Question

Find all the missing dimensions when a=9, b=13, c=64°

Answer 1

Part 1)
`c=12.14\ units`

Part 2)
`m\angle A=41.78^o`

Part 3)
`m\angle B=74.22^o`

Explanation:

step 1

Find the measure of side c

Applying the law of cosines

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

we have

`a=9\ units\\b=13\ units\\C=64^o`

substitute

`c^2=9^2+13^2-2(9)(13)cos(64^o)`

`c^2=81+169-234cos(64^o)`

`c^2=250-234cos(64^o)`

`c^2=147.4212`

`c=12.14\ units`

step 2

Find the measure of angle A

Applying the law of sines

`(a)/(sin(A))=(c)/(sin(C))`

substitute the given values

`(9)/(sin(A))=(12.14)/(sin(64^o))`

solve for sin(A)

`sin(A)=(sin(64^o))/(12.14)(9)`

`sin(A)=0.6663`

`m\angle A=sin^(-1)(0.6663)=41.78^o`

step 3

Find the measure of angle B

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

`m\angle A+m\angle B+m\angle C=180^o`

substitute the given values

`41.78^o+m\angle B+64^o=180^o`

`m\angle B+105.78^o=180^o`

`m\angle B=180^o-105.78^o`

`m\angle B=74.22^o`