Question

Consider a Triangle ABC like the one below. Suppose that C = 98, A = 74, and b = 11 (figure is not drawn to scale.) solve the triangle. Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth. if there is more than one solution, use the button labeled "or"

Answer 1

Answer:

$A=73.8\°$

$B=8.2\°$

$c=76.3\ units$

Step-by-step explanation:

step 1

Find the measure of side c

Applying the law of cosines

$c^{2}= a^{2}+b^{2}-2(a)(b)cos(C)$

substitute the given values

$c^{2}= 74^{2}+11^{2}-2(74)(11)cos(98\°)$

$c^{2}=5,823.5738$

$c=76.3\ units$

step 2

Find the measure of angle A

Applying the law of sine

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

substitute the given values

$\frac{74}{sin(A)}=\frac{76.3}{sin(98\°)}$

$sin(A)=(74)sin(98\°)/76.3$

$A=arcsin((74)sin(98\°)/76.3)$

$A=73.8\°$

step 3

Find the measure of angle B

we know that

The sum of the internal angles of a triangle must be equal to 180 degrees

so

$A+B+C=180\°$

substitute the given values

$73.8\°+B+98\°=180\°$

$171.8\°+B=180\°$

$B=180\°-171.8\°=8.2\°$