Application 1

A tunnel is to be made through mountain A and B. A point C, from which both A and B are visible, is 384.8 ft from A and 555.6 ft from B. How long is the tunnel if ÐACB = 35⁰ 30' ?

Solution

In triangle ABC, Ð ACB = 35⁰ 30', b = 384.8 and a = 555.6, using Law of cosines

i.e. c² = a² + b² - 2ab cos Ð ACB

\ c² = (555.6)² + (384.8)² - 2 (555.6) (384.8) cos (35⁰ 30')

\ c² = 109494.81 \ c = 330.9

\ The length of the tunnel is 330.9 ft.

Application 2

A plane I leaves an airport at 1 P.M. and flies a st. course at 400 miles per hour. Plane II leaves the same air port at 1.30 P.M. and flies at 300 miles per hour on a st. course which makes an angle of 78⁰ with that plane I. How far apart are the two planes at 3 P.M. ?

Solution

Distance covered = speed x time.

\ At 3 P.M. plane I will have covered a distance

(AB) = 400 x 2 = 800 miles and plane II will have

flown a distance (AC) = 300 x 2 = 600 miles.

The distance between two planes at 3 P.M. will be (BC).

Considering the triangle ABC for this problem we have

c = 800, b = 600, ÐA = 78⁰

then by Law of cosines, we have

\ a² = c² + b² - 2cb cos A

\ a² = (450)² + (800)² - 2 (450) (800) cos 78⁰ = 692803.58

\ a = 832.348 miles.

AAS : Two angles and a monoincluded side are given.

Index

3. 1 Solving Right Triangles
3. 2 Law of Cosines
3. 3 Law of Sines
3. 4 The Ambiguous Case of Law of Sines
3. 5 Areas of Triangles
Supplementary Problems

Chapter 4