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

Answer:

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.

Step-by-step explanation:

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

$\frac{6.92\,yd}{\sin 25.2^{\circ}} = \frac{4.82\,yd}{\sin B}$

$\sin B = \left(\frac{4.82\,yd}{6.92\,yd} \right)\cdot \sin 25.2^{\circ}$

$\sin B = 0.297$

$B \approx 17.278^{\circ}$

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:

$A = \sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}$, where $s = \frac{a+b+c}{2}$.

$s = \frac{14+30+40}{2}$

$s = 42$

$A = \sqrt{(42)\cdot (42-14)\cdot (42-30)\cdot (42-40)}$

$A \approx 168$

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