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

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 C 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, ACDACB = 24°. Hence, angle A, sides AC, DC, and the area, can be found.

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 63o. 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.

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.

By geometry, angle ACDACB = 11/24

Also, by geometry,

[ angle DAC = }

360° 180°

n

n

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.

2. The side of a regular pentagon is 24 ft. Find quantities as in Ex. 1. 3. The side of a regular octagon is 24 ft. Find quantities as in Ex. 1.

4. The radius of a circle is 24 ft. Find the lengths of the sides and apothems of the inscribed regular triangle, quadrilateral, pentagon, hexagon, heptagon, and octagon. Compare the area of the circle and the areas of these regular polygons; also compare the perimeters of the polygons and the circumference of the circle.

5. For the same circle as in Ex. 4, find the lengths of the sides of the circumscribing regular figures named in Ex. 4. Compare their areas and perimeters with the area and circumference of the circle.

6. If a be the side of a regular polygon of n sides, show that R, the 180° radius of the circumscribing circle, is equal to a cosec -; and that r, the 180° radius of the circle inscribed, is equal to a cot n

n

7. If r be the radius of a circle, show that the side of the regular inscribed 180° polygon of n sides is 2rsin -; and that the side of the regular circumn 180° scribing polygon is 2 r tan

n

8. If a be the side of a regular polygon of n sides, R the radius of the circumscribing circle, and r the radius of the circle inscribed, show that area of

34. Solution of oblique triangles. Since an oblique triangle can be divided into right-angled triangles by drawing a perpendicular from a vertex to the opposite side, it may be expected that knowledge concerning the solution of right-angled triangles will be of service in solving oblique triangles. This expectation will not be disappointed. An examination, which it is advisable for the student to make before proceeding farther, will show that all the sets of data from which a definite triangle can be drawn are those indicated in (1)-(4) below. The ability to make the following geometrical constructions is presupposed:

(1) To draw a triangle on being given two of its angles and a side opposite to one of them;

(2) To draw a triangle on being given two of its sides and an angle opposite to one of them;

(3) To draw a triangle on being given two of its sides and their included angle;

(4) To draw a triangle on being given its three sides.

In what follows, only the steps in the solutions will be indicated. The examples that are worked may be saved, so that the amount of labor required by the method of solution shown here can be compared with the amount required by another method which will be described later.

There are four cases in the solution of oblique triangles; these cases correspond to the four problems of construction stated above. CASE I. Given two angles and a side opposite to one of them.

A A

α

A

B

D

A

D

B

FIG. 30.

In ABC let A, B, a be known. Angle C and sides b, c are required. From C draw CD at right angles to AB or AB produced. In triangle CBD, angle CBD and side CB are known. .. BD and DC can be found.

Then, in triangle ACD, side DC and angle A are known. .. AC and AD can be found.

Side AB = AD

when B is acute.

BD when B is obtuse, and AB=AD+DB

Another method of solution is given in Art. 55.

1. Ex. 1, Art. 55.

3. Ex. 2, Art. 60.

EXAMPLES.

2. Ex. 2, Art. 55.

4. Other Exs. in Arts. 55, 60.

CASE II. Given two sides and an angle opposite to one of them. N.B. The first part of the text in Art. 56 should be read at this time.

Let (Fig. 30) AC, BC, angle A be known. [In a certain case, as shown in Art. 56, two triangles can be drawn which satisfy the given conditions.] From C draw CD at right angles to AB or AB produced.

In ACD, AC and A are known. .. AD, DC, angle ACD, can be found.

Then, in BCD, BC and CD are known. .. BD, angle DBC, can be found.

In one figure, AB

=

AD – BD, angle ABC = 180° – CBD. In other figure, AB = AD + DB. In both figures,

angle ACB = 180° — (CAB + ABC).

Another method of solution is given in Art. 56.

EXAMPLES.

1. Ex. 1, Art. 56.

3. Ex. 1, Art. 60.

2. Ex. 2, Art. 56.

4. Other Exs. in Arts. 56, 60.

CASE III. Given two sides and their included angle.

In ABC let b, c, A be known. Side a, B, C are required. From C draw CD at right angles to AB or AB produced.

In ACD, AC and angle A are known. .. CD and AD can be found.

Then, in triangle CDB, CD is now known, and BD-AD-AB or AB - AD. .. Angle CBD can be found. Angle ABC= 180° - CBD in figure on the left. Angle ACB=180° — (A + B). Another method of solution is given in Art. 57.

In ABC let a, b, c be known. The angles A, B, C are required.

From any vertex C draw CD at right angles to AB or AB pro

b2 + c2― a2

(1)

(2)