Therefore an angle has been drawn whose sine or whose cosine has the given value r.

Q.E.F. If r had been greater than 1, OM would have been greater than OA, M would have been outside the circle, and MP would not have cut the circle at all. Hence the method would have failed as we should expect.

99. Given the value of the sine or cosine of any angle, to find the other ratios.

In other words: To express the other ratios in terms of the sine or cosine.

As before, let r be the value of the given ratio.
Construct the figure of the last article.
Then, by Euc. I. 47, MP9+ OM' = 0P,

i.e. MP + pol = 1,
.. MP2= 1 -7

:. MP = N(1 – 7%). (i) Suppose that r is the sine of the required angle. Then MPO is the required angle, which we may call A.

.. p=sin A.

PM /(1 – 72) ; . cos A = cos MPO

PO

V(1 – sino A),

1 MO

sin A tan A = tan MPO =

PM 7(1 – 92) (1 - sin? A)' and so for the other ratios.

(ii) Suppose that r is the cosine of the required angle. Then MOP is the required angle, which we may call A.

.. p = cos A.

MP (1 – m2)
.. sin A = sin MOP=

J(1 - cosa ),
OP 1

MP_J(1 – nå) V (1 - coso A)
tan A tan MOP=

OM and so for the other ratios.

r

r

COS A

100. Given the value of the tangent or cotangent of any angle, to draw the angle. Let the given value be r, i.e. r : 1.

P Take OM=1: draw MP at rightangles to OM, and let MP=r. Join ОР.

(84+1)

r

Therefore an angle has been drawn whose tangent or whose cotangent has the given value r.

Q. E. F.

+

=

101. Given the value of the tangent or cotangent of any angle, to find the other ratios.

In other words : To express the other ratios in terms of the tangent or cotangent. From the figure of the last article, by Euc. I. 47, OP = OM? + MP2 = 1 + m2,

1. OP= 1(1+2).
(i) Suppose tan A so that MOP= A.

MP

tan A

OP 7 (1 + f2) 7(1 + tano A)' and so for the other ratios. (ii) Suppose r = cot A, so that MPO = A. MO 1

1 Then, sin A =sin MPO =

POJ(1+r2) (1 + cot’ A)? and so for the other ratios.

1 =

Then, sin A = sin MOP = OP =/(1+r)

=

=

=

102. Given the value of the secant or cosecant of any angle, to draw the angle.

Let the given value be r, i.e. r : 1.

Therefore an angle has been drawn whose secant or whose cosecant has the given value r.

Q. E. F.

=

=

sec A

103. Given the value of the secant or cosecant of any angle, to find the other ratios.

In other words: To express the other ratios in terms of the secant or cosecant. From the figure of the last article, by Euc. I. 47,

MP=OP – OM? = gol – 1,

. MP=1 (72 – 1).
(i) Suppose r = sec A, so that MOP= A.

MP _ (32 – 1) _ (sec? A – 1)
Then, sin A = sin MOP=

OP
and so for the other ratios.
(ii) Suppose r = cosec A, so that MPO= A.

OM 1 1
Then, sin A = sin MPO =

OP

cosec A' and so for the other ratios.

104. The results of Arts. 99, 101, 103, may also be obtained, without drawing a figure, from the general relations of equality established in the preceding chapter. Example 1. To express the tangent in terms of the cosecant.

1

1 tan A=

cot AJ(cosec? A – 1)

=

r

Example 2. To express the sine in terms of the tangent.

tan A tan A sin A=tan A x cos A=

(1+tan-A).

sec A

105. Other problems may be solved in a somewhat similar way. It is left to the student to prove the results of the following examples.

Example 1. To divide any angle AOB into two angles whose secants shall have a given ratio. Cut off OA and OB, so that OA : OB=given ratio.

: Join AB : and draw OC perpendicular to AB. OC shall divide AOB (internally or externally) as required.

Example 2. To divide any angle AOB into two angles whose tangents shall have a given ratio.

Let CD : DE=given ratio. Place CD and DE in a straight line. Upon CE describe a segment CFE of a circle containing an angle equal to AOB. Draw DF at right-angles to CE cutting the circumference in F.

DF shall divide CFE, which is equal to AOB, as required.

Example 3. To divide any angle AOB into two angles whose sines shall have a given ratio.

Produce A0 to A': and cut off OA' and OB so that OB :0A'=given ratio. Draw OC parallel to A'B.

OC shall divide AOB as required.

§ 2. THE ANGLE BEING GIVEN. 106. We now proceed to find the ratios of such angles as have simple geometrical relations by a method similar to the above. It is convenient to take the smallest line in our figure as the unit of length, in order to avoid unnecessary fractions. A second side is then found by some proposition connected with the particular angle in question. Lastly, a third side is always found by Euc, I. 47.

The student should in each case retain the figure with the values of its sides and angles marked, and not try at first to remember the ratios themselves.

107. The ratios of the half right-angle. [45°

, 50%, -2.]

=

Take AB = AC for unit length. Then

BC2 = ABP + AC2 = 1 + 1 = 2, [Euc. I. 47.

:. BC = 12,
AC 1

BA 1
.. sin 45°
BC72

BC
AC

BA
tan 45°

1

: cot 45° = 1,
BA

AC
BC

BC
2 : cox 45°

. BA

AC

: cos 45°

sec 45°

=

Since 45° is its own complement, any ratio of 45° is the same as its co-ratio.

108. The ratios of land of a right-angle.

On AB as base describe an equilateral triangle ABC, and bisect the angle ACB by CD.

Or briefly : Construct a bisected equilateral triangle. Then A BC being equilateral is also equiangular. (Euc. I. 5. :. each of its angles = of 2 right-angles = 60°, [Euc. I. 32.