A trigonometric solution to solve the trigonometric problem

Because there are two solutions, the value range of x*sin45°< 2 < x is (2,2 2).

In the above formula, both sides of the equation are eliminated at the same time a2*b 2

A 2*C 2-A 4=B 2*C 2-B 4 Treat b 4 as (b 2) 2

Using the squared difference, we get (a2-b 2)(c 2-a 2-b 2)=0

I saw this question yesterday, and I don't have an answer today.

x+2y-5=0

4x-3y+13=0

Simultaneous solution of this system of equations.

x=-1, y=3

So the intersection of two straight lines is (-1,3).

and L1 perpendicular.

l1 is x+2y-15=0

Perpendicular to him, then the x and y coefficients are reversed, and one of them becomes a sign so that it is 2x-y+a=0

Substitute the a coordinates.

2-3+a=0

a=5, so it's 2x-y+5=0

Solution: L1 is x+2y-5=0 (1), L2 is 4x-3y+13=0 (2), and (1) 3+(2) 2 gives 11x+11=0, xa=-1

Substituting xa=-1 into (1) gives ya=3, that is, the point a(-1,3) is perpendicular to the straight line L1 obtained by passing through point a

k×k1=-1，k=-1/k1

k1=-1/2∴k=2

The straight line is y-3=2[x-(-1)], i.e., 2x-y+5=0

This is a problem about the number of three angles, and the solution of this problem has a diagram.

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ab//cd

MB split the ABE

FD deuces

2+∠3+∠4+∠5+∠e=360

ab//cd

MB split the ABE

FD deuces

2+∠3+∠4+∠5+∠e=360

l₁: x+2y-5=0...1)l₂:

4x-3y+13=0...2) Find the coordinates of the intersection of l and l m: 4 (1)-(2) to get:

11y-33=0, so y=3;Substituting equation (1) yields x=5-2y=5-6=-1;

The coordinates of the intersection m are (-1,3);

y=-(1 2)x-(5 2) from (1); Therefore the slope of l k = -1 2;So the slope of the straight line perpendicular to it k = 2;

Therefore, the equation for a straight line that passes through the intersection point m and is perpendicular to l is: y=2(x+1)+3=2x+5;Or write it as a general formula: 2x-y+5=0;

Two straight lines are perpendicular, and the product of the slope of the straight line is equal to 1, so.

2/3)*(a/5)=1

Find a=-15 2

The straight line 2x-3y+6=0 and the straight line ax+5y-1=0 satisfy the one-time function y=kx+b

The position relationship of the straight line of the primary function is parallel and perpendicular, the two straight lines are parallel, the two k values are equal, k1=k2, the two straight lines are perpendicular, and the product of the two k values is equal to -1

2x-3y+6=0 deforms to y=2 3x+2 k, and the value is 2 3

ax+5y-1=0 deforms to y=-a 5 and x+1 k with a value of -a 5

2/3 x (-a/5)=-1

a/5=-3/2

a=15/2

You can set y=2x to be angular to the x-axis, y=-3x to be angledthen tan = 2, tan = 3 tan(α+=(2+3)/(1-2×3)=-1，tanθ=tan[π-=1，θ=π/4

Let the intersection point of two straight lines be o,be cd, aob= adc=73° aoe=180°-73°=107°

eao=180°-∠eao-∠aeb=180°－33°-107°=40°

AD is the EAC midline.

dac=∠eao=40°

acd=180°-73°-40°=67°

Two straight lines intersect on the y-axis, indicating that the intercepts of the two straight lines are equal, so.

6/2=3/b

Find b=-1

3x+2y+6=0 ..1)

2x+by-3=0 ..2)

Compared to the y-axis (i.e. x=0):

Substituting x=0 into (1) and (2) yields:

0+2y+6=0 ..3)

0+by-3=0 ..4)

3) +(4) 2 get:

2y+2by+6-6=0

2y+2by=0

b=-1

(1) Two straight lines calculate the coordinates of point A.

2) The slope of the straight line perpendicular to L1 is a negative reciprocal with L1, kl1=-1 2The slope of the straight line is 2

3) Find the straight line according to the slope and the coordinates of point A.

1. According to the law of sinusoid: a sina = b sinb sinb = c sinc = 2rsinc = c sinb b = (1 2 2 2) = 1 2c = 30 ° a 180 ° 30 ° 45 ° 105 ° sina = sin105 ° = sin (60 ° + 45 °) sin60 ° cos45 ° + cos 60 ° sin45 °

a＝c×sina/sinc=sin105°/sin30°＝[2/4×(√3+1)]/1/2)

2. C3B=2 Gen3 A sinb

a=3b/(2√3×sinb)

According to the sinusoidal theorem a sina=b sinb=c sincsina=a sinb b=[3b (2 3 sinb)] b= 3 2

a = 120° or 60°

If a 60°, b c, it is an equilateral triangle.

This question is a fill-in-the-blank question, which should be the most appropriate possibility, so choose C

c=60°

ab = 6 cosc = (a square plexus cons and cons + b - c square) 2aba square + b square = 73 infiltration soothsayer 4

a+b=11/2