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In (triangle)ABC, let angleA=45, side a=5sq2 and make side b=5sq3. Find angle b

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rapsplash | Student, Undergraduate | (Level 1) eNoter

Posted March 22, 2012 at 10:43 PM via web

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In (triangle)ABC, let angleA=45, side a=5sq2 and make side b=5sq3. Find angle b

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 22, 2012 at 10:57 PM (Answer #1)

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You should use the law of sines to find B angle's measure such that:

`a/(sinA) = b/(sinB)`

You need to substitute `5sqrt2`  for a, `45^o`  for A and `5sqrt3`  for b such that:

`(5sqrt2 )/(sin45^o) = (5sqrt3)/(sinB)`

You need to remember that `sin 45^o = sqrt2/2` , hence you need to substitute `sqrt2/2`  for `sin 45^o`  such that:

`(5sqrt2 )/(sqrt2/2) = (5sqrt3)/(sinB)`

Reducing by 5 both sides yields:

`(sqrt2 )/(sqrt2/2) = (sqrt3)/(sinB)`

`2 = (sqrt3)/(sinB) =gt sin B = sqrt3/2`

`B = sin^(-1) (sqrt3/2) =gt B = 60^o`

Hence, evaluating the measure of B angle yields `B = 60^o` .

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted March 24, 2012 at 1:01 PM (Answer #2)

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Note that this is the SSA case: there is the possibility that these values do not form a triangle, they form exactly one triangle, or two triangles could be formed using these values.

Proceeding as above, we use the Law of Sines:

`a/(sinA)=b/(sinb)`

`(5sqrt(2))/(sqrt(2)/2)=(5sqrt(3))/(sinB)`

`sinB=(5sqrt(6))/(10sqrt(2))`

`sinB=sqrt(3)/2`

`B=sin^(-1)(sqrt(3)/2)=>B=60^(circ)` or `B=120^(circ)` .

We check both answers to see if they are possible: if B is 60 degrees then C is 75 degrees; if B is 120 degrees then C is 15 degrees. Since both work there are two triangles with the given measures.

The measure of angle B is either `60^(circ)"or"120^(circ)`

** Another approach (in this case a little harder) is to use the Law of Cosines to find the missing side. Then you can choose either the Law of Sines or another application of the Law of Cosines to find the angle. If I were finding the missing side, I would use the Law of Cosines, the missing angle the Law of Sines.

The nice thing about the Law of Cosines approach is that the number of solutions is produced naturally -- no need to consider cases.

Here, if we label the missing side x we get :

`a^2=b^2+c^2-2bcCosA`

`(5sqrt(2))^2=(5sqrt(3))^2+x^2-2(5sqrt(3))(x)(sqrt(2)/2)`

`50=75+x^2-5sqrt(6)x`

`x^2-5sqrt(6)x+25=0`

`x=(5sqrt(6)+-sqrt(150-100))/2`

`x=(5sqrt(6)+-5sqrt(2))/2` hence there are two triangles formed with the given information.

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