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In the triangle ABC, ab = AC = 2A, ∠ ABC = ACB = 15 ° CD is the height on the waist ab
Extend Ba to D, so CD is the height of waist AB, ∠ ABC = ACB = 15 °, so, ∠ CAD = 30 °,
Delta CAD is right triangle ∠ CAD = 30 degree
So CD = 1 / 2Ac = a
As shown in the figure, in the triangle, ab = AC = 2A, angle ABC = angle ACB = 15 °, AC is the height of the waist, and the length of BD is calculated
∵∠ABC＝∠ACB＝15
∴∠BAD＝∠ABC+∠ACB＝30
∵BD⊥AD
∴BD＝AB/2
∵AB＝2a
∴BD＝a
In the triangle ABC, the angle ACB is equal to 90 °, the angle a is equal to 15 ° and ab = 20cm
AC=ABcos15°=20*cos15°
sin15°*cos15°=1/2sin30°
S△ABC=1/2 AB*AC*sinA=1/2 *AB² *(1/2)*sin30°=50cm²
Known triangle ABC, angle a is equal to 60 °, AB is equal to 1, AC is equal to 4, how to find the area of the triangle, the gods help
S = AB * ac * Sina / 2 = 1 * 4 * sin 60 ° / 2 = 4 * (√ 3 / 2) / 2 = √ 3 question: is s = AB * ac * Sina / 2 a triangle with an angle between two sides more useful? Answer: Well, if you add it to a parallelogram, it's not half of a parallelogram
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