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(2) – (3), then sin B – sin C = 2 cos B & C sin B;

and (2) + (3), then sin B + sin C = 2 sin —

B + C. sin B - sin C 2 cos2 sin

** sin B + sin C

2 sin B + C cos B-C

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Ex. VI. 1. If a = 5, C = 30°, sin A = ž, find c. 2. Find a, having given b = 12, c = 15, A = 60°. 3. Find tan , when a = 6,6 = 7,0 = 8. 14. What is the area of a triangle whose sides are 48, 52, 20?

5. Given two sides of a triangle to be 18 and 24, and the included angle 45°, find the area.

a

6. Two of the sides of a triangle are as 2 : 1, and the included angle is 60°, find the other angles. 7. In any triangle show thatb cos C - c cos B = Na? - 4 bc cos B cos C.

, sin B sin C 8. Show that the area of a triangle = 1, ap.

sin A 9. An object is observed from two stations 100 yards apart, and the angles subtended by the distance between the object and either station are 45° and 60° respectively. Find the distance of the object from each station.

10. An observation is made from a point known to be distant 120 and 230 yards respectively from two trees, and the angle which the trees subtend is found to be 120°. Find the distance between the trees. 11. If a sina 0 + b coso = m, )

. 1 1 1 1 b sino o + a cosa d = n, ļ then = + = = to a tan 0 = b tan d.

a b m n

) 12. If a', 6!, c' be the sides of the triangle formed by joining the feet of the perpendiculars from the angles A, B, C of the triangle ABC upon the opposite sides, then

al , b , c a + 6+ C2

az * 72 * 2 * 2 abc 13. A perpendicular AD is drawn from the angle A of a triangle, meeting the opposite side BC and D; and from Da perpendicular is drawn to AC, meeting it in E. Show that DE = b sin C cos C. 14. Show that the length of AD in the last examplebc sin A + ac sin B + ab sin C

3 a 15. Show that w 8 (s – a) (s ~ 6) (8 - 0) = 1 ab, when the triangle is right-angled at C. 16. Show that (a + b + c) sin A sin B

= (sin A + sin B + sin C)” ab. 17. Show that in any triangle

cos? A + cos' B + cos O + 2 cos A cos B cos C = 1.

18. The sides of a triangle ABC are in arithmetical progression ; show that its area = bay J2 a – b) (36 – 2a).

CHAPTER VIII.

SOLUTION OF RIGHT-ANGLED TRIANGLES.

37. A triangle can always be determined when any three elements, with the exception of the three angles, are given. In the latter case we have only the same data as when two angles are given, for the third can always be found by subtracting the sum of the other two from two right angles (Euc. I., 32).

Hence a right-angled triangle can always be determined when any two elements, other than the two acute angles, are given besides the right angle. And when one of the acute angles is given, the other may be obtained by subtracting it from a right angle.

We have the following cases :

CASE 1. When the two sides containing the right angle are given.

A We shall take C as the right angle | in every case.

Now tan A = ; or, L tan Å - 10 = log a = log b; or, L tan A = 10 + log a – log 6... (1):

This determines A, and we then have L a _ B = 90° - A...........................(2).

Also, o = sin A, or log a – log c = 'i sin A - 10. .. log c = 10 + log a – Lsin A ...........................(3): Hence the three elements, A, B, C; are determined.

a

CASE 2. When the hypothenuse and a side are given.
Let a be the given side.
We have sin A = or L sin A - 10 = log a – log c.

.. L sin A = 10 + log a – log C................... (1),

and B = 90° - A ...............................(2),
also 62 = co - a* = (c + a) (c – a);

.. log b = } {log (c + a) + log (c a)} .....(3)
CASE 3. When an acute angle and a side are given.
Let A, a be the given angle and side.
Then B = 90° - A............................................. (1
also -- = tan B, or log b = L tan B - 10 + log a.....(2),

and = sin A, or log a – log c = L sin A = 10, or log c = 10 + log rt - I sin A. CASE 4. When the hypothenuse and an acute angle are given. Let A be the given acute angle. We have B = 90° – A......................................(1). Also“ = sin A, or log a = log c + L sin A - 10... (2).

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And. = cos A, or log b = log c + L cos Å - 10.... (3).

It is evident, from Art. 30, that when the angles only of a triangle are known, we can determine the ratio only of the three sides of the triangle to each other:

Ex. 1. Given A = 23° 41', a = 35, solve the triangle.
This is an example of Case 3:
We have B = 90° - 23° 41' = 66° 19'.
Again, log b = L tan B - 10 - log a

== L tan 66° 19: – 10 – log 35
= 10:3579092 - 10 + 1.5440680

= 1.9019772 = log 79.795
.. b = 79.795.

Also log c = 10 + log a – L sin A

= 10 + log 35 - L sin 23° 411
= 10 + 1.5440680 - 9.6038817

= 1.9401863 = log 87.134 ...

..C = 87•134.
· Ex. 2. Given a = 214, 6 = 317, solve the triangle.

This falls under Case 1..
We have L tan A = 10 + log a – log b.

= 10 + log 214 – log 317
= 10 + 2.3304138 - log 2.5010593

= 9.8293545,
Next lower in tables is 9.8292599 = L tan 34° 1';
..d =

946. Also, by tables, D = 2724,

And

946 x 60":- 21" nearly. * 60" = "*2724

....Ltan A = L tan 34° 1' 21",

or A = 34° 1' 21". Hence B = 90 - 34° 1' 21'' = 45° 58' 39". And similarly may c be determined.

Ex. VII. 1. Given a = 32, A = 63° 45', find b.

Log 32 = 1:5051500, L cot 63° 45' = 9.6929750,

Log 15780 = 4:1981070, log 15781 = 4.1901345. 2. Given c = 151, A = 37° 42', find a.

Log 151 = 2.1789769, L sin 37° 421 = 9.7864157,

Log 92340 = 4.9653899, tab. diff. = 47. 3. Given a = 60, c = 65, find b, A.

Log 2 = •3010300, log 3 = .4771213,
Log 65 = 1.8129134, L sin 67° 22! = 9.9651953,

L sin 67° 23! = 9.9652480.
4. Given a = 73, 6 = 84, find A, C.

Log 73 = 1.8633229, L tan 40° 59; = 9.9389079,
Log 84 = 1.9242793, L tan 41° = 9.9391631,
L sin 40° 59; = 9.8167975, L sin 41° = 9.8169429,
Log 111.288 = 2:0464479,

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