(4) In securing a telephone pole, the guy wire should be fastened to the pole 12 feet from the ground, and secured to the ground 8 feet from the pole. Find the angle the guy wire makes with the pole and with the ground. How much wire is needed if 2 feet is allowed on each end?

(5) How far from the center of a circle of 12-inch radius will a tangent meet a diameter (extended) with which it makes an angle of, 18.4 ft. Answer.

10.7°? 56.31°? 33.69°?

70. Use of right triangle methods in solving obtuse triangles. a. When two sides and the included angle or the angles and a side of an obtuse triangle are known the rest of the triangle including the area may be found.

(1) Example: Find sides BD and AD of the obtuse triangle shown in figure 117, angle DAB being 30°, angle ABD 100°, angle BDA 50°, and side AB 9 inches.

FIGURE 117.-Reference triangle for solving obtuse triangle problems.

Solution: In this case, drop a perpendicular from the obtuse angle to the base forming two right triangles, ABC and BDC. From triangle ABC, obtain:

BC,

sin 30° =

and BC=(.5)(9)=4.5 in.

9

AC+CD=AD, and AD=7.7940+3.7760=11.5700 in.

Answer.

(2) Example: An airplane is flying a heading of 120° with an air speed of 240 mph. The wind velocity is 40 mph. from 90°. What are the ground speed, track, and angle of drift?

Solution: Problems of this nature are of frequent occurrence. Two sides of a triangle and the included angle BAC, figure 118, are known, since the direction of both the heading BA and the wind AC are given. To solve this type of triangle, extend the wind vector AC and drop a perpendicular to it from B. Note that the perpendicular is opposite the known angle. In the triangle ABD,

BD=AB sin 30°-240 (.5000)=120.0
AD AB cos 30°-240 (.8660)=207.8

=

ZABC= ZABD — ¿CBD=60°00′-54°26′=5°34′

Angle of drift.

The track is 180°- ZCBD=180—54°26′ =125°34' Track.

(3) Example: (a) In the heading-wind-track problem, figure 118, the known angle in the triangle, is frequently obtuse. In this case

some difficulty may be experienced in deciding how to draw the perpendicular necessary to form the right triangle required for the solution of the problem. Figure 119 shows the procedure to use when the angle is obtuse. The air speed is 200 mph. and the heading 75°. The wind is 50 mph. from 310°. The angle between the air speed vector and wind vector is 125°. If the air speed vector is extended

and a perpendicular dropped to it from C, a right triangle ADC is formed which can be easily solved for ground speed and track.

AC AD sec ZCAD, the ground speed

Track is simply heading (75°)+ ZCAD.

(b) If the wind vector had been extended and a perpendicular dropped from A to E, a different right triangle would have been formed which could be solved in the same manner.

b. Exercises. (1) An airplane at an altitude of 3,800 feet is sighted from two stations simultaneously. The angles of elevation are 58° and 47° respectively. Find the distance between the stations if the airplane is between the stations.

(2) By observing landmarks, a pilot calculates his ground speed as 188.5 mph. and his track as 221.5°. The wind is known to be 35.6 mph. from 211.5°. What should his air speed indicator read?

223.6 mph. Answer. 71. Miscellaneous exercises. (1) A target was sighted at an angle of depression of 18°. The bomber was flying at an altitude of 18,000 feet. How far will the plane have to travel to be directly over the target?

(2) Find the angles of a right triangle whose legs are 7 and 9 units long, respectively? 52.1° and 37.9°. Answer.

(3) An observer sighted a squadron of airplanes at an angle of elevation of 17.7°. They were flying at the time over a landmark that was known to be 5 miles away from the observer. What was the altitude of the airplanes in feet?

(4) In the right triangle shown in figure 120, find the other sides if—

B

(a) c=26, A=37.7°.

(b) b=42.4, A=32.3°.

(c) a=3.12, B=5.4°.

(5) In the right triangle shown in figure 121, find the missing sides and angles if—

(6) An airplane is directly over the middle of a lake and the angle

subtended by the lake (angular width) is 13.4°.

2 miles high, find the width of the lake in feet. (7) Find AD in figure 122.

B

If the airplane is

2,481 ft.

Answer.

(8) Find AC in figure 123.

1,682 ft. Answer. (9) An airplane flying a level course passes directly over a landmark. One minute later the landmark is observed again and has an angle of depression of 30°, the same size as the angle of elevation. If the plane is flying at the rate of 240 mph., find its altitude.