Let BP be the flagstaff on the top of the cliff BL, and let C be the place of observation.

BP 30 ft., LCB 38° 20', LCP = 40° 20'. Let CL = x,

=

4. At a point 180 ft. from a tower, and on a level with its base, the elevation of the top of the tower is found to be 65° 40.5'. What is the height of the tower?

5. From the top of a tower 120 ft. high the angle of depression of an object on a level with the base of the tower is 27° 43'. What is the distance of the object from the top and bottom of the tower?

6. From the foot of a post the elevation of the top of a column is 45°, and from the top of the post, which is 27 ft. high, the elevation is 30°. Find the height and distance of the column.

7. From the top of a cliff 120 ft. high the angles of depression of two boats, which are due south of the observer, are 20° 20′ and 68° 40'. Find the distance between the boats.

8. From the top of a hill 450 ft. high, the angle of depression of the top of a tower, which is known to be 200 ft. high, is 63° 20'. What is the distance from the foot of the tower to the top of the hill ?

9. From the top of a hill the angles of depression of two consecutive mile-stones, which are in a direction due east, are 21° 30′ and 47° 40'. How high is the hill ?

10. For an observer standing on the bank of a river, the angular elevation of the top of a tree on the opposite bank is 60°; when he retires 100 ft. from the edge of the river the angle of elevation is 30°. Find the height of the tree and the breadth of the river.

11. Find the distance in space travelled in an hour, in consequence of the earth's rotation, by an object in latitude 44° 20'. [Take earth's diameter equal to 8000 mi.]

12. At a point straight in front of one corner of a house, its height subtends an angle 34° 45', and its length subtends an angle 72° 30'; the height of the house is 48 ft. Find its length.

30. Problems requiring a knowledge of the points of the Mariner's Compass. The circle in the Mariner's Compass is divided into 32 equal parts, each part being thus equal to 360° ÷ 32, i.e. 114°. The points of division are named as indicated on the figure.

It will be observed that the points are named with reference to the points North, South, East, and West, which are called the cardinal points. Direction is indicated in a variety of ways. For instance, suppose C were the centre of the circle; then the point P in the figure is said to bear E.N.E. from C, or, from C the bearing of P is E.N.E. Similarly, C bears W.S.W. from P, or, the bearing of C from P is W.S.W. The point E.N.E. is 2 points North of East, and 6 points East of North. Accordingly, the phrases E. 221° N., N. 671° E., are sometimes used instead of E.N.E.

31. Mensuration. Let ABC be any triangle, and let the lengths of the sides opposite the angle A, B, C be denoted by a, b, c, respectively. From any vertex C draw CD at right angles to the opposite

side AB. It has been shown in arithmetic and geometry, that the area of a triangle is equal to one-half the product of the lengths of any side and the perpendicular drawn to it from the opposite vertex. [In (1) A is acute, in (2) ▲ is obtuse.]

It will be seen in Art. 45, that sin (180 — A) = sin A. Hence, the area of a triangle is equal to one-half the product of any two sides and the sine of their contained angle.

EXAMPLES.

1. Find the area of the triangle in which two sides are 31 ft. and 23 ft. and their contained angle 67° 30'.

2. Find area of triangle having sides 125 ft., 80 ft., contained angle 28° 35'.

3. Find area of triangle having sides 125 ft., 80 ft., contained angle 151° 25'. [Draw figures carefully for Exs. 2, 3.]

4. Find area of parallelogram two of whose adjacent sides are 243, 315 yd., and their included angle 35° 40'.

5. Find area of parallelogram two of whose adjacent sides are 14, 15 ft., and included angle 75o.

6. Find area of triangle having sides 40 ft., 45 ft., with an included angle 28° 57' 18".

7. Write two other formulas for area ABC, similar to that derived above. Also, derive them.

32. Solution of isosceles triangles. In an isosceles triangle, the perpendicular let fall from the vertex to the base bisects the base and bisects the vertical angle. An isosceles triangle can often be solved on dividing it into two equal right-angled triangles.

EXAMPLES.

1. The base of an isosceles triangle is 24 in. long, and the vertical angle is 48°; find the other angles and sides, the perpendicular from the vertex and the area. Only the steps in the solution will be indicated. Let ABC be an isosceles triangle having base AB = 24 in., angle = 48°. Draw CD at right angles to base; then CD bisects the angle ACB and base AB. Hence, in the right-angled triangle ADC, AD | AB = 12, ACD = ACB = 24°. Hence, angle A, sides AC, DC, and the area, can be found.

24

2. In an isosceles triangle each of the equal sides is 363 ft., and each of the equal angles is 75°. Find the base, perpendicular on base, and the area.

3. In an isosceles triangle each of the equal sides is 241 ft., and their included angle is 96°. Find the base, angles at the base, height, and area.

4. In an isosceles triangle the base is 65 ft., and each of the other sides is 90 ft. Find the angles, height, and area.

5. In an isosceles triangle the base is 40 ft., height is 30 ft. Find sides, angles, area.

6. In an isosceles triangle the height is 60 ft., one of equal sides is 80 ft. Find base, angles, area.

7. In an isosceles triangle the height is 40 ft., each of equal angles is 63°. Find sides and area.

8. In an isosceles triangle the height is 63 ft., vertical angle is 75°. Find sides and area.

33. Related regular polygons and circles. The knowledge of trigonometry thus far attained, is of service in solving many

problems in which circles and regular polygons are concerned. Some of these problems are:

(a) Given the length of the side of a regular polygon of a given number of sides, to find its area; also, to find the radii of the inscribed and circumscribing circles of the polygon;

(b) To find the lengths of the sides of regular polygons of a given number of sides which are inscribed in, and circumscribed about, a circle of given radius.

C

B

FIG. 29.

For example, let AB (Fig. 29) be a side, equal to 2 a, of a regular polygon of n sides, and let C be the centre of the inscribed circle. Draw CA, CB, and draw CD at right angles to AB. Then D is the middle point of AB.

360° 180°

By geometry, angle ACD = ACB = 1 · n

Also, by geometry,

n

[angle DAC = (24) 90° (2)90]

=

90°.

Hence, in the triangle ADC, the side AD and the angles are known; therefore CD, the radius of the circle inscribed in the polygon, can be found. On making similar constructions, the solution of the other problems referred to above will be apparent. The perpendicular from the centre of the circle to a side of the inscribed polygon is called the apothem of the polygon.

EXAMPLES.

1. The side of a regular heptagon is 14 ft.: find the radii of the inscribed and circumscribing circles; also, find the difference between the areas of the heptagon and the inscribed circle, and the difference between the area of the heptagon and the area of the circumscribing circle.