In ABC, AC = 15, BC = 13 and CD = 12, what is the area?

In ABC, AC = 15, BC = 13 and CD = 12, what is the area?

Because the Pythagorean theorem is satisfied (the square of a + the square of B = the square of C, the square of 12 + the square of 13 = the square of 15), the triangle is a right triangle, the right side is 12, the bottom of 13 multiplied by the height multiplied by half, and finally 78

It is known that the area of a (1.0,3) B (1.2.1) C (0.2.1) triangle ABC is

If we calculate the three sides, we know that ab = 2 √ 2, AC = 3, BC = 1, so a triangle is a right triangle with AB and BC as right angles and AC as hypotenuse, so s = √ 2

How does the formula Sina / (sinbsinc) = (1 / Sina) * (a ^ 2 / BC) come out?

Sine theorem
sinA=a/2R
sinB=b/2R
sinC=c/2R
Bring in left
2RA / BC = (2R / a) * a ^ 2 / (BC)
Because Sina = A / 2R
So 2R / a = 1 / Sina
Bring in the certificate

If the area of triangle is known as s = A & # 178; - (B & # 178; - C & # 178;) find Sina

The triangle area is s = A & # 178; - (B & # 178; - C & # 178;) = (1 / 2) bcsina,
∴sinA=2(a^+c^-b^)/bc,
According to the cosine theorem, a ^ + C ^ - B ^ = 2accosb,
∴sinA=4acosB/b=4sinAcosB/sinB,
∴tanB=4.
Can't find Sina

How to deduce the formula 1 / 2absinc of triangle area?

High ad for BC
AD/b=sinC
So ad = bsinc
S=0.5AD*BC=0.5abSinc

Is the area formula of triangle absinc or 1 / 2absinc
I'm working on the problem. I can't remember it clearly
By the way, what is the sin area of the parallelogram?

1/2absinc
The area of a parallelogram is absinc, and C is the angle between a and B

The graph of triangle area formula 1 / 2absinc

 
 
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In the triangle ABC, we prove that a ^ 2sin2b + B ^ 2sin2a = 2absinc
Don't use analysis

According to the sine theorem, we get a / Sina = B / SINB = C / sinc = 2R (R is the radius of triangle circumcircle)
So on the left
a^2sin2B+b^2sin2A
=4R^2(2sinA*sinAsin2B+sinB*sinB*sin2A)
=4R^2(2sinAsinAsinBcosB+2sinBsinBsinAcosA)
=4R^2sinAsinB(2sinAcosB+2cosAsinB)
=ab(2sin(A+B))
=ab(2sinC)
=2absinC
=Right

In △ ABC, it is known that a, B and C are the opposite sides of angles a, B and C respectively. It is proved that a2sin2b + b2sin2a = 2absin C

It is proved that: in ∵ ABC, Asina = bsinb = csinc = 2R (R is the radius of circumscribed circle) ∵ a = 2rsina, B = 2rsinb, C = 2rsinc, ∵ a2sin2b + b2sin2a = 2a2sinb · CoSb + 2b2sina · cosa = 8r2sina · SINB · (sinacosb + sinbcosa) = 8r2sina · SINB · sin (a + b) = 8r2sina · SINB · sin (π - C) = 8r2sina · SINB · sinc, and 2absinc = 2 · 2rsina · 2rsinb · sinc = 8r2sina · SINB · sinc nC，∴a2sin2B+b2sin2A=2absinC．

In the acute angle △ ABC, the opposite sides of ∠ a, B and C are a, B and C respectively. Let the area of △ ABC be s

Take AC as the base, make high, AC as D
sinA=BD/AB=BD/c
∴BD=csinA
S=1/2bcsinA