Solve angle abc by using the measurements angle abc = 90°, angle bac = 40°, and a = 10. round measures of sides to the
nearest tenth and measures of angles to the nearest degree.

∠ACB==50°

b=15.6 units

c=11.9 units

Step-by-step explanation:

step 1

Find the measure of angle BCA

we know that

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

∠ABC+∠BAC+∠ACB=180°

substitute the given values

90°+40°+∠ACB=180°

∠ACB=180°-130°=50°

step 2

Find the measure of side b

Applying the law of sines

a/sin(∠BAC)=b/sin(∠ABC)

substitute the given values

10/sin(40°)=b/sin(90°)

b=10/sin(40°)

b=15.6 units

step 3

Find the measure of side c

Applying the law of sines

c/sin(∠ACB)=a/sin(∠BAC)

substitute the given values

c/sin(50°)=10/sin(40°)

c=[10/sin(40°)]*sin(50°)

c=11.9 units

idk lol xd

step-by-step explanation:

ez do it yourself lazy

is there more info provided? there is not enough information given in the problem