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As shown in △ ABC, the angle ACB = 90 °, the angle a = 60 ° and the height of the hypotenuse CD = the root sign 3. Find the length of ab D is on AB, CD is perpendicular to AB, AC is perpendicular to CB
Angle B = 30 degrees
CB=2CD=2√3
AC=1/2AB
According to Pythagorean theorem, ab = 4
As shown in the figure, in RT △ ABC, ∠ ACB = 90 ° and CD is the height above AB edge. If ad = 8 and BD = 2, find CD
∵ in RT △ ABC, ∵ ACB = 90 ° and CD is the height of AB edge
∴∠BDC=∠ACB=90°
∵∠B=∠B
∴△ABC∽△CBD
∴CD2=AD•BD，
∵AD=8，BD=2，
∴CD=
8×2=4．
As shown in the figure, in the right triangle ABC, the angle ACB is equal to 90 degrees, ab = 5cm, BC = 3cm, CD is perpendicular to point D. find: the length of CD
Right angle △ ACB is similar to right angle △ CDB (∠ BAC = ∠ BCD, the same is the remainder angle of ∠ ABC),
And AC = 4cm (Pythagorean theorem)
So CD / 3 = 4 / 5 (the corresponding sides of similar triangles are proportional)
CD＝12／5
＝2.4cm.
As shown in the figure, in △ ABC, ad ⊥ BC, the perpendicular foot is D, ab = 2 root sign 2, AC = BC = 2 root sign 5, find the length of AD
Six fifths root five
In the equal right triangle ABC, ∠ BAC = 90 °, P is a point in △ ABC, PA = 1, Pb = 3, PC = radical 7, and find the size of ∠ CPA
It is suggested that ∠ CPA = 135 °; simple explanation is as follows: rotate △ ABP anticlockwise about point a to ⊿ ACQ (q is outside ⊿ ABC, right side of AB and AC), and connect PQ; then PQ = √ (AP ⊿ AQ ⊿ AQ ⊿ 2) = √ (1 1 Apq = 45 °; CQ = BP = 3 ∵ 3 3 = (√
In △ ABC, ∠ cab = 90 degrees, AC = AB, P is a point in △ ABC, which satisfies the quadratic root of PA = 1, Pb = 3, PC = 7, and calculates the degree of angle CPA
Let P turn to Q. AQ = AP = 1, BQ = PC = √ 7, ∠ PAQ = 90 °. PQ ^ 2 = 2, QB ^ 2 = 7, Pb ^ 2 = 9, which satisfies PQ ^ 2 + QB ^ 2 = Pb ^ 2
As shown in the figure, in the isosceles right triangle ABC, ∠ a = 90 °, P is a point in △ ABC, PA = 1, Pb = 3, PC= 7, then ∠ CPA=______ Degree
Rotate △ ABP anticlockwise about point a by 90 ° and then connect PQ, then AQ = AP = 1, CQ = Pb = 3, ∠ QAC = ∠ PAB, ∵ QAP = 90 ° and ∵ QPA = 45 ° and ? PAB + ∠ PAC = 90 ° so ? PAQ = ∠ QAC + ∠ cap = ∠ PAB + ∠ PAC = 90 ° so pq2 = aq2 + ap2 = 2, and ∠ QPA = 45 ° in △
As shown in the figure, in the isosceles right triangle ABC, ∠ a = 90 °, P is a point in △ ABC, PA = 1, Pb = 3, PC= 7, then ∠ CPA=______ Degree
Rotate △ ABP anticlockwise about point a by 90 ° and then connect PQ,
Then AQ = AP = 1, CQ = Pb = 3, ∠ QAC = ∠ PAB,
∵∠QAP=90°，
∴∠QPA=45°，
And ? PAB + ∠ PAC = 90 °,
So ∠ PAQ = ∠ QAC + ∠ cap = ∠ PAB + ∠ PAC = 90 °,
Therefore, PQ2=AQ2+AP2=2, and ψ QPA=45 °,
In △ CPQ, PC2 + pq2 = 7 + 2 = 9 = CQ2
∴∠QPC=90°，
∴∠CPA=∠QPA+∠QPC=135°．
So the answer is 135 degrees
As shown in Fig. 3, in the isosceles right angle △ ABC, ∠ BAC = 90 °, P is a point in △ ABC, PA = 1, Pb = 3, PC = 7 under the radical sign, and calculate the degree of ∠ CPA
As shown in the figure,
∵AB=AC,∠BAC=90°,
﹤ ABP ﹤ 90 ° to make the point of ABP △ 90
∴AP'=AP=1,∠PAP'=90°,P'C=PB=3,
∴PP'=√2,∠APP'=45°,
∴P'P²+PC²=P'C²,
∴∠P'PC=90°,
∴∠APC=135°
In △ ABC, if AB = 3, BC = radical 13, AC = 4, then the height of edge AC is
By cosine theorem
cosA=(b^2+c^2-a^2)/(2bc)
=(9+16-13)/(2*3*4)
=1/2
It is known that a = π / 3
Then Sina = √ 3 / 2
S△ABC=1/2bcsinA=bh
1/2*3*4*√3/2=1/2*4*h
2h=3√3
h=3√3/2
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