In the triangle ABC, the root number 2, tanA3, tanB2 of c 2, then the area of the triangle is

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In the triangle ABC we know that c is equal to 2, a+b is equal to double the root number, and tanatanb = 3, find the area of the triangle.

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a+b is equal to twice the number of roots.

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In the triangle abc it is known that c is equal to 2, a+b is equal to the double root number, and tanatanb = 3, then the area of the triangle is s abc = 1 2absinc = 15 5

Because tan a=3, tan b=2

So sina=3 10, sinb=2 5tanc=-tan(a+b).

(tana+tanb)/(1-tanatanb)c=45°,sinc=√2/2

a=csina/sinc=(2√2)*(3/√10)/(2/2)=6√10/5

b=csinb/sinc=(2√2)*(2/√5)/(2/2)=8√5/5

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Why tana 3, tanb 2, get sina 3 root number 10, sinb 2 root number 5? Because cos a+sin a=1 tana=sina cosa bimodal simultaneous sina , sinb can be obtained in the same way,

A 2=B 2+C 2+BC, B 2+C 2-A 2=-BC, cosa=-1 2,B=3C 2,By the cosine theorem of the rotten hole plexus,117=9C 2 4+C 2+3C 2 Hungry Sakura2=19C 2 4, C 2=468 19,Sina= 3 2, S abc=(1 2)BCSINA=(3 4)*468 19* 3 2=351 Trembling judgment 3 38

Solution: Make use of the sine theorem.

a/sina=b/sinb

2√3/(1/2)=6/sinb

sinb=√3/2

So, b = 60° or b = 120°

1）b=60°，c=90°，c=4√3，s=(1/2)absinc=(1/2)*2√3*6*1=6√3

2）b=120°，c=30°,c=a=2√3，s=(1/2)absinc=(1/2)*2√3*6*(1/2)=3√3

A = 2 root number 3, b = 6, a = 30

a/sina=b/sinb

sinb = bsina a = root number 3 2

b=60c=90

c&2=a&2+b&2-2abcosc

c = 4 root number 3

s = 1 2bcsina = 6 root number 3

According to the sine theorem: a sina = c sinc

Get: sinc = 2 out of 2 root number 3

So c = 60 degrees, or c = 120 degrees.

When c = 60 degrees, ABC is a right triangle.

s abc = ac 2 = 6 root number 3

When c = 120 degrees, b 180 30 120 30 degrees, abc is an isosceles triangle.

s abc = acsinb 2 = 3 root number 3

You can set the height to h.

Such axhx1 2 = root number 3

The solution yields h=1 high bisected bc edge.

1 2bc is root number 3

So from the Pythagorean theorem, we can know that c=2

Or it can be proved by the appeal answer that ABC is an isosceles triangle.

b=c=2