Question

1. Find C if A=25.2°, a=6.92 yd, and b=4.82 yd

2. Find c if C=118°, a=75.0 km, and b=131 km

3. Find B if a=17.3ft, b=22.6ft, c=29.8ft

4. Find the area of triangle ABC if a=14, b=30, and c=40

5. Find the area of triangle XYZ (shown with pic) (number inside triangle is 30°)​

Answer 1

1)
`C = 137.522^(\circ)`, 2)
`c \approx 178.917\,km`, 3)
`B \approx 49.004^(\circ)`, 4)
`A \approx 168`, 5) Unsolvable due to lack of information.

Explanation:

1) The angle B is determined by the Law of Sines:

`(6.92\,yd)/(\sin 25.2^(\circ)) = (4.82\,yd)/(\sin B)`

`\sin B = \left((4.82\,yd)/(6.92\,yd) \right)\cdot \sin 25.2^(\circ)`

`\sin B = 0.297`

`B \approx 17.278^(\circ)`

The angle C is:

`C = 180^(\circ) - 25.2^(\circ) - 17.278^(\circ)`

`C = 137.522^(\circ)`

2) The side c is computed by the Law of Cosine:

`c = \sqrt{(75\,km)^(2)+(131\,km)^(2)-2\cdot (75\,km)\cdot (131\,km)\cdot \cos 118^(\circ)}`

`c \approx 178.917\,km`

3) The angle B is determined with the help of the Law of Cosine:

`(22.6\,ft)^(2) = (17.3\,ft)^(2) + (29.8\,ft)^(2) - 2\cdot (17.3\,ft)\cdot (29.8\,ft)\cdot \cos B`

`510.76 = 1187.33 - 1031.08\cdot \cos B`

`1031.08\cdot \cos B = 676.57`

`\cos B = 0.656`

`B \approx 49.004^(\circ)`

4) The formula for the surface of a triangle by just knowing their sides is:

`A = √(s\cdot (s-a)\cdot (s-b)\cdot (s-c))`, where
`s = (a+b+c)/(2)`.

`s = (14+30+40)/(2)`

`s = 42`

`A = √((42)\cdot (42-14)\cdot (42-30)\cdot (42-40))`

`A \approx 168`

5) The statement is not complete and there is no possibility to find the figure).