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88. The ambiguities which occur in the last case in the solution of oblique-angled triangles (Art. 85) may be discussed in the same manner as those in Art. 86; or, by means of the polar triangle, the last case may be deduced from that of Art. 86.


1. The sides of a triangle are 105°, 90°, and 75° respectively: find the sines of all the angles.

sin (8-0).

Solve a triangle 2. Shew that tan} A tan | B=

sin s when a side, an adjacent angle, and the sum of the other two sides are given.

3. Solve a triangle having given a side, an adjacent angle, and the sum of the other two angles.

4. A triangle has the sum of two sides equal to a semicircumference: find the arc joining the vertex with the middle of the base.

5. If a, b, c are known, c being a quadrant, determine the angles: shew also that if & be the perpendicular on c from the opposite angle, cos® 8 = cos'a + cos b.




6. If one side of a spherical triangle be divided into four equal parts, and 0,, 0,, 03, 0, be the angles subtended at the opposite angle by the parts taken in order, shew that

sin (6, +0.) sin 0, sin 0, = sin (0, +0.) sin 0, sin 03.


In a spherical triangle if A =B= 2C, shew that

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8. In a spherical triangle if A = B= 2C, shew that

( (cos 8+ sin



= 1. COS a


с 8 sin

2 9. If the equal sides of an isosceles triangle ABC be bisected by an arc DE, and BC be the base, shew that


2 2 10. If c,, c, be the two values of the third side when A, a, 6 are given and the triangle is ambiguous, shew that

tantan = tan } (- a) tan} (6+a).





89. To find the angular radius of the small circle inscribed in a given triangle.

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Let ABC be the triangle; bisect the angles A and B by arcs meeting at P; from P draw PD, PE, PF perpendicular to the sides. Then it may be shewn that PD, PE, PF are all equal; also that AE=AF, BF = BD, CD=CE. Hence BC + AF = half the sum of the sides = 8; therefore AF = 8 - a. Let PF=r. Now tan PF = tan PAF sin A F (Art. 62);

A thus tan r=tan a

(1). 2

sin (8 – a) .....

The value of tan y may be expressed in various forms; thus from Art. 45, we obtain

А (sin (8

(8-5) sin (8 — c) tan

sin 8 sin (8-a) s

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sin (8 – a) = sin {I (6+c) – 1 a}

= sin } (b + c) cos }a-cos 1 (6+c) sina
sin ţa cos } a {cos ļ (B-C) –cos } (B+C)}(Art. 54)

1 , )
sin ŽA
sin a sin 1 B sin } C

sin 1 B sin C therefore from (1) tan r =

cos. A hence, by Art. 51,


sin a


W{-cos Scos (S-A) cos (S-B) cos (S-C)} tan p=

2 cos ? A cos B cos

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4 cos 1 A cos } B cos1C = cos S + cos (S – A) + cos(S – B)+cos(S-C);

hence we have from (4)

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cos S S cos cos

+cos (– 4) + C08 (S B) + COB (S– 09}...(5).


90, To find the angular radius of the small circle described 80 as to touch one side of a given triangle, and the other sides produced,

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Let ABC be the triangle; and suppose we require the radius of the small circle which touches BC, and AB and AC produced. Produce AB and AC to meet at A'; then we require the radius of the small circle inscribed in A'BC, and the sides of A'BC are a, T-6, 7b

c respectively. Hence if r, be the required radius, and & denote as usual 1 (a+b+c), we have from Art, 89,

А tan

(1), 2


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sin s



tan 7,

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From this result we may derive other equivalent forms as in the preceding Article; or we may make use of those fornis immediately, observing that the angles of the triangle A'BC are A, п– , п — 7-B, -Crespectively. Hence : being 1 (a + b + c) and S being 3 (A + B + C) we shall obtain (sin 8 sin (8 b) sin (s—c)

(2), sin (8 – a)

sin (8 – a)
cos 1 B cos1 C

tanr, =

cos. A
{-cos S cos (S – A) cos (S B) cos (S – C)}
2 cos. A sin i B sin C

.(4), 2 cos . A sin B sin 10 1

cos S- cos (S – ) + cos (S B) + cos (S




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These results may also be found independently by bisecting two of the angles of the triangle A'BC, so as to determine the pole of the small circle, and proceeding as in Art. 89.

91. A circle which touches one side of a triangle and the other sides produced is called au escribed circle ; thus there are three escribed circles belonging to a given triangle. We may denote the radü of the escribed circles which touch CA and AB respectively by r, and rg, and values of tanr, and tanr, may be found from what has been already given with respect to tanr, by appropriate changes in the letters which denote the sides and angles.

In the preceding Article a triangle A'BC was formed by producing AB and AC to meet again at A'; similarly another triangle may be formed by producing BC and BA to meet again, and another by producing CA and CB to meet again. The original triangle ABC and the three formed from it have been called associated triangles, ABC being the fundamental triangle. Thus the inscribed and escribed circles of a given triangle are the same as the circles inscribed in the system of associated triangles of which the given triangle is the fundamental triangle,

92. To find the angular radius of the small circle described about a given triangle.



Let ABC be the given triangle; bisect the sides CB, CA at D and E respectively, and draw from D and E arcs at right angles to CB and CA respectively, and let P be the intersection of these

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