1. Answer:
A right triangles is a triangle having a 90 degree side. According to the figure, the sides of this triangle are expressed in inches. Therefore, we can find the missing sides and angles as follows:
m∠B:
The sum of the three interior angles of any triangle is 180°, therefore:
[tex]m \angle B + 51^{\circ} + 90^{\circ} = 180^{\circ} \\ \\ \boxed{m \angle B = 39^{\circ}}[/tex]
CA and AB:
We must use the law of sines as follows:
[tex]\frac{CA}{sin39^{\circ}}=\frac{9}{sin51^{\circ}} \\ \\ \therefore CA=\frac{9sin39^{\circ}}{sin51^{\circ}} \\ \\ \therefore \boxed{CA=7.3in}[/tex]
[tex]\frac{AB}{sin90^{\circ}}=\frac{9}{sin51^{\circ}} \\ \\ \therefore AB=\frac{9sin90^{\circ}}{sin51^{\circ}} \\ \\ \therefore \boxed{AB=11.6in}[/tex]
2. Answer:
According to the figure, the sides of this triangle are expressed in meters. Therefore, we can find the missing sides and angles as follows:
m∠A:
The sum of the three interior angles of any triangle is 180°, therefore:
[tex]m \angle A + 53^{\circ} + 90^{\circ} = 180^{\circ} \\ \\ \boxed{m \angle A = 37^{\circ}}[/tex]
CA and CB:
We must use the law of sines as follows:
[tex]\frac{CA}{sin53^{\circ}}=\frac{5}{sin90^{\circ}} \\ \\ \therefore CA=\frac{5sin53^{\circ}}{sin90^{\circ}} \\ \\ \therefore \boxed{CA=4.0m}[/tex]
[tex]\frac{CB}{sin37^{\circ}}=\frac{5}{sin90^{\circ}} \\ \\ \therefore CB=\frac{5sin37^{\circ}}{sin90^{\circ}} \\ \\ \therefore \boxed{CB=3.0m}[/tex]
3. Answer:
According to the figure, the sides of this triangle are expressed in miles. Therefore, we can find the missing sides and angles as follows:
m∠B:
The sum of the three interior angles of any triangle is 180°, therefore:
[tex]m \angle A + 28^{\circ} + 90^{\circ} = 180^{\circ} \\ \\ \boxed{m \angle B = 62^{\circ}}[/tex]
CB and AB:
We must use the law of sines as follows:
[tex]\frac{CB}{sin28^{\circ}}=\frac{29.3}{sin62^{\circ}} \\ \\ \therefore CB=\frac{29.3sin28^{\circ}}{sin62^{\circ}} \\ \\ \therefore \boxed{CA=15.6mi}[/tex]
[tex]\frac{AB}{sin90^{\circ}}=\frac{29.3}{sin62^{\circ}} \\ \\ \therefore AB=\frac{29.3sin90^{\circ}}{sin62^{\circ}} \\ \\ \therefore \boxed{AB=33.2mi}[/tex]
4. Answer:
According to the figure, the sides of this triangle are expressed in miles. Therefore, we can find the missing sides and angles as follows:
m∠A:
The sum of the three interior angles of any triangle is 180°, therefore:
[tex]m \angle A + 24^{\circ} + 90^{\circ} = 180^{\circ} \\ \\ \boxed{m \angle A = 66^{\circ}}[/tex]
CA and CB:
We must use the law of sines as follows:
[tex]\frac{CA}{sin66^{\circ}}=\frac{14}{sin90^{\circ}} \\ \\ \therefore CA=\frac{14sin66^{\circ}}{sin90^{\circ}} \\ \\ \therefore \boxed{CA=12.8mi}[/tex]
[tex]\frac{CB}{sin24^{\circ}}=\frac{14}{sin90^{\circ}} \\ \\ \therefore CB=\frac{14sin24^{\circ}}{sin90^{\circ}} \\ \\ \therefore \boxed{CB=5.7mi}[/tex]
