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As shown in the figure, in triangle ABC, D is a point on the edge of BC, angle 1 = angle 2, angle 3 = angle 4, angle BAC = 63 ° to find the degree of angle DAC
prove:
∵∠3=∠1+∠2,∠3=∠4,∠1=∠2
∴∠4=2∠2
∵∠BAC =63°
∴3∠2+63°=180°
∴∠2=39°
∴∠3=78°
∴∠DAC=180-78-78=24°
In the triangle ABC, D is a point on the edge of BC, angle one equals angle two, angle three equals angle four, angle BAC equals 63 ° and finds the degree of ∠ DAC
prove:
∵∠3=∠1+∠2,∠3=∠4,∠1=∠2
∴∠4=2∠2
∵∠BAC =63°
∴3∠2+63°=180°
∴∠2=39°
∴∠3=78°
∴∠DAC=180-78-78=24°
In the triangle ABC, D is a point on the edge of BC. Angle 1 equals angle 2, angle 3 equals angle 4, and angle BAC equals 63 degrees. Find the degree of angle DAC
∵ 3 is the external angle of ∵ abd,
∠ 3 = ∠ 1 × 2,
∵∠1=∠2,∠3=∠4
∴∠4=2∠2
∠2＋∠4=180°-∠BAC=180°-63°=117°
Ψ 1 = ∠ 2 = 117 ° / (1 + 2) = 39 °
∠DAC=∠BAC-∠1=63°-39°=24°.
As shown in the figure, △ ABC is an equilateral triangle, CD ⊥ BC, and BC = CD
∵△ ABC is an equilateral triangle, ACB = 60 °, AC = BC,
As shown in the figure, the known point D is in the triangle ABC. It is proved that the angle ADB = angle DAC + angle DBC + angle C
Because ∠ DAB + ∠ DBA + ∠ ADB = 180 °∠ a + ∠ B + ∠ C = 180 °
So ∠ DAB + ∠ DBA + ∠ ADB = ∠ a + ∠ B + ∠ C
∠DAB+∠DBA+∠ADB=∠DAB+∠DAC+∠DBA+∠DBC+∠C
So ∠ ADB = ∠ DBC + ∠ DBC + ∠ C
As shown in the figure, △ ABC is an equilateral triangle, CD ⊥ CB, and BC = CD. Find the size of ∠ DAC and ∠ ADB
∠DAC=∠ADB=15°
Because ∠ ACD = 150 ° AC = CD, ACD is isosceles triangle, and BC is located between AC and CD
There is another case where DAB = 30 ° and
At this time, AC is located between BC and CD
As shown in the figure, in △ ABC, D is a point on BC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ∠ BAC = 120 °, then ∠ DAC=______ Degree
∵∠BAC=120°，
∴∠2+∠3=60°①
∵∠1=∠2，
∴∠4=∠3=∠1+∠2=2∠2②
Substituting (2) into (1) gives: 3 ∠ 2 = 60 °,
∠2=20°，
∴∠1=20°．
∴∠DAC=120°-20°=100°．
So the answer is: 100
As shown in the figure, in △ ABC, D is a point on BC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ∠ BAC = 120 °, then ∠ DAC=______ Degree
∵∠BAC=120°，
∴∠2+∠3=60°①
∵∠1=∠2，
∴∠4=∠3=∠1+∠2=2∠2②
Substituting (2) into (1) gives: 3 ∠ 2 = 60 °,
∠2=20°，
∴∠1=20°．
∴∠DAC=120°-20°=100°．
So the answer is: 100
In RT △ ABC, ∠ C = 90 °, BC = 5, Sina = 0.7, find the value of cosa and Tana
∵∠C=90°，
∴sinA=BC
AB，
∵BC=5，sinA=0.7，
∴BC
AB=0.7，
∴AB=50
7，
According to Pythagorean theorem, AC = 5
Fifty-one
7，
∴cosA=AC
AB=
Fifty-one
10，tanA=BC
AC=7
Fifty-one
51．
In RT △ ABC, ∠ C = 90 degrees, BC = 5, sin = 0.7, find the values of COS A and Tan A. in RT △ ABC, ∠ C = 90 degrees, BC = 5, sin = 0.7, calculate the values of COS A and Tan a Wrong number, sin a = 0.7
Because in RT △ ABC, ∠ C = 90 degrees, BC = 5, Sina = 0.7
So Sina = BC / AB = 0.7
So AB = 50 / 7, AC = 5 / 7 times root number 51
Cosa = AC / AB = (Radix 51) / 10
Tana = CB / AC = (7 times root number 51) / 51
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