B

a

104. Escribed Circle. – To find the radii of the escribed circles of a triangle.

A circle, which touches one side of a triangle and the other two sides produced, is called an escribed circle of the triangle.

В.

C Let O be the centre of the escribed circle which touches the side BC and the other sides pro

E duced, at the points D, E, and F, respectively, and let the radius of this circle be ri. We then have from the figure A ABC = A AOB + A AOC – A BOC.

bri ari + 2 2 2 =4710b+c-a)=r(s – a). (Art. 99)

S ..=

(1)

a

Similarly it may be proved that if r, rg are the radii of the circles touching AC and AB respectively,

s

S
73
b'

12 =

=

105. To find the Distance between the Centres of the Inscribed and Circumscribed Circles * of a Triangle.

Let I and be the incentre and circumcentre, respectively, of E the triangle ABC, IA and IC B bisect the angles BAC and BCA;

AM

F

therefore the arc BD is equal to the arc DC, and DOH
bisects BC at right angles.
Draw IM perpendicular to AC. Then

A+C
ZDIC =

- BCD + BCI = DCI.
2
=

A
2

=

... DI= DC=2R sin

1. The sides of a triangle are 18, 24, 30; find the radii of its inscribed, escribed, and circumscribed circles.

Ans. 6, 12, 18, 36, 15.

2. Prove that the area of the triangle ABC is

1 ca
2 cot A + cot B

3. Find the area of the triangle ABC when (1) a=4, b= 10 ft., C= 30°.

Ans. 10 sq. ft. (2) b = 5, c= 20 inches, A = 60°.

43.3 sq. in. (3) a = 13, b= 14, c= 15 chains.

chains.

84 sq:

=

6. Prove that the area of the triangle ABC is represented by each of the three expressions :

2 R2 sin A sin B sin C,

rs, and

=

Rr(sin A + sin B + sin C). 7. If A = 60°, a=v3, b=v2, prove that the area

= |(3+3). 8. Prove R(sin A + sin B + sin C)= 8.

9. Prove that the bisectors of the angles A, B, C, of a triangle are, respectively, equal to

A
B

С
2 bc cos
2 ca cos

2 ab cos
2
2

2
b+c cta

a+b

106. To find the Area of a Cyclic* Quadrilateral.

А.
Let ABCD be the quadrilateral, and
a, b, c, and d its sides. Join BD.
Then, area of figure = S

= 4ad sin A + bc sin C
= {(ad + bc)sin A

(1)
Now in A ABD, BD' = a+ d— 2 ad cos A,

A , and in A CBD,

BD' = 62 +62-2 bc cos C

c
=b' + 0-2 bc cos A.
a’ - h - 02 + d?

ca
.. cos A

2 ad + bc) a’ - - 0 + 47

- 62 ... sin A =

cé
V
1

2 ad + bc)
V (2 ad + 2 bc)? – (a’ – 12 – ? + da)?

2 (ad + bc)

a

V[(a + d)?—(6 — c)?][(b + c)2 – (a – d)?]

2 (ad + bc)

✓(a+d+b-c)(a + d-b+c) (b+c+a-d)(6+c-a+d)

2(ad + bc) 2v (s - a)(8 – b) (8 - 0) (s – d)

ad + bc

S

S

(where 2 s= a +b+c+d).

Substituting in (1), we have

S=V (s – a)(s — b)(8 — c)(s – d).

S

The more important formulæ proved in this chapter are summed up as follows:

s s . 6. tanA=V

(8 – 6)(8 —c).

s(s-a) 7. sin A = 2 V8(8 – a) (8 — b) (s — c)

• (Art. 100) bc 1

V2bc? + 2 ca’ + 2 a2b2 - a* - 64 - c. 2bc

8. Area of A= V8 (8-a) (8 - b)(8 -c).

(Art. 102)

9. Area of A = (a +b+c)=rs

Ź

(s — a) (s —b) (s — c).

S

10.

1 =

In a right triangle ABC, in which C is the right angle, prove the following:

sin? A - sina B 1. cos 2B

sin? A + sin’B

9. (sin A - sin B)' + (cos A + cos B)' = 2.

In any triangle ABC, prove the following statements :

С

A - B 11. (a + b) sin

=CCOS