In △ ABC, a, B and C are the opposite sides of angle ABC respectively, and the area of △ ABC is s = 1 / 4 (b ^ 2 + C ^ 2), then B = ()

In △ ABC, a, B and C are the opposite sides of angle ABC respectively, and the area of △ ABC is s = 1 / 4 (b ^ 2 + C ^ 2), then B = ()

S = 1 / 4 (b ^ 2 + C ^ 2) > = (1 / 2) BC
And the triangle area s = (1 / 2) bcsina, because Sina

In Δ ABC, if a = 2, a = 30 ° and C = 45 °, what is the area s Δ ABC equal to?
This is a problem of sine theorem in mathematics, which is compulsory for senior two. I don't know much about it. Please write it in detail,

Because a = 30 degrees, C = 45 degrees
So: A / Sina = C / sinc, 4 = C / (√ / 2), C = 2 √ 2
B = 105 ° again
So the area of △ ABC is s
＝1/2acsinB
=2√2*(√6+√2）/4
＝√3+1
Glad to answer for you/~

As shown in the figure, in △ ABC, ∠ C = 90 °, D and E are the points on AC and ab respectively, and ad = BD, AE = BC, de = DC

As shown in the figure, in △ ABC, D is the point on BC, ab = 20, BD = 16, DC = 9, ad = 12. Please find all the right angles in the figure and explain the reasons

There are three right triangles, they are: Triangle abd, triangle ACD, triangle BCA
The reason is: ab = 20, BD = 16, ad = 12,
So AB ^ 2 = BD ^ 2 + ad ^ 2,
So the triangle abd is a right triangle, angle ADB = 90 degrees
So the angle ADC is 90 degrees,
So ACD is a right triangle,
Because ad = 12, DC = 9,
So AC = 15,
So AC ^ 2 + AB ^ 2 = BC ^ 2
So the triangle BCA is a right triangle

As shown in the figure, in △ ABC, ad is perpendicular to BC and D, AB + BD = DC. Please tell us why ∠ ABC = 2 ∠ C
Extend CB to e so that be = ba

Certification:
Extend CB to e, make be = Ba, connect AE
∵BE=BA
∴∠E=∠BAE
∴∠ABC=∠E+∠BAE=2∠E,
∵AB+BD=CD
∴BE+BD=CD
∴ED=CD
∵AD⊥BC
The ad is the vertical bisector of CE
∴AE=AC
∴∠E=∠C
∴∠ABC=2∠E=2∠C

In triangle ABC, ad is the bisector of angle BAC. It is proved that AB / AC = BD / DC by sine formula

Let angle DAB = angle CAD = @
AB/BD= sin (angle ADB) / sin @,AC/DC=sin (angle ADC) / sin@
Because sin (angle ADB) =sin (angle ADC)
So AB/BD=AC/CD
So AB/AC=BD/DC.

As shown in the figure, in △ ABC, ad is the bisector of ∠ BAC, and BD: DC = AB: AC

It is proved that: as shown in the figure, the parallel line of AD through C intersects the extension line of BA at the point E, ∠ DAC = ∠ ace, ∠ bad = ∠ e, ∵ ad is the bisector of ∠ BAC, ｜ bad = ∠ DAC. ｜ ace = ∠ e, ∵ AC = AE, ∵ CE ‖ ad, ∵ BD: DC = Ba: AE, ∵ BD: DC = AB: AC

For example, in △ ABC, ad bisects ∠ BAC, then BD / DC = AB / AC, how to prove?

It should be an auxiliary line
Make the extension line of CE parallel to ab intersection ad at point E
So BD / DC = AB / CE
Because it can be proved that ACE is isosceles triangle, so CE = AC
Namely: BD / DC = AB / AC

A proof of cosine theorem
Triangle angles a, B and C correspond to sides a, B and C respectively. It is proved that a ^ 2-B ^ 2 / C ^ 2 = sin (a-b) / sinc

A topic of cosine theorem
In the triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively. If a, B and C are in equal proportion sequence, try to find:
(1) The range of angle B;
(2) The range of T = SINB + CoSb

(1) ∵ B & sup2; = AC
cosB=(a²+c²-b²)/(2ac)=(a²+c²-ac)/(2ac)≥(2ac-ac)/(2ac)=1/2
∵ CoSb is a decreasing function, ∵ 0 ＜ B ≤ 60 °
(2)t=sinB+cosB=>t²=(sinB+cosB)²=>t²-1=sin2B
Because 0 ＜ B ≤ 60 ° therefore, 0