In the RT triangle ABC, ∠ C = 90 ° C = 34. If a to B is equal to 8 to 15, then a = b =

In the RT triangle ABC, ∠ C = 90 ° C = 34. If a to B is equal to 8 to 15, then a = b =

Let a = 8x, B = 15x
8x²+15x²=34²
The solution is x = 2
So a = 16, B = 30

As shown in the figure, in the RT triangle ABC, the angle c = 90 °, BC = AC, D is the midpoint of AC, and the value of Tan angle abd is calculated

Because the triangle is isosceles,
So AB = √ 2Ac = 2 √ 2ad,
DF is perpendicular to ab through point D,
Then AF = DF = √ 2 / 2ad,
Ad = √ 2AF, so AB = 4AF
tanABD=DF/BF=DF/(AB-AF)=1/3.
complete!

As shown in the figure, in the triangle ABC, ab = AC, e is a point on the CA extension line, ED is perpendicular to BC at D, and ab is at p. it is proved that the triangle AEF is an isosceles triangle

It is proved that: ∵ AB = AC,  B = ∠ C
∵ED⊥BC,∴∠E+∠C=90º,∠BFD+∠B=90º
So ∠ e = 90 ° - ∠ C, ∠ BFD = 90 ° - ∠ B,  e = ∠ BFD
And ∠ AFE = ∠ BFD, ﹥ AFE = ∠ E
The △ AEF is an isosceles triangle

As shown in Figure RT, in triangle ABC, angle ACB is equal to 90 degrees, and angle B is equal to 30 degrees (1) CD = 4cm, find the length of AB? (2) AB = 12cm, find the length of ad? If it can be done with Pythagorean, what about it? Point D is on AB, CD is vertical to AB, AC is vertical to CB

(1)bc=cd/sin30=4*2=8
AC = CD / cos30 = 8 / 3
Ac * BC = AB * CD * 0.5 = = > CD = 32 / 3
(2)
ac=ab*sin30=0.5*12=6
Angle ACD = angle B = 30 degrees
So ad = AC * sin30 = 0.5 * 6 = 3

As shown in the figure, in the RT triangle, angle ABC equals 90 degrees, angle a equals 30 degrees, and BC is perpendicular to point D

In RT △ ABC
∵∠A=30
∴BC=1/2AC
In RT △ BCD
∵∠CBD=30
∴CD=1/2BC
∴CD=1/4AC

In RT triangle ABC, the angle ACB is equal to 90 degrees, angle a is equal to 30 degrees, AB is equal to 12, CD is the height of triangle ABC, and the length of CD is calculated

∵ angle ACB equals 90 degrees, angle a equals 30 degrees, and ab equals 12 degrees
∴BC=6 AC=6√3
Let ad = X
Then (6 √ 3) 2 - x? 2 = 6? 2 - (12-x) 2
X=9
∴AD=9

As shown in the figure, in △ ABC, ab = BC = 12cm, ∠ ABC = 80 °, BD is the angular bisector of ∠ ABC, de ‖ BC, find the length of de

∵ AB = BC, BD is the angular bisector of ∵ ABC,
∴AD=CD，
∵DE∥BC，
∴AE=BE=1
2AB=6cm，
BD is the horizontal angle of ABC,
∴∠ABD=∠CBD，
∵DE∥BC，
∴∠CBD=∠BDE，
∴∠ABD=∠BDE，
∴DE=BE=6cm．

As shown in the figure, in RT △ ABC, ab = AC, ∠ BAC = 90 °, 1 = ∠ 2, CE ⊥ BD is extended to e. it is proved that BD = 2ce

It is proved that: extended CE and Ba intersect at point F, as shown in the figure, ∵ be ⊥ EC,  bef = ∠ CEB = 90 °. ? BD bisection ? ABC, ? 1 = ∠ 2, ∵ f = ∠ BCF, ? BF = BC, ∵ be ⊥ CF, ? CE = 12CF, ? ABC, AC = AB, ∵ a = 90 °, CBA = 45 °, f = (180-45) ⊥ 2 = 67.5 °

As shown in the figure, in RT △ ABC, ∠ ABC = 90 ° CD ⊥ at point D, if BC = 5cm, AC = 12cm, find the length of AD

You don't know the title?
I tried to guess, is it CD ⊥ AB cross AB extension line and point D is like this?
If so, then BC = 5, AC = 12, so AB = √ 119
Triangle ABC is similar to triangle ACD
The corresponding edge is proportional, so ad / AC = AC / ab
Ad = 144 / √ 119

Given that in the RT triangle ABC, the angle c = 90 degrees, the angle a = 30 degrees, B = 6, find the length and area of the other two sides of the RT triangle All right, score

In RT △ ABC, ∠ a = 30 °
The relationship ratio of the three sides of a triangle is a: B: C = 1: √ 3:2
∵b=6
∴a=2√3 c=4√3
The area of △ ABC is s = 1 / 2 * a * b = 1 / 2 * 2 √ 3 * 6 = 6 √ 3