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The construction of the wonderful canon of logarithms

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66 Trigonometrical Propositions.<br />

produced ; its arc C D subtract from, or add to, <strong>the</strong><br />

given side B D, and you have B C. <strong>The</strong>n multiply<br />

<strong>the</strong> sine <strong>of</strong> <strong>the</strong> complement <strong>of</strong> A D by <strong>the</strong> sine <strong>of</strong><br />

<strong>the</strong> complement <strong>of</strong> B C ; divide <strong>the</strong> product by <strong>the</strong><br />

sine <strong>of</strong> <strong>the</strong> complement <strong>of</strong> C D, and <strong>the</strong> sine <strong>of</strong> <strong>the</strong><br />

complement <strong>of</strong> A B will be produced; hence you<br />

have A B itself.<br />

Given A D, & <strong>the</strong> angle D with <strong>the</strong> side B D, to<br />

find <strong>the</strong> angle A.<br />

This follows from <strong>the</strong> above, but <strong>the</strong> problem<br />

would require <strong>the</strong> " Rule <strong>of</strong> Three " to be applied<br />

thrice. <strong>The</strong>refore substitute A for B and B for A,<br />

and <strong>the</strong> problem will be as follows :<br />

Given B D c2f D, with <strong>the</strong> side A D, to find <strong>the</strong><br />

angle B.<br />

This is exactly <strong>the</strong> same as <strong>the</strong> sixth problem, and<br />

is solved by <strong>the</strong> " Rule <strong>of</strong> Three " being applied<br />

twice only.<br />

8. Given A D, cSr" <strong>the</strong> angle D with <strong>the</strong> side A B, to find<br />

tlie angle B.<br />

Multiply <strong>the</strong> sine <strong>of</strong> A D by <strong>the</strong> sine <strong>of</strong> D ; divide<br />

<strong>the</strong> product by <strong>the</strong> sine <strong>of</strong> A B, and <strong>the</strong> sine <strong>of</strong> <strong>the</strong><br />

angle B will be produced.<br />

9. Given K Vi, & <strong>the</strong> angle D with <strong>the</strong> side A B, to find<br />

<strong>the</strong> side B D.<br />

Multiply radius by <strong>the</strong> sine <strong>of</strong> <strong>the</strong> complement <strong>of</strong><br />

D, divide <strong>the</strong> product by <strong>the</strong> tangent <strong>of</strong> <strong>the</strong> complement<br />

<strong>of</strong> A D, and <strong>the</strong> tangent <strong>of</strong> <strong>the</strong> arc C D will be<br />

produced. <strong>The</strong>n multiply <strong>the</strong> sine <strong>of</strong> <strong>the</strong> complement<br />

<strong>of</strong> C D by <strong>the</strong> sine <strong>of</strong> <strong>the</strong> complement <strong>of</strong> A B,<br />

divide <strong>the</strong> product by <strong>the</strong> sine <strong>of</strong> <strong>the</strong> complement <strong>of</strong><br />

A D, and you have <strong>the</strong> sine <strong>of</strong> <strong>the</strong> complement <strong>of</strong><br />

B C. Whence <strong>the</strong> sum or <strong>the</strong> difference <strong>of</strong> <strong>the</strong> arcs<br />

B C and C D will be <strong>the</strong> required side B D.<br />

10. Given

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