Solve the following triangle. Given A=51 degrees b=40 c=45

Question

asked Oct 9, 2020 233k views

1 Answer

Black Swan · Answer 1 · 2020-10-14T09:20:40+0000

Answer:

$a=36.87\ units$

B=57.47^o

C=71.53^o

Explanation:

step 1

Find the length side a

Applying the law of cosines

a^2=b^2+c^2-2(b)(c)cos(A)

substitute the given values

a^2=40^2+45^2-2(40)(45)cos(51^o)

a^2=1,359.4466

$a=36.87\ units$

step 2

Find the measure of angle B

Applying the law of sines

(a)/(sin(A)) =(b)/(sin(B))

substitute the given values

(36.87)/(sin(51^o)) =(40)/(sin(B))

$sin(B)=(sin(51^o))/(36.87){40}$

$B=sin^(-1)((sin(51^o))/(36.87){40})=57.47^o$

step 3

Find the measure of angle C

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

so

A+B+C=180^o

substitute the given values

51^o+57.47^o+C=180^o

108.47^o+C=180^o

C=180^o-108.47^o=71.53^o