Elements of Geometry And Trigonometry From the Works of Am Legendre Adapted
Elements of Geometry And Trigonometry From the Works of Am Legendre Adapted
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In our eReader you can find the full English version of the book. Read Elements of Geometry And Trigonometry From the Works of Am Legendre Adapted Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Elements of Geometry And Trigonometry From the Works of Am Legendre Adapted
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), and the angles CBP and CEP will be equal to the angles A and B respec- tively. Draw BL and PQ, the one perpendicular, and the other parallel to OB. "We then have, OE = cos a, DC =. Sin 5, We have from the figure, OE = OL + QP OD cos h.
(1. ) In the light-angled triangle OLD, OL = OD cos DOL = cos b cos c.
The right-angled triangle PQD has its sides respectively perpendicular to those of ODD ; it is, therefore, similar to it, and the angle QDP is equal to c, and we have, QP = PD sin QDP = PD ...sin c • • rhe right-angled triangle CPD gives, PD = CD cos CDP = sin 5 cos ^ ; substituting this value in ( 2 ), we have, QP = sin 5 sin c cos A ; (2. ) TRIGONOMETRY. 87 and now substituting these values of OE^ OL, and QP^ in ( 1 ), we have, cos a = cos h cos c + sm J sin c cos A • (3. ) Id the same way, we may deduce, cos b =. Cos a cos c + sin a sin c cos jB • • (4. ) cos c = cos a cos J + sin a sin h cos C • • (5. ) That is, the cosine of either side of a spherical triangle is equal to the rectangle of the cosines of the other two sides plus the rectangle of the sines of these sides into the cosine of their included angle.
What to read after Elements of Geometry And Trigonometry From the Works of Am Legendre Adapted?
You can find similar books in the "Read Also" column, or choose other free books by Charles Davies to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
), and the angles CBP and CEP will be equal to the angles A and B respec- tively. Draw BL and PQ, the one perpendicular, and the other parallel to OB. "We then have, OE = cos a, DC =. Sin 5, We have from the figure, OE = OL + QP OD cos h.
(1. ) In the light-angled triangle OLD, OL = OD cos DOL = cos b cos c.
The right-angled triangle PQD has its sides respectively perpendicular to those of ODD ; it is, therefore, similar to it, and the angle QDP is equal to c, and we have, QP = PD sin QDP = PD ...sin c • • rhe right-angled triangle CPD gives, PD = CD cos CDP = sin 5 cos ^ ; substituting this value in ( 2 ), we have, QP = sin 5 sin c cos A ; (2. ) TRIGONOMETRY. 87 and now substituting these values of OE^ OL, and QP^ in ( 1 ), we have, cos a = cos h cos c + sm J sin c cos A • (3. ) Id the same way, we may deduce, cos b =. Cos a cos c + sin a sin c cos jB • • (4. ) cos c = cos a cos J + sin a sin h cos C • • (5. ) That is, the cosine of either side of a spherical triangle is equal to the rectangle of the cosines of the other two sides plus the rectangle of the sines of these sides into the cosine of their included angle.
What to read after Elements of Geometry And Trigonometry From the Works of Am Legendre Adapted?
You can find similar books in the "Read Also" column, or choose other free books by Charles Davies to read onlineMoreLess
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