Solution: It is very simple to solve with sine theorem.
S=(ab*sina)/2
=(2A*2A*sin 150 )/2
=A^2
2. Half of the vertex angle
3. The hypotenuse is 5 degrees
The height on the hypotenuse is 02 of 65438+5.
4. In triangle ABC, angle C is a right angle, CD is perpendicular to AB, angle A is equal to 30 degrees, and angle DCE is equal to.
45 60 50 65
Solution: It is 45.
Analysis: According to angle A = 30, we can know that angle B = 60, because BC=BD, so angle CDB= angle BCD= angle B = 60.
Because AE=AC, angle A = 30, angle AEC= angle ACE = 75, there is an angle DCE =180-75-60 = 45 in the triangle DCE according to the theorem of the sum of internal angles of the triangle.
5. It is known that the high line on one waist of an isosceles triangle is half the waist length, so a base angle of this isosceles triangle is ().
Solution: It can be calculated that the top angle is 150, and the low angle is 15.
If you make a standard drawing, you can know that it should be an obtuse triangle, and the height of the waist is outside the triangle. According to the fact that the right-angled side opposite to the angle of 30 in the right-angled triangle is equal to half of the hypotenuse, we can know that the complementary angle of the top angle of the isosceles triangle is 30, so the top angle is 150.