Determine whether the given triangle has no solution, one solution or two solutions. then solve the triangle. round measures of sides to the nearest tenth and measures of angles to the nearest degree
a = 119°, a=7, b=4
question 7 options:
one solution; c ≈ 7; b = 30°; c = 119°
no solution
one solution; c ≈ 4; b = 31°; c = 30°
one solution; c ≈ 4; b = 30°; c = 31°

The triangle has one solution. The remaining side c ≈ 4 and remaining angles B = 30°; C = 31°.

Option D is correct.

Step-by-step explanation:

if angle A is obtuse and if a > b then the triangle has one solution

We are given ∠ = 119° which is obtuse and side a= 7 and side b - 4 i.e 7>4 so, the triangle has one solution.

Finding remaining side c and ∠B and ∠C

Using Law of sines to find ∠B

a/sin A = b/sin B

7/sin 119° = 4/sin B

7 * sin B = 4 * sin 119

7*sin B = 4(0.874)

sin B = 3.496/7

B = sin^-1(0.4994)

B = 29.96 = 30°

We know that sum of angles of triangle = 180°

So, 180° = 119° + 30° +∠C

180° = 149° + ∠C

=> ∠C = 180° - 149°

∠C = 31°

Now finding c

b/sin B = c /sin C

4/Sin 30 = c/sin 31

4* sin 31 = c*sin 30

4*0.515 = c* 0.5

=> c =  4*0.515/0.5

c = 4.12 ≈ 4

So, Option D one solution; c ≈ 4; B = 30°; C = 31° is correct.

is there any photos or whatever?