In RT triangle ABC, angle c = 90 °, a: C = 3:5, B = 12, find a and C?

In RT triangle ABC, angle c = 90 °, a: C = 3:5, B = 12, find a and C?

Let a = 3x, C = 5x. Then: C ^ 2 = a ^ 2 + B ^ 2 25X ^ 2 = 9x ^ 2 + 144 16x ^ 2 = 144 x ^ 2 = 9 x = 3, so: a = 3x = 3 * 3 = 9, C = 5x = 5 * 3 = 15

In RT triangle ABC, if a: B = 3:4, C = 20, then =?, B =?

Let a = 3x, then B = 4x, from Pythagorean theorem we can get (3x) ^ 2 + (4x) ^ 2 = 20 ^ 2, then x = 4, so B = 4 * 4 = 16

In RT △ ABC, ∠ C = 90 °, a = 30 ° and BC = 4, we solve this right triangle

In the right angle △ ABC, ∠ B = 90 - ∠ a = 60 °, ∵ Tana = BCAC = tan30 ° = 33, ∵ AC = 433 = 43, ∵ Sina = BCAC = 12, ∵ AC = 8

Given the line segment a.b.h, find the triangle ABC so that BC = a, AC = B, CD = h on the side of ab

Method:
1. Draw a straight line L and take any point D on the line L;
2. Make DC vertical L through D, and intercept DC = H;
3. Take C as the center of the circle and a as the radius to make the arc intersection line L at point B;
4. Take C as the center of the circle and B as the radius to make the arc intersection line L at point a;
5. Connect BC and AC, then the triangle ABC is the triangle
&There are two such triangles, namely triangle ABC and triangle AB & # 39; C

As shown in the figure, in △ ABC, ∠ C = 90 ° and AC = BC, make the vertical bisectors of AB, BC and Ca respectively. What do you find

The three vertical bisectors intersect at the midpoint of ab

If △ ABC is known, find the middle line on the edge of BC. Draw with ruler and gauge, and keep the trace of drawing

1. Take a as the center of the circle, any length as the radius, draw an arc, intersect AB with D, AC with E
2: Take D.F as the radius and the length greater than 1 / 2DE as the radius to draw an arc, intersecting at point o
3: Connect a as ray Ao, intersect BC at point P, and AP is obtained

It is known that in the triangle ABC, ab = m, AC = n, and the angle BAC = Alpha

B as BD ⊥ AC to D
sinα=BD/AB
BD=AB×sinα=m×sinα
S△ABC=1/2ACxBD=1/2mn×sinα

Right angle MCN, line segment a, and acute angle alpha. Find triangle ABC, so that angle c equals 90 degrees, angle B = angle alpha, BC = a

The first step is to make a right angle, angle MCN, or angle C
Secondly, the length of a line segment is intercepted on a delay line of the angle, and the acute angle alpha is made at the end point
Third, the intersection of the extension of angle alpha and another extension of angle c is angle A
complete

In the triangle ABC, AC = B, BC = a, acute angle a = Alpha, angle B = beta
Verification: sin alpha fraction a = sin beta fraction B

Let AB = C
S△ABC=1/2absinC=1/2acsinβ=1/2bcsinα
asinβ=ab/csinC
bsinα=ab/csinC
asinβ=bsincα
a/sinα=b/sinβ

Given the line segment a and the acute angle ∠α, calculate RT △ ABC so that one side of it is a and the other acute angle is ∠α. If the above conditions are satisfied, the triangle ()
A. 1 B. 2 C. 3 d. 4

① Take the length of line OA as a on one side of ∠ α, and make the vertical line of OA cross the other side of ∠ α through point a to get △ AOB, then a is the right angle side. ② take the length of line OA as a on one side of ∠ α, and make the vertical line cross the other side of ∠ α through point a to make the other side of ∠ α, then a is the hypotenuse. ③ the opposite side of ∠ α is ∠ α, and a is the right angle side