Question

How do you determine the number of possible triangles and find the measure of the three angles given `b=12, c=10, mangleB=49`?

Answer 1

`:. A=92^@02', B=49^@, and C=38^@58'`

Explanation:

Since the given information is for a SSA triangle it is the ambiguous case. In the ambiguous case we first find the height by using the formula `h=bsin A`.

Note that A is the given angle and its side is always a so the other side will be b.

So if `A < 90^@` and if

1. `h < a < b` then then there are two solutions or two triangles.

2. `h < b < a` then there is one solution or one triangle.

3. `a < h < b` then there is no solution or no triangle.

If `A >=90^@` and if

1. `a > b` then there is one solution or one triangle.

2. `a <=b` there is no solution

Now let's use the Law of Cosine `a^2 =b^2+c^2-2bc cos A` and the

quadratic formula `x=(-b+-sqrt(b^2-4ac)) /(2a)`to figure out our solutions.

That is,

`h=10 sin 49^@~~7.55` and since `7.55 < 10 < 12` we have

`h < b < a` so we are looking for one solution. Hence,

`b^2 =a^2+c^2-2ac cos B`

`12^2=a^2 +10^2-2(10)(a) cos 49^@`

`144=a^2+100-(20 cos49^@) a`

`0=a^2-(20 cos49^@) a-44`

`a=((20 cos49^@) +-sqrt((20 cos49^@) ^2-4(1)(-44) ))/2`

`a=((20 cos49^@)+sqrt((20 cos49^@)^2+176 ))/2` or

`a=((20 cos49^@)-sqrt((20 cos49^@)^2+176 ))/2`

`a~~15.89 or cancel(a~~-2.77`

`:. a~~15.89`

To find the measure of angle A we use the law of cosine and solve for A. That is,

`A=cos^-1 [(12^2+10^2-a^2)/(2*12*10)]=92^@02'`

and therefore

`C=180^@-49^@-92^@ 02'=38^@58'`