45. CASE I. To solve a right-angled triangle when two sides are given.

Let ABC be a right-angled triangle, of which A is the right angle, and suppose that any two sides are given;

then the third side may be found from the equation

whence C and B may be obtained.

Example. Given B=90°, a=20, b=40, solve the triangle.

The solution of a trigonometrical problem may often be obtained in more than one way. In the present case the triangle can be solved without making use of Euc. 1. 47.

Another solution may be given as follows:

46. CASE II. To solve a right-angled triangle when one side and one acute angle are given.

Let ABC be a right-angled triangle of which A is the right angle, and suppose one side b and one acute angle Care given; then

whence B, a, c may be determined.

Example 1. Given B=90°, A =30°, c=5, solve the triangle.
We have C=90° - A=90° - 30° 60°.

NOTE. The student should observe that in each case we write down a ratio which connects the side we are finding with that whose value is given, and a knowledge of the ratios of the given angle enables us to complete the solution.

Example 2. If C=90°, B=25° 43′, and c= 100, solve the triangle, having given tan 25° 43′ =·482 and cos 25° 43′ = .901.

.. b 90·1 x tan 25° 43′ = 90·1 x ·482=43.4282.

EXAMPLES. V. a.

Solve the triangles in which the following parts are given:

1. A=90°, a=4, b=2/3. 3. C-90°, b=12, a=4,/3. 5. a=20, c=20, B=90°. 7. b=c=2, A =90°.

9. C=90° ̊, a=9.√3, A=30°. 11. A=60°, c=8, C'=90°. 13. B=90°, C'=60°, b=100. 15. B=C=45°, c=4,

2. c=6, b=12, B=90°.

4.

a=60, b=30, A=90°. 6. a=5,√3, b=15, C=90°. x 8. 2c=b=6/3, B=90°. 10. A=90°, B=30°, a=4. 12. A 60°, C=30°, b=6. 14. A=30°, B=60°, b=20√3. X 16. 2B-C=60°, a=8.

17. If C=90°, cot A='07, b=49, find a. 18. If C-90°, A=38° 19′, c=50, find a; given sin 38° 19'='62.

19. If a=100, B=90°, C=40° 51′, find c; given tan 40° 51'='8647.

20. If b=20, A=90°, C=78° 12′, find a;

21. If B=90°, A =36°, c=100, solve the triangle; tan 36°=73, sec 36° = 1.24.

given

22. If A=90°, c=37, a=100, solve the triangle; sin 21° 43′ = 37, cos 21° 43'-'93.

given

23. If A 90°, B=39° 24′, b=25, solve the triangle; given cot 39° 24′ = 1.2174, cosec 39° 24′ = 1.5755. 24. If C=90°, a=225, b=272, solve the triangle; given tan 50° 24'-1.209.

47. It will be found that all the varieties of the solution of right-angled triangles which can arise are either included in the two cases of Arts. 45 and 46, or in some modification of them. Sometimes the solution of a problem may be obtained by solving two right-angled triangles. The two examples we give as illustrations will in various forms be frequently met with in subsequent chapters.

Example 1. In the triangle ABC, the angles A and B are equal to 30° and 135° respectively, and the side AB is 100 feet; find the length of the perpendicular from C upon AB produced.

Thus the distance required is 50 (√3+1) feet.

Example 2. In the triangle ABC, AD is drawn perpendicular to BC; solve the triangle, having given

AD=5, LABD=60°, LACD=45°.

In the right-angled triangle ABD,

1. ABC is a triangle, and BD is perpendicular to AC produced: find BD, given

A=30°, C=120°, AC=20.

2. If BD is perpendicular to the base AC of a triangle ABC, find a and c, given

A=30°, C-45°, BD=10.

3. In the triangle ABC, AD is drawn perpendicular to BC making BD equal to 15 ft.: find the lengths of AB, AC, and AD, given that B and C are equal to 30° and 60° respectively.

4. In a right-angled triangle PQR, find the segments of the hypotenuse PR made by the perpendicular from Q; given

QR=8, 4 QRP=60°, 4 QPR=30°.

5. If PQ is drawn perpendicular to the straight line QRS, find RS, given

PQ=36, RPQ=30°, 4 SPQ=60°.

6. If PQ is drawn perpendicular to the straight line QRS, find RS, given

PQ-20, PRS=135°, PSR=30°.

7. In the triangle ABC, the angles B and C are equal to 45° and 120° respectively; if a=40 find the length of the perpendicular from A on BC produced.

8. If CD is drawn perpendicular to the straight line DBA, find DC and BD, given

AB=59, 4 CBD=45°, ▲ CAB=32° 50′, cot 32° 50′ =1·59.