Selected quad for the lemma: end_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
end_n	arch_n	line_n	right_a	1,244	5	9.1480	5	false

Text snippets containing the quad

ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A52264	Institutio mathematica, or, A mathematical institution shewing the construction and use of the naturall and artificiall sines, tangents, and secants in decimal numbers, and also of the table of logarithms in the general solution of any triangle, whether plain or spherical, with their more particular application in astronomie, dialling, and navigation / by John Newton.	Newton, John, 1622-1678.	1654 (1654)	Wing N1061; ESTC R20441	128,229	417

contrary_o 25._o theor._n if_o the_o radius_fw-la of_o a_o circle_n be_v divide_v in_o extreme_a and_o mean_a proportion_n the_o great_a segment_n shall_v be_v the_o side_n of_o a_o decangle_n in_o the_o same_o circle_n these_o foundation_n be_v lay_v we_o will_v proceed_v to_o the_o make_n of_o the_o table_n whereby_o any_o triangle_n may_v be_v measure_v chap._n iii_o of_o trigonometria_fw-la or_o the_o measure_n of_o all_o triangle_n the_o dimension_n of_o triangle_n be_v perform_v by_o the_o golden_a rule_n of_o arithmetic_n which_o teach_v of_o four_o number_n proportional_a one_o to_o another_o any_o three_o of_o they_o be_v give_v to_o find_v out_o a_o four_o therefore_o for_o the_o measure_n of_o all_o triangle_n there_o must_v be_v certain_a proportion_n of_o all_o the_o part_n of_o a_o triangle_n one_o to_o another_o and_o these_o proportion_n must_v be_v explain_v in_o number_n 2._o and_o the_o proportion_n of_o all_o the_o part_n of_o a_o triangle_n one_o to_o another_o 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fraction_n still_o remain_v to_o it_o and_o therefore_o the_o table_n of_o sines_n tangent_n and_o secant_v can_v be_v exact_o make_v by_o any_o mean_n and_o yet_o such_o may_v and_o aught_o to_o be_v make_v wherein_o no_o number_n be_v different_a from_o the_o truth_n by_o a_o integer_fw-la of_o those_o part_n whereof_o the_o radius_fw-la be_v take_v as_o if_o the_o radius_fw-la be_v take_v of_o ten_o million_o no_o number_n of_o these_o table_n ought_v to_o be_v different_a from_o the_o truth_n by_o one_o of_o ten_o million_o that_o you_o may_v attain_v to_o this_o exactness_n either_o you_o must_v use_v the_o fraction_n or_o else_o take_v the_o radius_fw-la for_o the_o make_n of_o the_o table_n much_o great_a than_o the_o true_a radius_fw-la but_o to_o work_v with_o whole_a number_n and_o

make_v your_o plain_n declination_n not_o to_o exceed_v a_o quadrant_n or_o 90_o de_fw-la and_o as_o when_o it_o decline_v nothing_o it_o be_v a_o full_a south_n or_o north_n plain_a so_o if_o it_o decline_v just_a 90_o it_o be_v then_o a_o full_a east_n or_o west_n plain_a these_o precept_n be_v sufficient_a to_o find_v the_o declination_n of_o any_o plain_a howsoever_o situate_v but_o that_o there_o may_v be_v no_o mistake_n we_o will_v add_v a_o example_n 1_o example_n now_o because_o the_o line_n of_o shadow_n agnostus_n fall_v between_o p_o the_o pole_n of_o the_o plain_n horizontal_a line_n and_o s_o the_o south_n point_n therefore_o according_a to_o the_o former_a direction_n i_o add_v the_o horizontal_a distance_n pg_n 24_o deg_n to_o the_o sun_n azimuth_n g_v 40_o deg_n and_o their_o aggregate_v be_v ps_n 64_o deg_n the_o declination_n seek_v and_o in_o this_o case_n it_o be_v upon_o the_o same_o coast_n with_o the_o sun_n that_o be_v west_n according_a to_o the_o rule_n 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min._n and_o the_o angle_n at_o p_o be_v give_v one_o hour_n distance_n from_o the_o meridian_n who_o measure_n in_o the_o equinoctial_a be_v 15_o deg_n &_o the_o angle_n at_o n_o be_v always_o right_a that_o be_v 90_o deg_n wherefore_o by_o