Question

How do you solve the triangle given `C=15^circ15', a=6.25, b=2.15`?

Answer 1

Covert minutes to decimal degrees
Use The Law of Cosines to find the length of side c
Use The Law of Sines to find `B`
Use the sum of the interior angles equals `180^@` to find `A`

Explanation:

Use the factor `1^@/(60')` to convert the minutes to decimal degrees:

`(15')/1(1^@)/(60')= (15/60)^@ = 0.25^@`

`C = 15.25^@`

Use The Law of Cosines to find the length of side c:

`c=sqrt(a^2+b^2-2(a)(b)cos(C)`

`c=sqrt((6.25)^2+(2.15)^2-2(6.25)(2.15)cos(15.25^@)`

`c~~4.21`

Use The Law of Sines to find B:

`sin(B)/b = sin(C)/c`

`B = sin^-1(sin(C)b/c)`

`B = sin^-1(sin(15.25^@)2.15/4.21)`

`B~~ 7.72^@`

To find A use the some if the interior angle equal `180^@`:

`A + B + C = 180^@`

`A + 7.72^@ + 15.25^@ = 180^@`

`A = 157.03^@`