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In the right angle △ ABC, the opposite sides of ∠ C = 90 °, a, B and C are a, B and C respectively, and the perimeter of △ ABC is 23 + 5, and the hypotenuse C = 4. Calculate the area of △ ABC and the height h on the hypotenuse
A + B + 4 = 23 + 5A2 + B2 = 16, ab = 43 − 32, then the area of △ ABC is 12ab = 43-3, and 12ab = 12CH, then H = ABC = 43 − 38
What triangle is sin ^ 2A + sin ^ 2B + sin ^ 2C less than 2
I want a specific process
sin^2A+sin^2B+sin^2C < 2∴1 < 3 -(sin^2A+sin^2B+sin^2C)∴2 < 3 +[1-2(sinA)^2]+[1-2(sinB)^2]+[1-2(sinC)^2]∴2 < 3 + [cos(2A)+cos(2B)+cos(2C)]∴2 < 3 + [2cos(A+B)cos(A-B) + cos(2C)] ∴0 < 2cos(A+B)cos(A...
sin^2A+sin^2B+sin^2C-2cosAcosBcosC=2
sina^2+sinb^2+sinc^2-2cosacosbcosc
=3-(cosa^2+cosb^2+cosc^2+2cosacosbcosc)
=3-{cosa*[cosa+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)+2]}
=3-{-cos(b+c)*[-cos(b+c)+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)]+1}
=3-{-cos(b+c)*cos(b-c)+cos(b+c)*cos(b-c)+1}
=2
Is there any other way besides this?
It can be proved that sin2b + sin2c is moved to the other side and two are simultaneously transformed into cosine of angle B and C by the basic relation of trigonometric function, and then cosa is transformed into angle B and C according to a = π - b-c
It is proved that: (1) sin ^ 2A + sin ^ 2B + sin ^ 2c-2cosacosbcosc = 2 holds
That is to say, sin2a = 2-sin2b-sin2c + 2cosacosbcosc holds
Because 2-sin2b-sin2c + 2cosacosbcosc = cos2b + cos2c + 2cos (π - B-C) cosbcosc
=cos2B+cos2C-2cos（B+C）cosBcosC=cos2B+cos2C-2（cosBcosC-sinBsinC）cosBcosC
=cos2B+cos2C-2cos2Bcos2C+2sinBsinCcosBcosC
=（cos2B-cos2Bcos2C）+（cos2C-cos2Bcos2C）+2sinBsinCcosBcosC
=cos2Bsin2C+cos2Csin2C+2sinBsinCcosBcosC
=（cosBsinC+cosCsinC）2
=sin2（B+C）=sin2（π-A）=sin2A
That's proof. It's a bit messy. Take a closer look --
If you have any questions, please ask or hi me
In the triangle ABC, the changes of angle ABC are ABC. If (2b-c) cosa = acosc, what is angle a equal to
The first question is about set, and the second question is "x greater than 1 is the condition that x square is greater than X"
(2b-c)cosA-acosC=0
Then the sine theorem is used to get the following results
(2sinB-sinC)cosA-sinA*cosC=0
2sinBcosA-(sinCcosA+sinAcosC)=0
2sinBcosA-sin(A+C)=0
2sinBcosA-sinB=0
So cosa = 1 / 2
So a = 60 degrees
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