Sine Law

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  • Topic: Triangle, Length, Quadrilateral
  • Pages : 7 (1099 words )
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  • Published : March 12, 2013
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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/sin⁡50 = a/sin⁡90

a = (150 sin⁡90)/sin⁡50

a = 195.81 m

Solved for b

195.81/sin⁡20 = b/sin⁡30

b = (195.81 sin⁡30)/sin⁡20

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/sin⁡72.7 = a/sin⁡42.5

a = (122 sin⁡42.5)/sin⁡72.7

a = 86.33 m

Solved for b

122/sin⁡72.7 = b/sin⁡64.8

b = (122 sin⁡64.8)/sin⁡72.7

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/sin⁡27 = 5/sin⁡A

sin A = (5 sin⁡27)/3

A = 49.17o

Solved for c

c/sin⁡103.83 = 3/sin⁡27

c = (3 sin⁡103.83)/sin⁡27

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 meters from the base of the tower. If the base of the tower and points X and Y are on the same level, find the approximate distance from X and Y.

SOLUTION

Solved for a

30/sin⁡35 = a/sin⁡90

a = (30 sin⁡90)/sin⁡35

a = 52.3 m

Solved for b

53.3/sin⁡40.5 = b/sin⁡14.5

b = (52.3 sin⁡14.5)/sin⁡40.5

b = 20.16 m

Therefore, the distance between points X and Y is 20.16 m.

The vertex of an isosceles triangle is 40o. If the base of the triangle is 18 cm. find its perimeter.

SOLUTION

Solved for a

a/sin⁡70 = 18/sin⁡40

a = (18 sin⁡70)/sin⁡40

a = 26.31 cm

Solved for perimeter of the triangle

P = 2a + 18
= 2(26.31) + 18
= 70.62 cm

Therefore, the perimeter of the triangle is 70.62 cm.

An 80-meter building is on top of a hill. From the top of the building, a nipa hut is sighted with an angle of depression of 54o. From the foot of the building the same nipa hut was sighted with an angle of depression of 45o. Find the height of the hill and the horizontal distance from the building to the nipa hut.

SOLUTION

Solved for c

c/sin⁡36 = 80/sin⁡9

c = (80 sin⁡36)/sin⁡9

c = 300.59 m

Solved for b

300.59/sin⁡90 = b/sin⁡45

b = (300.59 sin⁡45)/sin⁡90

b = 212.55 m

Since a and b are the legs of an isosceles right triangle then their measures are equal.
Therefore, the height of the hill is 212.55 m.

The lengths of the diagonals of a parallelogram are 32.5 cm and 45.2 cm. The angle between the longer side and the longer diagonal is 35o.Find the length of the longer side of the parallelogram.

SOLUTION

Solved for A

22.6/sin⁡A = 17.75/sin⁡35

sin A = (22.6 sin⁡35)/17.75

A = 46.910

Solved for b

b/sin⁡98.09 = 17.75/sin⁡35

b = (17.75 sin⁡98.09)/sin⁡35

b = 30.62 cm

Therefore, the...
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