the_o first_o case_n of_o right_n angle_v spherical_a triangle_n the_o perpendicular_a n_o 11_o may_v thus_o be_v find_v as_o radius_fw-la 90_o 10.000000_o to_o the_o tangent_fw-la of_o np11_n 15d_o 9.428052_o so_o be_v the_o sine_fw-la of_o pn_n 51.53_o 9.893725_o  _fw-fr  _fw-fr to_o the_o tangent_fw-la of_o n11_n 11.85_o 9.321777_o which_o be_v the_o distance_n of_o the_o hour_n of_o 1_o and_o 11_o on_o each_o side_n of_o the_o meridian_n thus_o in_o all_o respect_n must_v you_o find_v the_o distance_n of_o 2_o and_o 10_o of_o clock_n by_o resolve_v the_o triangle_n np10_n and_o of_o 3_o and_o 9_o of_o clock_n by_o resolve_v the_o triangle_n np9_n and_o so_o of_o the_o rest_n in_o which_o as_o the_o angle_n at_o pincreaf_v which_o for_o 2_o 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the_o hour-line_n will_v do_v upon_o the_o plane_n itself_o and_o as_o it_o do_v appear_v by_o the_o figure_n set_v at_o the_o end_n of_o every_o hour_n line_n in_o the_o scheme_n now_o have_v already_o the_o pole_n elevation_n give_v as_o be_v in_o the_o horizontal_a there_o be_v nothing_o else_o to_o be_v do_v but_o to_o calculate_v the_o true_a hour-distance_n upon_o the_o line_n ezw_n from_o the_o meridian_n szn_n and_o then_o to_o proceed_v as_o former_o and_o note_n that_o because_o the_o hour_n equidistant_a on_o both_o side_n the_o meridian_n be_v equal_a upon_o the_o plane_n the_o one_o half_a be_v find_v the_o other_o be_v also_o have_v you_o may_v therefore_o begin_v with_o which_o side_n you_o will_n in_o the_o triangle_n zp11_n right_o angle_v at_o z_o i_o have_v zp_n give_v the_o compliment_n of_o the_o height_n of_o the_o pole_n 38_o deg_n 47_o min._n the_o which_o be_v also_o the_o height_n of_o the_o stile_n to_o this_o dial_n and_o the_o angle_n at_o p15_n degree_n one_o hour_n distance_n from_o the_o meridian_n upon_o the_o equator_fw-la to_o find_v the_o side_n z11_n for_o which_o by_o the_o first_o case_n of_o right_n angle_v spherical_a triangle_n the_o proportion_n be_v as_o before_z as_o the_o radius_fw-la 90_o 10.000000_o to_o the_o sine_fw-la of_o pz_n 38.47_o 9.793863_o so_o be_v the_o tangent_fw-la of_o zp11_n 15d_o 9.428052_o  _fw-fr  _fw-fr to_o the_o tangent_fw-la of_o z11_n 9.47_o 9.221915_o and_o thus_o in_o all_o respect_n must_v you_o find_v the_o distance_n of_o 2_o and_o 10_o of_o 3_o and_o 9_o and_o so_o forward_o as_o be_v direct_v for_o the_o hour_n in_o the_o horizontal_a plane_n the_o north_n plane_n be_v but_o the_o back_n side_n of_o the_o south_n lie_v in_o the_o same_o azimuth_n with_o it_o

fraction_n be_v in_o the_o calculation_n very_o tedious_a beside_o here_o no_o fraction_n almost_o be_v exquisite_o true_a therefore_o the_o radius_fw-la for_o the_o make_n of_o rhese_fw-mi table_n be_v to_o be_v take_v so_o much_o the_o more_o that_o there_o may_v be_v no_o error_n in_o so_o many_o of_o the_o figure_n towards_o the_o left_a hand_n as_o you_o will_v have_v place_v in_o the_o table_n and_o as_o for_o the_o number_n superfluous_a they_o be_v to_o be_v cut_v off_o from_o the_o right_a hand_n towards_o the_o left_a after_o the_o end_n of_o the_o supputation_n thus_o to_o find_v the_o number_n answer_v to_o each_o degree_n and_o minute_n of_o the_o quadrant_n to_o the_o radius_fw-la of_o 10000000_o or_o ten_o million_o i_o add_v eight_o cipher_n more_o and_o then_o my_o radius_fw-la do_v consist_v of_o sixteen_o place_n this_o do_v you_o must_v next_o find_v out_o the_o right_a sin_n of_o all_o the_o arch_n less_o than_o a_o quadant_n in_o the_o same_o part_n as_o the_o radius_fw-la be_v take_v of_o whatsoever_o bigness_n it_o be_v and_o from_o those_o right_a sin_n the_o tangent_n and_o secant_v must_v be_v find_v out_o 21._o the_o right_a sin_n in_o make_v of_o the_o table_n be_v either_o primary_n or_o secondary_a