2

+

=

Example 3. Solve tan2 6+sec 0=5.
(1) Since tan-A=sec 0 – 1, .. seca – 1+sec =5.
(2)
... sec 0 + sec =6,

1 25
... sec A+sec A+ +6+

4 4
1 5
.. sec A+

2 2'

4 6
.. sec A or =2 or – 3.

2 2 (3) Reference to Art. 108 shows that sec 8=2 is satisfied by 0=60°. But negative values of the ratios have not yet been treated. Hence for the present we may discard the solution sec 0= -3.

Example 4. Solve 2 sin A=/3 tan A.
Method by factors. Here
2 sin A - J3 tan A=0 gives sin A (2-13 sec A)=0,

.: either sin A=0 or 2-3 sec A=0.
Now
sin A=0 gives A=0.

2 And 2-/3 sec A=0 gives sec A=

13'

.: A=30°. .. A=0 or 30°.

EXAMPLES IV.

5

139

1. Draw the angle A and find all its ratios in each of the following cases :(1) sin A = 1 (2)

tan A = 3,

(3) sec A = 14, (4) cos A = .6, (5) cot A = .3, (6) cosec A= 34. 2. Express tan A in terms of sin A and of cos A. 3. Express sec A in terms of sin A and of cot A. 4. Express cosec A in terms of sec A and of cos A. 5. Express sin A in terms of tan A and of sec A.

6. Divide an angle into two angles whose cosines shall have a given ratio.

7. Divide an angle into two angles whose cosecants shall have a given ratio.

=

=

=

=

=

=

.

=

Find the acute angles including 0 and 90° which satisfy each of the following equations (8—25).

8. 4 + 4 cos 0 = 3 sec 0. 17. 4 sin 0 = cosec 0 – 2 /2.
9. 2 sin 0 + 2 cosec 0=5. 18. 3 tano 0 + 2 = cot.
10. tan 0= cot .

19. 2 cosA = 2 sino 0 + 1.
11. 2 sin 0
= cosec 0.

20. 4 sin A = cosec A - 2.
12. secs A + 3 cosec = 8. 21. tan 0 + 4 cosa 0 = 3.
13. 2 sin 0 tan A.

22. sec 0 2 + 2 tan 0. 14. tan A + cot A = 2.

23. 2 sin Otan 0+1=tan 0 + 2 sin 0.
15. 12. cos 0 - cot 0 = 0. 24. 2 cosa 0 + /3. sin 0 = 2.
16. 2 sino A = 3 cos 0.

25. tan 0 + cotA= 4.
Prove the following identities : (26—32).
26. (sin 30° + sin 60°) (cos 30° - cos 60°) = sin 30o.
27. (cos 60° + cos 30°) (sin 60° – sin 30°) = cos 60°.
28. sin 30° + sin 60° = 12.cos 15°.
29. cos 30o – cos 60° = 12.sin 15°.
30. (sin 60° – sin 45°) (cos 45o + cos 30%) = sino 30°.
31. cos 36o – sin 18° = 1.
32. 4 sin 18° cos 36° = 1.

1 33. If the sine of an angle be > and the cosine of that

12 angle be > ], between what limits does the angle lie?

1 34. If the tangent of an angle be > 13 and the cotangent of

> that angle be > 1/3, between what limits does the angle lie ?

35. Trace the changes in value of (1) 1 - cos 0; (2) sec 0 - 1; (3) sin 0 + cos 6; (4) sec 0 – tan 6 : as 0 changes from 0 to 90°.

Solve the following simultaneous equations :
36.

cosa B, and 4 cot a = sec B.
1

/3 37. sin (A + B)

A 38. In Example I. 19, show that the angles ODC and DOC are, respectively, 60° and 30°.

sino a=

and cos (1 – B) =

CHAPTER V.

RATIOS OF COMPOUND ANGLES.

116. The following two propositions will be found useful in many applications.

I. If a magnitude AB be bisected at C and divided at D,

then (1) AD + DB= 2. AC, and (2) AD - DB = 2.CD.

For (1) AD+DB=AB= 2. AC. and (2) AD-DB=(AC + CD)-(CB-CD)=2.CD.

II. If a magnitude AB be bisected at C and extended to D,

=

=

then (1) AD-BD=2. AC, and (2) AD+BD= 2.CD.

For (1) AD-BD = AB = 2. AC. and (2) AD+BD=(AC + CD)+(CD-CB)=2.CD.

These propositions may be applied to straight lines, angles, triangles, &c., as in Arts. which follow.

117. To express the area of a triangle, when two sides and the included angle are given.

We have shown (Art. 22) that the area of a triangle is measured by 1 (base x altitude).

A A

С
B

B
с
B

LC
Let ABC be any triangle. Draw AL perpendicular to the
base BC produced if necessary.
Then
A ABC = { BC. LA.

(1) The angle ACB is either acute, obtuse, or right.

In each case, let C denote that angle at the point C, which is not obtuse.

LA
Then, in each case, sin C =

CA
.. CA.sin C= LA.
Substituting for LA in (1)
A ABC 1 BC.CA.sin C.

(2) Hence, the area of a triangle is measured by half the product of two sides and the sine of the (non-obtuse) angle contained by them.

118. To express the sine of the internal or external angle of a triangle in terms of the area of the triangle and the sides containing the angle.

By the last article we have, if C be the non-obtuse angle at C, which is either internal or external to the triangle ABC,

BC.CA sin C= A ABC,

2. Δ ABC .. sin C=

BC.CA Hence, the sine of the internal or external angle of a triangle

twice area of the triangle

product of sides containing the angle The student should observe that both the numerator and the denominator of the above fraction represent an area. Hence the sine is given correctly as an abstract number.

119. We proceed to find the ratios of angles formed by the addition or subtraction of two angles.

We shall adopt a method which in each case involves the application of the preceding article. But alternative proofs are given in Chapter XIII., which the student may use in preference to these.

The beginner is very liable to fall into the mistake of supposing that sin (A + B) must equal "sin A +sin B, that sin 2A must equal 2 sin #, and

He must remember, as was pointed out before, that sin is not a multiplier which can be separated from its accompanying symbols. Thus, sin (A+B) means “The sine of the angle formed by adding the two angles A and B”; while sin A+sin B means “The number formed by adding the two sines of the angles.”

In what follows all the angles whose ratios are involved are ordinary (i.e. positive) acute angles.

SO on.

12C. To prove the formula sin (A+B)=sin A cos B+cos A sin B..(1).

B

A
А

B

N

=

Let A, B be the angles at the base of a triangle, whose vertical angle at C is obtuse. Draw CN perpendicular to AB.

Then, producing AC, the exterior angle at C is acute and interior LA + interior .B= exterior .C.

[Euc. I. 32.

2 A ABC NC. BA .. sin (A + B) = sin (exterior 2 C)

AC.BC AC. BC
NC (BN+NA)_NC BN NA

NC
AC.BC АС BC AC BC

sin A. cos B + cos A .sin B. 121. To prove the formulasin (A-B)=sin Acos B-cos A sin B..(2).

+

Let A be an (acute) angle exterior, and B an angle interior, to the triangle A BC. Draw CN perpendicular to BÀ produced. Then exterior 2A - interior , B=interior - C. [Euc. I. 32.

2Δ ABC NC.BA .. sin (A - B) = sin interior 2 C) =

AC. BC AC. BC
NC(BN – AN) NC BN AN NC
AC. BC AC BC AC BC

= sin A.cos B-COS A . sin B.

=