the opposite sides of angles A, B and C are A, B and C, respectively, and the side c=, Tana+Tanb = Tana Tanb-, and the area of △ABC is S△ABC=. Find the value of a+B.
Reference answer:
Classic example: solution: (1)√.
∫2R(sin2A-sin2C)=(a-b)sinB
∴ 2R〔()2-()2〕=(a-b) ∴ a2-c2=ab-b2
∴ ∴ cosC=,∴ C=30
(2)S = absinC = 2r Sina 2r sinb sinC = R2 Sina sinb
=-〔cos(A+B)-cos(A-B)〕=〔cos(A-B)+cosC〕
= [cos (a-b)+] When cos (a-b) = 1, s has a maximum value.
Classroom exercises:
1.d； 2.a； 3.d； 4.b； 5.c； 6.c； 7.b； 8.a； 9.60 or120; 10.4cm； and 4cm; 1 1.50; 12.2 or;
13, solution: from 2 sin (a+b)-= 0, we get that sin(A+B)=, ∫△ABC is an acute triangle.
∴ A+B = 120, C = 60, and ∵a and b are two roots of the equation x2-2x+2 = 0, ∴a+b=2,
a b=2，∴c2=a2+b2-2a bcosc =(a+b)2-3ab = 12-6 = 6，
∴c=，S△ABC=absinC=×2×=。
14. solution: from =, =, we can get =, and the deformation is sinAcosA=sinBcosB.
∴sin2A=sin2B, and ∵ a ≠ b, ∴ 2a = π-2b, ∴ a+b =. ∴△ ABC is a right triangle.
A2+b2= 102 and =, a = 6, b = 8, and the radius of the inscribed circle is r===2.
15、
Solution: Let the degrees of A, B, C and D be 3x, 7x, 4x, 10x respectively. According to the sum of the internal angles of the quadrilateral, there are 3x+7x+4x+ 10x = 360. The solution is x = 15 ∴ A = 45.
Connect BD to get two triangles △BCD and △ABD.
In △BCD, it is obtained by cosine theorem
BD2 = BC2+DC2-2BC DC cosC = a2+4a 2-2a 2a = 3 a2，
∴ BD = A, when DC2=BD2+BC2, we can get that △BCD is a right triangle with DC as the hypotenuse. ∴ CDB = 30, so ∠ ADB = 120.
In △ABD, AB = = = comes from sine theorem.
∴ The length of AB is
16, solution: Tana+Tanb = Tana Tanb-Available =-, that is, Tan (a+b) =-
∴tan(π-C)= -，∴-tanC=-，∴tanC=∵C∈(0，π)，∴C=
The area of △ABC is S△ABC=, ∴absinC= that is, ab×=, ∴ab=6.
From the cosine theorem, we can get C2 = A2+B2-2abckos ∴ () 2 = A2+B2-2abckos ∴ () 2 = A2+B2-AB = (a+b) 2-3ab.
∴(a+b)2=，∫a+b & gt； 0，∴a+b=
Thirdly, the solution of m=2 or m=
But the sum of 2 does not satisfy the above formula, so such m does not exist.