In triangle ABC, given sin (a + b) = 0.6, sin (a-b) = 0.2 and ab = 3, find the area of triangle ABC

In triangle ABC, given sin (a + b) = 0.6, sin (a-b) = 0.2 and ab = 3, find the area of triangle ABC

sin（A+B）=sinAcosB+sinBcosA=3/5...(1)
sin（A-B）=sinAcosB-sinBcosA=1/5...(2)
(1)=3*(2)
sinAcosB+sinBcosA=3sinAcosB-3sinBcosA
2sinBcosA=sinAcosB
tanA/tanB =2.(3)
Sin (a + b) = 3 / 5, so sinc = 3 / 5, COSC = 4 / 5, Tanc = 3 / 4, so tan (a + b) = - 3 / 4 = (Tana + tanb) / (1-tanatanb), substitute into equation (3)
So Tana = 2 + root 6, tanb = 1 + root 6 / 2
Set the height as H and draw: ab = H / Tana + H / tanb
So h = 2tamb * AB / 3 = 2 + root 6
Area s = 1 / 2Ab * H = 1 / 2 * 3 * (2 + 6) = 3 / 2 (2 + 6)

An organic substance A is composed of C, h and O. under certain conditions, the transformation relationship between a, B, C, D and E is as follows: It is known that the vapor density of C is 22 times that of hydrogen under the same conditions, and silver mirror reaction can occur (1) The name of the functional group contained in substance D is __, The structural formula of substance e is ___, (2) Write the chemical equation to achieve the following conversion: ①A→B______， ②A+O 2 →C______ ③ C equation for reaction with silver ammonia solution ___

(I) because sin (C-A) = 1, C − a = π 2, and C + a = π - B, ‡ a = π 4 − B2, ‡ Sina = sin (π 4 − B2) = 22 (cosb2 − sinb2), ‡ sin2a = 12 (1 − SINB) = 13, and Sina > 0, ‡ Sina = 33 (II) as shown in the figure, acsinb = bcsina ‡ BC = a

An organic substance A is composed of C, h and O. under certain conditions, the transformation relationship between a, B, C, D and E is as follows: It is known that the vapor density of C is 22 times that of hydrogen under the same conditions, and silver mirror reaction can occur (1) The name of the functional group contained in substance D is __, The structural formula of substance e is ___, (2) Write the chemical equation to achieve the following conversion: ①A→B______， ②A+O 2 →C______ ③ C equation for reaction with silver ammonia solution ___

(I) since sin (C-A) = 1, C − a =
π
two
, and C + a = π - B,
∴A＝
π
four
−
B
two
，
∴sinA＝sin(
π
four
−
B
two
)＝
two
two
(cos
B
two
−sin
B
two
)，
∴sin2A＝
one
two
(1−sinB)＝
one
three
，
And Sina > 0, ‡ Sina =
three
three
(II) as shown in the figure, it is obtained from the sine theorem
AC
sinB
＝
BC
sinA
∴BC＝
ACsinA
sinB
＝
six
•
three
three
one
three
＝3
two
，
Sinc = sin (a + b) = sinacosb + cosasinb=
three
three
×
two
two
three
+
six
three
×
one
three
＝
six
three
∴S△ABC＝
one
two
AC•BC•sinC＝
one
two
×
six
× three
two
×
six
three
＝3
two

In triangle ABC, (1) Sina = sin (B + C) (2) cosa = - cos (B + C)

Proof: in △ ABC, a + B + C = 180 °
Namely: a = 180 ° - B-C
So:
sinA=sin(180°-B-C)=sin[180°- (B+C)]=sin(B+C)
and
cosA=cos(180°-B-C)=cos[180°- (B+C)]=-cos(B+C)
The equation is proved!

In △ ABC, cosa = 3 is known 5， (I) calculation 2 − cos (B + C); (II) if the area of △ ABC is 4 and ab = 2, find the length of BC

(I) sin2a2 − cos (B + C) = 1 − cosa2 + cosa = 1 − 352 + 35 = 45. (II) in △ ABC, ∵ cosa = 35, ∵ Sina = 45. From s △ ABC = 4, 12bcsina = 4, BC = 10, ∵ C = AB = 2,

In triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively, and C is the largest side. Find cosa by sin (a + π / 4) + cos (a + π / 4) = (3 times root sign 2) / 4 If a = 4 and B = 5, find the length of C

Sin (a + π / 4) + cos (a + π / 4) = (3 √ 2) / 4sin π / 4cosa + sinacos π / 4 + cosacos π / 4-sinasin π / 4 = (3 √ 2) / 4cosa + Sina + cosa Sina = 3 / 22cosa = 3 / 2cosa = 3 / 4. Cosa = (b ^ 2 + C ^ 2-A ^ 2) / 2bc3 / 4 = (25 + C ^ 2-16) / 10C is simplified to 4C ^ 2-30c + 36 = 0 (4