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The sides of plane triangles are to each other as the sines of the opposite angles.
From A, any angle of the triangle A B C, (see fig. p. 80.) let AD be drawn perpendicular to B C.
Then as the triangle A B D is right angled at D, we have A B. sin B = RIADand for the same reason A C. sin C =R.AD. Hence A B . sin B = AC. sin C, and consequently A B : sin C :: AC : sin B. or A B : AC:: sin C : sin B.,
In every plane triangle if a perpendicular be drawn from any of the angles on the opposite side, the segments of that side are to one another as the tangents of the parts into which the opposite angle is divided by the perpendicular.
For (see fig. p. 80.) R.DB = AD. tan D AB; and R.DC = AD, tan DA C. Therefore R. BD: R. DC:: AD. tan DAB:AD. tan D AC, or BD:DC:: tan DAB : tan DAC.
Cor. B D:DC :: cot B : cot C; for tan D AB = cot B, and tan DAC = cot C.
PROPOSITION VI. The sum of any two sides of a triangle is to their difference, as the tangent of half the sum of the angles opposite to those sides is to the tangent of half their difference.
Let A B C be any plane triangle. Then A B : AC :: sin C : sin B; whence A B + AC: A B – A C: sin C + sin B : sin C. sin B. But (Prop. 2. Trig. Formula 4.) C
C + B
tan sin C + sin B
Or sin C + sin B; sin C sin B
C B* tan
Therefore A B + AC:
C + B С B sin C - sin B::tan
: tan 2
C + B С B A B - AC:: tan
: tan 2
PROPOSITION VII. If a perpendicular be drawn from any angle of a plane triangle to the opposite side, or base, the sum of the segments of the base is to the sum of the other two sides, as the difference of those sides is to the difference of the segments of the base.
For (Geo. Theo. 48. Cor.) the rectangle under the sum and difference of the segments of the base is equal to the rectangle under the sum and difference of the sides ; therefore the sum of the segments of the base is to the sum of the other two sides, as the difference of those sides is to the difference of the segments of the base.
PROPOSITION VIII. It is proposed to investigate the relation between the sides of a plane triangle, and the cosine of any of its angles, to a given radius.
Let A B C be any plane triangle, and AD a perpendicular from one of its angles, as A, on the opposite
A B C, or on B C produced. . Then when
A A D falls within the triangle we have (Geo. Theo. 47. Cor.) 2 .BC.CD = B C + A C2 BA?. But (Trig.
B C D Prop. 3. Cor. 2.) rad . CD= AC. cos AC . cos C
2 BC.AC. cos C C, or CD=
= BC+ AC rad
B C + A Co - A B - A BP, and consequently cos C = rad .
2 BC.AC B C2 + A C2 A B2 cos C
to radius unity. 2 BC, AC Again, when A D falls without the triangle, we have (Geo. Theo.46. Cor.) 2 BC.CD= B A - (B C + A C.) But (Trig. Prop. 3. Cor. 2.) rad .CD= AC cos A CD= AC cos ACB; for ACB is an obtuse angle, and supplemental to A CD. Hence - AC. cos' ACB
2 BC.AC.cos ACB C D =
and therefore rad
rad 2 BC.AC.
cos A B C = A B – (B C2 + AC%); or
= B C2
rad + A C2 – A BR; whence we have, exactly as in the former case,
BC% + A C - A B cos A CB=rad
; B C2 + A C2 АВ? or cos A CB =
to radius unity.
2 BC, AC Cor. If the angles of a plane triangle be represented by the capital letters A, B, and C, and the sides opposite those angles respectively by the small letters a, b, and c, the above expressions, when radius is unity, give us
12 + c cos A
2 b c
a” + c2 .62 cos B =
It is proposed to investigate the relation between the sides of a plane triangle, and the sine and cosine of the half of any of its angles.
Adopting the notation in the corollary to the preceding proposition, we have 62 + c - a
? cos A =
But (Form. 8. Schol. Prop. 2. Trig.) 2 b c
12 + c a? 1 + cos A = 2 cos2- therefore 2 cos? = 1 +
2 b c 62 + c a2 2 b c + 32 + c a
But (Geo. Theo.41.) 2 b c 2 b c
2 b c
a ? 6+6, therefore 2 cose 2 b c + 32 + c = bt
2 b c
2 b c
A b+c+ a . b + c - a btcta.b + c
a. Hence cos?
4 b c b+c+ a btc-a btcta (b + c + a
a 2 2 2
; for the bc subtraction of the half of any quantity, is an operation equivalent to adding the half of it, and then subtracting the whole of the quantity from the sum. btcta
be put = $ the above expression is cose 2 S.S
S.s to radius unity, or cos? = rad?
to a given bc
S.S radius. Hence cos
Again, (Form. 8. 2 A
A Schol. Prop. 2. Trig.) 1 - cos A = 2 sine ; ; therefore 2 sine.
2 b2 + c -a? 2 b c
73 + c ao 1
a” - 12 c2 + 2 b c 2 b c 2 b c 2 b c
2 b c a? (62 + ca
But b' + c? — 2bc= b (Geo.
A to a given radius. Hence sin
A A 4 sin? cos? 2
Hence substituting the above values of rad A
A sin? and cos?
we have 2
Cor. 2. As by analogous processes we would deduce like expressions for the angles B and C; we may arrange the resulting formulæ as under, viz.
Sb. S rad”.