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It is known that in RT △ ABC ∠ ACB = 90 ° CD ⊥ AB, the perpendicular foot is d BC = 2, BD = radical 3, and the acute angles in △ ABC △ ACDC △ BCD are calculated respectively
∵CD⊥AB
BD=√3,BC=2
According to Pythagorean theorem, CD = 1
∴∠B=30°
∴∠BCD=60°
∵∠ACB=90°
∴∠A=60°,∠ACD=30°
As shown in the figure, in the RT triangle ABC, the angle ACB = 90 ° CD is perpendicular to D, ab = 2, root sign 6, AC = 2 times root sign 6 Find the value of COS angle BCD
∠ACB=90°,
The diagonal edge AB and right angle edge AC are both 2 √ 6,
This condition is contradictory
As shown in the figure, in RT △ ABC, CD ⊥ AB is in D, AC = 10cm, ad = 6cm, BC = radical 73 (1) Find the angle of CD and BD (2) Find the area of △ ABC
(1) ∵ CD ⊥ ab ᙽ the angle between Cd and BD = 90 ° (2) ∵ ad ⊥ CD ∵ CD = 8 ∵ ∵ ACD ∵ abcac / AB = CD / bcab = AC * BC / CD = 10 * √ 73 / 8 = 5 / 4 √ 73s △ ABC = AB * CD / 2 = 5 / 4 √ 73 * 8 / 2 = 5 √ 73
As shown in the figure, in RT △ ABC, ∠ ACB = 90 ° and point D is the midpoint of AB, and CD= Five 2, if the area of RT △ ABC is 1, then its circumference is () A. 5+1 Two B. 5+1 C. 5+2 D. 5+3
As shown in the figure, ∵ in RT △ ABC, ∵ ACB = 90 ° and point D is the midpoint of AB, and CD=
Five
2，
∴AB=2CD=
5．
∴AC2+BC2=5
The area of ∵ RT △ ABC is 1,
∴1
If 2Ac · BC = 1, then AC · BC = 2
∴（AC+BC）2=AC2+BC2+2AC•BC=9，
/ / AC + BC = 3 (minus the negative value),
∴AC+BC+AB=3+
The circumference of △ ABC is 3+
5．
Therefore, D
In RT △ ABC, ∠ C = 90 ° a = radical 6, ∠ a = 60 ° find B, C Pythagorean theorem
a/sinA=b/sinB=c/sinC
√6/sin60=b/sin30=c/sin90=2√2
b=√2,c=2√2
If the angle c = 90 degrees, C = root 11, a + B = root 5 + root 6, then the area of RT triangle is（
Radical 30 / 2
In triangle ABC, is it known that point D is a point on edge BC, and angle DAC = angle B, angle BAC = angle ADC?
Yes
Angle BAC = bad + DAC
ADC=B+DA
So equal
In the triangle ABC, D is a point on the edge of BC, and the angle bad = angle B, the angle ADC = angle c, the angle BAC = 63 ° to find the degree of angle DAC
Let B = X
Angle bad = angle B = x
Then the angle ADC = angle c = 2x
X+2X=180-57=123
X=41
Angle DAC = 57-41 = 16
Known: D is a point on the BC side of the triangle ABC, and the angle ADC = angle BAC. Do you think that angle DAC = angle B? Explain your reasons
Equal, angle ADC + angle ACD + angle DAC = 180, angle BAC + angle ACD + angle B = 180, angle ADC = angle BAC, angular DAC = angle B
Known: as shown in the figure, ∠ DAC = ∠ B, verification: ∠ ADC = ∠ BAC
It is proved that: ∵ DAC = ∠ B, ∠ C = ∠ C,
∴∠ADC=180°-∠C-∠DAC，∠B=180°-∠C-∠BAC，
∴∠ADC=∠BAC．
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