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(2.) Let the value of sin B be > 1.
We then have sin A > 1, which is impossible.

Hence, in this case, it is impossible to form a triangle with the given elements.

(3.) Let the value of a sin B be < 1.

Then, since, Art. 15, the sine of an angle is the same as the sine of its supplement, there are two values of A which satisfy the equality, sin A = g sin B, and these values are supplementary.

Let A, A' be the two values, then the relation between them is A + A' = 180°.

If a is not greater than b, then A is not greater than B, and there is no doubt as to which value of A is to be taken. If, however, a is greater than that is, if the given angle is opposite to the less of the given sides, we must have A greater than B, and both values of A may satisfy this condition. This particular case, when the given angle is opposite to the less of the given sides, is called the ambiguous case. We will illustrate this geometrically.

42. The ambiguous case.
Let a, b, B be given to construct the triangle.

Draw the line BC equal to the given side a, and draw BA making an angle B with BC.

Then with centre C and radius
CA equal to b describe an arc
meeting BA or BA produced in
A and A', and join CA and CA.

Each of the triangles ABC,
A 'BC satisfies thegiven conditions. B.

For, in the triangle ABC, we have BC a, AC = 6, and LABC B; and in the triangle ABC, we have BC a, AC = b, and L ABC B.

Again, the sides BA and BA correspond to the two valuer

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of c which are obtained from the two values of A in the equality sin A = 7 sin B (see last Art.).

And the angles ACB and A'CB correspond to the two values of C which would also be found.

COR. If a perpendicular CD be drawn from C upon AA', and if c' and c be the lengths of BA' and BA respectively, it may be easily shown that c' + 6 = 2 a cos B, and cred C = 2 6 cos A.

12,

60°, C

Ex. VIII. Solve the following triangles, having given1. b = 12, c = 6, A = 60°. 2. a = 18, 6 = 18 2, A = 30°.

. 3. a = 5 73,6 = 5 12,c = f (6 + 12). 4. a = B

15° 5. a = 3 12 + 16,6 = 6, C = 45°. 6. a =

10 13,6 = 15/2, A = 45°. Given7. b = 251, c = 372, A = 40° 32', find B and C. Log 121 = 2.0827854, L cot 20° 16' = 10-4326795,

2:7944880, L tan 27° 44' = 9.7207827,
L tan 27° 45' = 9.7210893.

b
341, B

28° 24', find A.
Log 237 23747483, L sin 28° 24' 9.6772640,
Log 341 = 2.5327544, L sin 19° 18' = 9.5191904,

L sin 19° 19' = 9.5195510.
9. C 26° 32', and a : 6:: 3:5, find A, B.

Log 2 = -3010300, L cot 13° 16' = 10-6275008,
L tan 46° 40' = 10·0252805, L tan 46° 41' = 10·0255336.

Log 623

8. a =

237,

14, 6 = 16, c = 18, find A, B. log 2 = •3010300, log 3 -4771213, L tan 24° 5' = 9.6502809, L tan 24° 6' 9.6506199, L tan 29° 12' = 9.7473194, L tan 29° 13' = 9.7476160.

10. a =

12. a =

5,6

11. a = 3,6 2 A = 60°, find B, C, and c.

Log 2 = 3010300, log 3 = .4771213,
L sin 35° 15' = 9.7612851, L sin 35° 16' = 9.7614638,
Log 1.3797 = .1397847, log 1.3798 = 1398161.

6, c = 7, find A.
Log 2 = 3010300, L tan 22° 12' = 9.6107586,

,
Log 3

4771213, L tan 22° 13' = 9.6111196. 13. If c, c' be the two values of the third side in the ambiguous case when a, b, A are given, show that,

(c - c)2 + (c + c')? tano A 4a. 14. If a, b, A are given, show from the equation

62 + ca 2 bc cos A a”; that if c and c' be the two values of the third side

cc' 62 - a-, and c + c' 2 6 cos A. 15. Show also from the same equation that there is no ambiguous case when a = b sin A, and that c is impossible when a <b sin A.

16. Having a - b, A, B, solve the triangle.

17. Given the ratios of the sides, and the angle A, solve the triangle.

Φ

A a B)

cot

b cos ♡

= C.

CHAPTER X.

HEIGHTS AND DISTANCES.

43. We shall now show how the principles of the previous chapters are practically applied in determining heights and distances.

We have not space here to describe the instruments by which angles are practically measured, but we shall assume that they can be measured to almost any degree of accuracy.

B В

tan A.

a

= 30°.

=

200.

44. To find the height of an accessible object.

Let AC be the object, and let 1A any distance BC from its foot be measured.

At B let the angle of elevation ABC be observed.

Suppose BC = a, 4 ABC = 0.

Then, we have
AC

AC
tan ABC or

BC : AC = a tan e, the height required. Ex. Let a = 200, and o

1 200 V3. Then AC = 200 tan 30°

13

3 45. To find the height of an inaccessible object.

A At any point B in the

horizontal plane of the base let the angle of elevation ABC be observed.

Measure a convenient distance BD in the straight line CB produced, and observe

the angle of elevation ADC. Let BD A, LABC

0, ZADB φ. Then, Euc. I. 32, 2 BAD = 0 - . AB sin ADB AB

sin o Now, BD sin BAD

sin (0 - 0)' :: AB

(1). AC Again,

sin ABC AB

= sin e, .. AC AB sin o,

sin o sin or, from (1) AC = a

sin (0 - 0) Ex. Let BD 120, 0 =

45°.
Then, AC = 120. sin 60° sin 45°

sin (60° - 459)

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=

or

a

= a.

sin •
sin (0-0)

60°, 0

120.

B, 2 ADB

J31 sin 60° sin 45°

2 2 120

= 120. sin 15°

13 1

22 3

60 (3 + 3).

13 1 46. To find the height of an inaccessible object when it is not convenient to measure any distance in a line with the base of the object.

Let a distance BD be measured in any direction in the same horizontal plane as BC, and let the angles ABC, ABD, ADB be observed. Let BD = a, ABC = a LABD

=y Then, Euc. I., 32, BAD 108° (8 + x). AB sin ADB

sin

y
sin BAD sin {180° - (8 + x)}
AB
sin (8 + x)

sin (B + )
AC
Again, = sin ABC = sin :: AC = AB.sin
AB

sin or from (1), AC

y D

sin (8 + x) 47. To find the distance of an object by observation from the top of a tower whose height is known.

Let B be the object in the same BS horizontal plane with c the foot of the tower, and let the angle of depression DAB be observed.

Let AC = h, 2 DAB = 0.

Now, BD

or

'

sin y

sin ช.

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AB = 0.

•(1).

a

a,

a

sin a.

= a.

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