In solving the following Examples the Tables will be required:

21. From a window it is observed that the angle of elevation of the top of a house on the opposite side of the street is 29o, and the angle of depression of the bottom of the house is 56°; determine the height of the house, supposing the breadth of the street to be 80 feet.

22. The elevations of two mountains in the same straight line with an observer are 9° 30′ and 18° 20′: on approaching four miles nearer they have both an elevation of 37o. Find the heights of the mountains in yards.

23. P and Q are two inaccessible objects; a straight line AB in the same plane as P and Q is measured and found to be 280 yards; the angle PAB is 95o, the angle QAB is 47° 30', the angle QBA is 110°, and the angle PВA is 52° 20'. Determine the length of PQ.

24. A, B, C are three objects at known distances apart; namely AB=1056 yards, AC=924 yards, and CB=1716 yards. An observer places himself at a station S from which C appears directly in front of A, and observes the angle CSB to be 14° 24'. Find the distance CS.

25. A, B, C are three objects at known distances apart; namely AB=320 yards, AC-600 yards, BC=435 yards. From a station Š it is observed that ASB=15o, and BSC 30°. Find the distances of S from A, B, and C; the point B being nearest to S, and the angle ASC being the sum of ASB and BSC.

XIII. Geometrical Solutions.

123. The present Chapter will consist of Geometrical Solutions of some Trigonometrical Problems.

124. To construct an angle with a given sine or cosine.

Suppose we require an angle the sine of which is a given quantity a.

Describe a circle with unity for its diameter, and draw any diameter AB of this circle. With centre B, and radius equal to a, describe a circle; let C be one of the points where the circumferences of the two circles intersect: join AC and BC.

Then ACB is a right angle, by Euclid m. 31; and

therefore the sine of BAC is

BC

AB'

BAC is such an angle as is required.

that is a. Therefore

If we require an angle the cosine of which is a given quantity a, then the same construction may be made; and ABC will be such an angle as is required.

125. If an angle is required to have a given cosecant, then since the cosecant is the reciprocal of the sine the angle must have a known sine, and therefore may be found by the preceding Article.

Similarly, if an angle is required to have a given secant or a given versed sine, the angle must have a known cosine, and therefore may be found by the preceding Article.

126. To construct an angle with a given tangent or cotangent.

Suppose we require an angle the tangent of which is a given quantity a.

If the tangent of an angle is a the cosine is

a

and the sine is

√(a2+1)

1

√(a2 + 1) '

Therefore since these quanti

ties are known, we can construct the angle by Art. 124. Or we may proceed independently thus:

C

B

Take a straight line AB the length of which is unity; draw BC at right angles to AB and equal to a, and join CA,

Then the tangent of BAC is

BC

AB'

that is a. Therefore

BAC is such an angle as is required.

If we require an angle the cotangent of which is a given quantity a, then the same construction may be made; and ACB will be such an angle as is required.

127.

To divide a given angle into two parts which shall have their sines in a given ratio.

Let BAC be the given angle, and let it be required to divide it into two parts, such that the sine of one part may be to the sine of the other as m is to n.

Draw a straight line KP parallel to AB, and at a distance m from it; draw a straight line LP parallel to

AC, and at a distance r from it. Let P be the point of intersection of these straight lines; join AP. Then AP shall divide the angle BAC in the required manner.

For draw PM perpendicular to AB, and PN perpendicular to AC. Then

128.

To divide a given angle into two parts which shall have their cosines in a given ratio.

Let BAC be the given angle, and let it be required to divide it into two parts, such that the cosine of one part may be to the cosine of the other as m is to n.

On AB take AM=m, and on AC take AN=n; draw MP at right angles to AB, and NP at right angles to AC. Let P be the point of intersection of these straight lines; join AP. Then AP shall divide the angle BAC in the required manner.

If the point P does not fall within the angle BAC, we conclude that the angle cannot be divided into two parts in the proposed manner.

129. To divide a given angle into two parts which shall have their tangents in a given ratio.