fiora

I know. You know. I know that you know. I know that you know that I know.

The quick brown fox jumps over a lazy dog.

(This sentence have 26 English alphabets.)

 Never trouble trouble till trouble troubles you.

sin(x+y)=sinx*cosy+cosx*siny

sin(x-y)=sinx*cosy-cosx*siny

so, sin(x+y)+sin(x-y)=sinx*cosy+cosx*siny+sinx*cosy-cosx*siny=2sinx*cosy

    sinx*cosy=1/2[sin(x+y)+sin(x-y)]

orginal qustion=6*sin(47y)*cos(21y)=（1/2）*6*[sin(47y+21y)+sin(47y-21y)=3*[sin68y+sin26y]

                     =3sin68y+3sin26y

Yes it is.

PACAFIC+BALTIC+ARCTIC=CCCCCCC

4658985+260785+635785=5555555

0=C

2=B

R=3

4=P

5=C

6=A

7=T

8=I

9=F

LOL! You changed your question!.....

Here is my graph from desmos.

https://www.desmos.com/calculator/mf39mbkv1i

Is it cool?and Thank You CPhill for correted my mistook.

Hello everybody.I am fiora.And I going to do this problem with you all.

First,I think you guys only get part of the answers right.

Now,let me repeat the question, The angles: B = 35o,  A1 = 23o  and   h = 3.5cm ; (The angle  A1 is located between  b and  h ) Find the area of a triangle  ABC.

At first, what is b? b in this question can be the base of the triangle or the opposite side of the angle B.

However, if b is the base of the triangle ,then b is perpendicular (90 degrees) to the h (height),but according to given,the angle between b and h is 23 degrees,so b can not be the base.

Now,b can only be the opposite side of angle B.

Here is my picture of this situations.

  In triangle ABC, the opposite side of angle B is AC

and in triangle ABE,the opposite side of angle B is AE

Look at triangle ABC,area of triangle ABC=1/2(AD*BC)=1/2[AD*(BD-CD)]

BD=AD/tan(7pi/36)  

AD=h is the height of the triangle,so angle ADC=90 degrees

Threfore angle ACD=180 degrees -angle CAD- angle ADC=180 degrees -23 degrees -90 degrees=67 degree

so CD=AD/tan(67pi/180)

so area fo triangle ABC=1/2[AD*(BD-CD)]=

1/2[3.5cm*（3.5cm/tan(7pi/36)-3.5cm/tan67(pi/180)=1/2[3.5cm*3.5128561668618683cm]

=6.14749829200826952 cm^2

In triangle ABE,h (AD) is the height of the triangle 

so angle ADE=90 degrees ,and angle DAE=angle CAD= 23 degree,plus segment AD=segment AD

so triangle CAD is congruent to triangle EAD.

so CD=ED=3.5/tan(7pi/36) (the corresopding sides of congruent triangles are equal) 

area of triangle ABE=1/2*[AD*BE]=1/2[AD*(BD+DE)]=1/2[3.5 cm*(3.5 cm/tan7(pi/36)+3.5 cm/tan(67pi/180)]=11.347314790575381975 cm^2

Thank you CPhill,I get it!

When a Quadrilateral is  a parallelogram ,the two pairs of opposite sides are equal.

so we can set up the equation like 7=8y-1 and 2x+3=9 , then we get y=1 and x=3

so the awnser  is x+y=3+1=4