# Math

In ΔABC, a = 7.2cm, b = 9.3cm, ∠A = 35° (6)

• Determine the number of possible solutions.
• Determine the measure of ∠B if it exists.

We are given `Delta ABC,a=7.2,b=9.3 ` (where we assume the triangle is named in standard fashion, with capital letters signifying the vertices and corresponding lower case letters signifying the sides opposite the angles). We are asked to determine the number of possible solutions, and to find the measure of angle B if it exists.

Here we are given two side lengths and a non-included angle (SSA). Two typical methods for solving a triangle are to use the Law of Sines or the Law of Cosines. In the SSA case, either method will work.

(A) Here we will use the Law of Sines; `(sinA)/a=(sinB)/b=(sinC)/c `

Substituting the known values we get the proportion: `(sin35^@)/7.2=(sinB)/9.3 `

Thus we get `sinB=(9.3 sin35^@)/7.2~~.7409 `

Then we get `B~~sin^(-1)(.7409)=47.8^@ " or " 132.2^@ `

(Note that when using the Law of Sines, we need to check both the acute and obtuse case when given SSA.)

Both of these yield a triangle, so there are two possible answers with angle B given as above.

(B) We could also use the Law of Cosines:`a^2=b^2+c^2-2bc*cosA `

`7.2^2=9.3^2+c^2-2(9.3)c*cos35^@ `

This gives a quadratic in c:

`c^2-15.236c+34.65=0 `

Using the quadratic formula we get:

`c~~(15.236 +- 9.671)/2=12.454 " or " 2.783 `

So there are two possible solutions. We can use these values for c to calculate angle C and thus angle B.

(e.g. if c=2.78 then ` (sin35^@)/7.2=(sinC)/2.78 --> sinC=.2215 --> 12.8^@ ` so angle B has measure 132.2 degrees.)

See attachment for an illustration of the two possibilities.

** In any SSA case, there are three possibilities: the givens cannot be part of a triangle, exactly one triangle is formed (including the special case of a right triangle), or two triangles can be formed. Using the Law of Cosines, the number of real solutions to the quadratic reveals the number of triangles formed. Using the Law of Sines, we must investigate the possible cases.**

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