Sine Law
APPLICATION OF SINE LAW
The shorter diagonal of a parallelogram is 5.2 m. Find the perimeter of the parallelogram if the angles between the sides and the diagonal are 40o and 30o10’.
From the top of a 150 m lighthouse, the angles of depression of two boats on the shore are 20o and 50o, respectively. If they are due north of the observation point, find the distance between them.
SOLUTION
Solved for a
150/sin50 = a/sin90
a = (150 sin90)/sin50
a = 195.81 m
Solved for b
195.81/sin20 = b/sin30
b = (195.81 sin30)/sin20
b = 286.26 m
Therefore, the distance between the two ships is 286.26 m. Two policemen 122 meters apart are looking at a woman on top of a tower. One cop is on the east side and the other on the west side. If the angles of elevation of the woman from the cops are 42.5o and 64.8o, how far is she from the two cops?
SOLUTION
Solved for a
122/sin72.7 = a/sin42.5
a = (122 sin42.5)/sin72.7
a = 86.33 m
Solved for b
122/sin72.7 = b/sin64.8
b = (122 sin64.8)/sin72.7
b = 115.62 m
Therefore, the distances of the woman from the two cops are 86.33 m and 115.62 m.
The angle between Rizal St. and Bonifacio St. is 27o and intersect at P.Jose and Andres leaves P at the same time. Jose jogs at 10 kph on Rizal St. If he is 3 km from Andres after 30 minutes, how fast is Andres running along Bonifacio St?
SOLUTION
Solved for A
3/sin27 = 5/sinA
sin A = (5 sin27)/3
A = 49.17o
Solved for c
c/sin103.83 = 3/sin27
c = (3 sin103.83)/sin27
c = 6.42 km
Solved for Andres’ Rate
R= (6.42 km)/(.5 hh) = 12.84 kph
Therefore, Andres running at Bonifacio Street at the rate of 12.84 kph.
The angle of elevation of the top of a tower is 40o30’ from point X and 55o from another point Y. Point Y is 30
The shorter diagonal of a parallelogram is 5.2 m. Find the perimeter of the parallelogram if the angles between the sides and the diagonal are 40o and 30o10’.
From the top of a 150 m lighthouse, the angles of depression of two boats on the shore are 20o and 50o, respectively. If they are due north of the observation point, find the distance between them.
SOLUTION
Solved for a
150/sin50 = a/sin90
a = (150 sin90)/sin50
a = 195.81 m
Solved for b
195.81/sin20 = b/sin30
b = (195.81 sin30)/sin20
b = 286.26 m
Therefore, the distance between the two ships is 286.26 m. Two policemen 122 meters apart are looking at a woman on top of a tower. One cop is on the east side and the other on the west side. If the angles of elevation of the woman from the cops are 42.5o and 64.8o, how far is she from the two cops?
SOLUTION
Solved for a
122/sin72.7 = a/sin42.5
a = (122 sin42.5)/sin72.7
a = 86.33 m
Solved for b
122/sin72.7 = b/sin64.8
b = (122 sin64.8)/sin72.7
b = 115.62 m
Therefore, the distances of the woman from the two cops are 86.33 m and 115.62 m.
The angle between Rizal St. and Bonifacio St. is 27o and intersect at P.Jose and Andres leaves P at the same time. Jose jogs at 10 kph on Rizal St. If he is 3 km from Andres after 30 minutes, how fast is Andres running along Bonifacio St?
SOLUTION
Solved for A
3/sin27 = 5/sinA
sin A = (5 sin27)/3
A = 49.17o
Solved for c
c/sin103.83 = 3/sin27
c = (3 sin103.83)/sin27
c = 6.42 km
Solved for Andres’ Rate
R= (6.42 km)/(.5 hh) = 12.84 kph
Therefore, Andres running at Bonifacio Street at the rate of 12.84 kph.
The angle of elevation of the top of a tower is 40o30’ from point X and 55o from another point Y. Point Y is 30