the_o primarie_a sin_n be_v those_o by_o which_o the_o rest_n be_v find_v and_o thus_o the_o radius_fw-la or_o whole_a sine_fw-la be_v the_o first_o primary_n sine_fw-la the_o which_o how_o great_a or_o little_o soever_o be_v equal_a to_o the_o side_n of_o a_o sixangled_n figure_n inscribe_v in_o a_o circle_n that_o be_v to_o the_o subtense_n of_o 60_o degree_n the_o which_o be_v thus_o demonstrate_v out_o of_o the_o radius_fw-la or_o subtense_n of_o 60_o degree_n the_o sine_fw-la of_o 30_o degree_n be_v easy_o find_v the_o half_a of_o the_o subtense_n be_v the_o measure_n of_o a_o angle_n at_o the_o circumference_n opposite_a thereunto_o by_o the_o 19_o of_o the_o second_o if_o therefore_o your_o radius_fw-la consist_v of_o 16_o place_n be_v 1000.0000.0000.0000_o the_o sine_fw-la of_o 30_o degree_n will_v be_v the_o one_o half_a thereof_o to_o wit_n 500.0000.0000.0000_o 22._o the_o other_o primary_n sin_n be_v the_o sin_n of_o 60_o 45_o 36_o and_o of_o 18_o degree_n be_v the_o half_a of_o the_o subtense_n of_o 120_o 90_o 72_o and_o of_o 36_o degree_n 23._o the_o subtense_n of_o 120_o degree_n be_v the_o side_n of_o a_o equilateral_a triangle_n inscribe_v in_o a_o circle_n and_o may_v thus_o be_v find_v the_o rule_n subtract_v the_o square_a of_o the_o subtense_n of_o 60_o degree_n from_o the_o square_n of_o the_o diameter_n the_o square_a root_n of_o what_o remain_v be_v the_o side_n of_o a_o equilateral_a triangle_n inscribe_v in_o a_o circle●_n or_o the_o subtense_n of_o 120_o degree_n the_o reason_n of_o the_o rule_n the_o subtense_n of_o a_o arch_n with_o the_o subtense_n of_o the_o compliment_n thereof_o to_o 180_o with_o the_o diameter_n make_v in_o the_o meeting_n of_o the_o two_o subtense_n a_o right_a angle_a triangle_n as_o the_o subtense_n ab_fw-la 60_o degree_n with_o the_o subtense_n ac_fw-la 120_o degree_n and_o the_o diameter_n cb_n make_v the_o right_a angle_a triangle_n abc_n right_o angle_v at_o a_o by_o the_o 19_o of_o the_o second_o and_o therefore_o the_o side_n include_v the_o right_a angle_n be_v equal_a in_o power_n to_o the_o three_o side_n by_o the_o 〈◊〉_d of_o the_o second_o therefore_o the_o square_a of_o ab_fw-la be_v take_v from_o the_o square_n of_o cb_n there_o remain_v the_o square_a of_o ac_fw-la who_o squar_fw-la root_n be_v the_o subtense_n of_o 〈◊〉_d degree_n or_o the_o side_n of_o a_o equilateral_a triangle_n inscribe_v in_o a_o circle_n example_n let_v the_o diameter_n cb_n be_v 2000.0000_o 0000.0000_o the_o square_a thereof_o be_v 400000._o 00000.00000.00000.00000.00000_o the_o subtense_n of_o ab_fw-la be_v 100000.00000.00000_o the_o square_a thereof_o be_v 100000.00000.00000_o 00000.00000.00000_o which_o be_v substract_v from_o the_o square_n of_o cb_n the_o remainder_n be_v 300000.00000.00000.00000.00000.00000_o who_o square_a root_n 173205.08075.68877_o the_o subtense_n of_o 120_o degree_n consectary_n hence_o it_o follow_v that_o the_o subtense_n of_o a_o arch_n less_o than_o a_o semicircle_n be_v give_v the_o subtense_n of_o the_o compliment_n of_o that_o arch_n to_o a_o semicircle_n be_v also_o give_v 24._o the_o subtense_n of_o 90_o degree_n be_v the_o side_n of_o a_o square_n inscribe_v in_o a_o circle_n and_o may_v thus_o be_v find_v the_o rule_n multiply_v the_o diameter_n in_o itself_o and_o the_o square_a root_n of_o half_a the_o product_n be_v the_o subtense_n of_o 90_o degree_n or_o the_o side_n of_o a_o square_n inscribe_v in_o a_o circle_n the_o reason_n of_o this_o rule_n the_o diagonal_a line_n of_o a_o square_n inscribe_v in_o a_o circle_n be_v two_o diameter_n and_o the_o right_a angle_a figure_n make_v of_o the_o diagonal_o be_v equal_a to_o the_o right_a angle_a figure_n make_v of_o the_o opposite_a side_n by_o the_o 20_o the_o of_o the_o second_o now_o because_o the_o diagonal_a line_n