In the RT triangle ABC, ∠ C = 90 ° a - ∠ B = 22 ° to find the degree of ∠ a ∠ B

In the RT triangle ABC, ∠ C = 90 ° a - ∠ B = 22 ° to find the degree of ∠ a ∠ B

∵∠A-∠B=22°①
And ∠ a + ∠ B = 90 ° 2 (two acute angles in a right triangle are complementary)
ν ① + ② gives 2 ∠ a = 112 °
∴∠A=56°,∠B=∠A-22°=34°

In the RT triangle ABC, the angle c is equal to 90 degrees, and the angle a is twice the angle B, then the degree of angle a is?

60°
According to the sum of internal angles, a + B + C = 180 °;
If C = 90 ° then a + B = 90 °;
If a = 2B, then a + A / 2 = 90 ° and a = 60 °

In RT △ ABC, ∠ C = 90 °, given C = 26, B = 24, find the length of a and the degree of ∠ B (the result is accurate to 1 °)

∵∠C=90°，
∴a2+b2=c2，
∵c=26，b=24，
∴a=10，
∴sinB=b
c=24
26=12
13，
∴∠B=67°．

It is known that a right angle side and an oblique side of the RT triangle ABC are 24 and 25 respectively. The shortest side of a'b'c 'similar to the RT triangle a'b'c' is 21. Find the perimeter and the height on the hypotenuse of the RT triangle a'b'c ' Right now,

According to the Pythagorean theorem, the length of another right angle side of RT triangle ABC is: √ (25? - 24?) = 7;
Because the two triangles are similar, let the length of the other right angle side and the hypotenuse of RT triangle a'b'c 'are a and B respectively. It is better to set the height of the hypotenuse
Therefore, 7 / 21 = 24 / a = 25 / B, a = 72; b = 75
The circumference of RT triangle a'b'c 'is 21 + 72 + 75 = 168;
According to the area formula, 1 / 2 × 21 × 72 = 1 / 2 × 75 × h, H = 20.16

In the RT triangle ABC, the angle c = 90 ° if a + B = 4, C = 3, what is the area of the RT triangle ABC?

a+b=4
Square on both sides
a²+2ab+b²=16
Pythagorean theorem
a²+b²=c²=9
Substitution
a²+2ab+b²=16
9+2ab=16
ab=7/2
So area = AB / 2 = 7 / 4

3、 In the RT triangle ABC. Angle c = 90 degrees, a = 3 √ 7, B = 3 √ 21, find the size of angle A 4、 In the RT triangle ABC, the angle c = 90 degrees, BC ∶ AC = 3 ∶ 4, find the three trigonometric function values of angle A 5、 There are hidden reefs within 19 nautical miles around an island. When a cargo ship sails from west to East, the island moves 60 degrees to the north and runs 40 √ 3 nautical miles It's urgent to sign up

3、 According to the title BC: AC: ab = 3:4:5 ﹥ Sina = BC: ab = 0.6, cosa = AC: ab = 0.8, Tana = BC: AC: AC = 3: 7 / 3 √ 21 = √ 3 / 3 ﹥ a = 30o4

As shown in the figure, in the RT triangle ABC, the angle BAC = 90 degrees, ab = AC, P is any point on the extension line of BC, and the perpendicular lines of AP are be, CF, e and F are perpendicular feet respectively. (1) verification: be + CF = ef (2) if P is any point on line BC, other conditions remain unchanged: is there any definite quantitative relationship between the lengths of segments be, CF and ef? Please draw a graph and prove your conclusion

Please draw the pictures yourself
It is proved that: (1) from the vertical lines be, CF, e and F of straight line AP passing through B and C respectively are perpendicular feet, we can know that: ∠ bea = ∠ AFC = 90 °
∵ ∠BAC＝90°
∴ ∠CAF+∠EAB ＝180°－∠BAC ＝180°－90°＝90°
In the AEB of the right triangle ∠ EBA + ∠ EAB = 180 ° - ∠ bea = 90 °
∴ ∠CAF＝∠EBA
In △ EBA and △ ACF, ab = AC (known), ∠ bea = ∠ AFC = 90 ° and ∠ CAF = ∠ EBA
∴ △EBA ≌ △ACF
　　　　　　∴　BE = AF,EA＝CF
∴ BE+CF＝EA+AF ＝EF
(2) If point P is close to point B, there is cf-be = EF; if point P is close to point C, there is be-cf = EF
It is proved as follows: the perpendiculars be, CF, e and F of straight line AP through B and C are perpendicular feet respectively, ∠ bea = ∠ AFC = 90 °
∵ ∠BAC＝90°
.∴ ∠BAE + ∠FAC＝90°
In the right triangle Abe, ∠ EBA + ∠ BAE = 90 °
∴ ∠EBA ＝ ∠FAC
In △ EBA and △ AFC, ab = AC, ∠ bea = ∠ AFC = 90 ° and ∠ EBA = ∠ fac
∴ △EBA　≌　△AFC
　　　　　∴ AE＝CF,AF＝BE
∴ CF－BE ＝ AE －AF =　EF　　
Similarly, be-cf = EF can be proved

In the RT triangle ABC, the opposite sides of ∠ C = 90 °, a, ∠ B, ∠ C are respectively a, B, C. If a ratio B = 3 to 4, C = 75cm, find a, B

RT triangle ABC: A ^ 2 + B ^ 2 = C ^ 2
Because a / b = 3 / 4, that is, a = (3 / 4) * B
So ((3 / 4) * b) ^ 2 + B ^ 2 = C ^ 2
b^2=16/25*c^2
a. C, B are greater than 0
b=4/5c=(4/5)*75=60
a=(3/4)*60=45

Pythagorean Law: if we know that in RT △ ABC ∠ C = 90? B = 14cm, C = 10cm, then the area of RT △ ABC?

A2 + B2 = C2 = 100; (a + b) 2 = 142 = 196, (a + b) 2 - (A2 + B2) = 2Ab = 196-100 = 96 s = AB / 2 = 2Ab / 4 = 96 / 4 = 24 select a

[Pythagorean theorem] it is known that in RT △ ABC, the sum of two right angles is 14 cm, and the length of inclined side is 10 cm

Let the right angle side be a and B, and the equation can be listed by the problem
a+b=14 ①
a²+b²=10² ②
① 2-2
2ab＝14²－10²＝96
Therefore, ab = 96 / 2 = 48
S△ABC＝ab/2＝48/2＝24cm²
Therefore, the area of the triangle is 24cm
Solution 2: because 6,8,10 are Pythagorean numbers, 6 ﹥ 8 ﹣ 2 = 10  2
Therefore, the two right angles of RT △ ABC are 6 and 8, respectively
S△ABC＝ab/2＝(6×8)/2＝24cm²
Therefore, the area of the triangle is 24cm