Question

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

Answer 1

Answer:

$a=36.87\ units$

$B=57.47^o$

$C=71.53^o$

Step-by-step 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

$\frac{a}{sin(A)} =\frac{b}{sin(B)}$

substitute the given values

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

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

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

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

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

$108.47^o+C=180^o$

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