ab_fw-la and_o cd_o be_v equal_a it_o be_v all_o one_o whether_o i_o multiply_v ac_fw-la by_o itself_o or_o by_o the_o other_o diagonal_a cd_o the_o product_n will_v be_v still_o the_o same_o then_o because_o the_o side_n ab_fw-la ac_fw-la and_o bc_n do_v make_v a_o right_a angle_a triangle_n right_o angle_v at_o c_o by_o the_o 〈◊〉_d of_o the_o second_o &_o that_o the_o 〈◊〉_d ac_fw-la and_o ●b_n be_v equal_a by_o the_o work_n the_o half_a of_o the_o square_n of_o ab_fw-la must_v needs_o be_v the_o square_n of_o ac_fw-la or_o cb_n by_o the_o 17_o the_o of_o the_o second_o who_o square_a root_n the_o subtense_n of_o cb_n the_o side_n of_o a_o square_a or_o 90_o degree_n example_n let_v the_o diameter_n ab_fw-la be_v 200000.00000_o 00000_o the_o square_a thereof_o be_v 400000.00000_o 00000.00000.00000.00000_o the_o half_a whereof_o be_v 200000.00000.00000.00000.00000_o 00000._o who_o square_a root_n 14142●_n 356●3_n 73095._o be_v the_o subtense_n of_o 90_o degree_n or_o the_o side_n of_o a_o square_n inscribe_v in_o a_o circle_n 25._o the_o subtense_n of_o 36_o degree_n be_v the_o side_n of_o a_o decangle_n and_o may_v thus_o be_v find_v the_o rule_n divide_v the_o radius_fw-la by_o two_o then_o multiply_v the_o radius_fw-la by_o itself_o and_o the_o half_a thereof_o by_o itself_o and_o from_o the_o square_a root_n of_o the_o sum_n of_o these_o two_o product_n subtract_v the_o half_a of_o radius_fw-la what_o remain_v be_v the_o side_n of_o a_o decangle_n or_o the_o subtense_n of_o 36_o degree_n the_o reason_n of_o the_o rule_n for_o example_n let_v the_o radius_fw-la ebb_n be_v 100000.00000.00000_o then_o be_v bh_n or_o the_o half_a thereof_o 500000._o 00000.00000_o the_o square_a of_o ebb_n be_v 100000_o 00000.00000.00000.00000.00000_o and_o the_o square_a of_o bh_n 250000.00000.00000.00000_o 00000.00000.00000_o the_o sum_n of_o these_o two_o square_n viz_o 125000.00000.00000_o 00000_o 00000._o 00000_o be_v the_o square_a of_o he_o or_o hk_n who_o square_a root_n be_v 1118033●_n 887●9895_n from_o which_o deduct_v the_o half_a radius_fw-la bh_n 500000000000000_o and_o there_o remain_v 618033988749895_o the_o right_a line_n kb_n which_o be_v the_o side_n of_o a_o decangle_n or_o the_o subtense_n of_o 36_o degree_n 26_o the_o subtense_n of_o 72_o degree_n be_v the_o side_n of_o a_o pentagon_n inscribe_v in_o a_o circle_n and_o may_v thus_o be_v sound_a the_o rule_n subtract_v the_o side_n of_o a_o decangle_n from_o the_o diameter_n the_o remainder_n multiply_v by_o the_o radius_fw-la shall_v be_v the_o square_n of_o one_o side_n of_o a_o pentagon_n who_o square_a root_n shall_v be_v the_o side_n itself_o or_o subtense_n of_o 72_o degree_n the_o reason_n of_o the_o rule_n in_o the_o follow_a diagram_n let_v ac_fw-la be_v the_o side_n of_o a_o decangle_n equal_a to_o cx_o in_o the_o diameter_n and_o let_v the_o rest_n of_o the_o semicircle_n be_v bisect_v in_o the_o point_n e_o then_o shall_v either_o of_o the_o right_a line_n ae_n or_o ebb_n represent_v the_o side_n of_o a_o equilateral_a pentagon_n for_o ac_fw-la the_o side_n of_o a_o decangle_n subtend_v a_o arch_n of_o 36_o degree_n the_o ten_o part_n of_o a_o circle_n and_o therefore_o aeb_fw-mi the_o remain_a arch_n of_o a_o semicircle_n be_v 144_o degree_n the_o half_a whereof_o ae_n or_o ebb_n be_v 72_o degree_n the_o five_o part_n of_o a_o circle_n or_o side_n of_o a_o equilateral_a pentagon_n the_o square_a whereof_o be_v equal_a to_o the_o oblong_v make_v of_o db_n and_o bx_n demonstration_n draw_v the_o right_a line_n exit_fw-la ed_z and_o aec_fw-la then_o will_v the_o side_n of_o the_o angle_n ace_n and_o ecx_n be_v equal_a because_o cx_o be_v make_v equal_a to_o ac_fw-la and_o aec_fw-la common_a to_o both_o and_o the_o angle_n themselves_o be_v equal_a because_o they_o be_v in_o equal_a segment_n