Question

How do you determine the number of possible triangles and find the measure of the three angles given `a=8, b=10, mangleA=20`?

Answer 1

`A=20^@,B_1= 25^@19', C_1 = 134^@41'` and

`A=20^@,B_2= 154^@41'', C_2 = 5^@19'`

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=10sin20^@~~3.42`, since `3.42 < 8 < 10` we have

`h < a < b` so we are looking for two solutions. Hence,

`a^2 =b^2+c^2-2bc cos A`

`8^2=10^2 +c^2-2(10)(c) cos 20^@`

`64=100+c^2-(20cos20^@)c`

`0=c^2-(20cos20^@)c+36`

`c=((20cos20^@)+-sqrt((-20cos20^@)^2-4(1)(36) ))/2`

`c=((20cos20^@)+sqrt((-20cos20^@)^2-144 ))/2` or

`c=((20cos20^@)-sqrt((-20cos20^@)^2-144 ))/2`

`:.c_1~~16.63 or c_2~~2.16`

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

`B_1=cos^-1 [(8^2+c_1^2-10^2)/(2*c_1*8)]=25^@19'`

and therefore

`C_1=180^@-20^@-25^@ 19'=134^@41'`

`B_2=cos^-1 [(8^2+c_2^2-10^2)/(2*c_2*8)]=154^@41'`

and therefore

`C_2=180^@-20^@-154^@41'=5^@19'`