 | Charles Hutton - 1811 - 404 páginas
...c, the measure of ACE, supposing there to be • known АРБ = p, UPC = p, CP =: d, BC = L, AC = R. Since the exterior angle of a triangle is equal to the sum of the two interior opposite angles (th. 16 Geom.), we have, with respect to the triangle IAP, AIB = p -f- IAP ; and with... | |
 | Charles Hutton - 1812 - 624 páginas
...to find c, the measure of ACB, supposing there to be known APB = r, BPC =fi, CP = d, BC = L, AC = R. Since the exterior angle of a triangle is equal to the sum of the two interior opposite angles (th. 1 6 Geom.), we have, •with respect to the triangle IAP, AIB = P+IAP; and with... | |
 | John Gummere - 1814 - 398 páginas
...took the angle of elevation BAG = 26° 30'. Required the height of the tree. Calculation. 1. Because the exterior angle of a triangle is equal to the sum of the two interior and opposite ones, the angle BDC = DAC + ACD ; therefore ACD - BDC — DAG = 25° : now in the triangle ADC we have... | |
 | Olinthus Gregory - 1816 - 276 páginas
...staof the centre of 'r the sutnmit of ;Ie APB : it is re* the measure of here to be known p, ce = d,BQ Since the exterior angle of a triangle is equal to the sum of the two interior opposite angles, we have, with respect to the triangle IAP, AIB = P + IAP; and with regard to the triangle... | |
 | John Farrar - 1822 - 270 páginas
...another distance CB = 60 feet in the same direction, and take the angle ABD = 23° 45'. Now, sinre the exterior angle of a triangle is equal to the sum of the interior opposite angles, we have that is 4l° — 23° 45' =17° 15'. Hence in the triangle BAC.... | |
 | John Farrar - 1833 - 274 páginas
...measure another distance CB = 60 feet in the same direction, and take the angle ABD = 23° 45'. Now, since the exterior angle of a triangle is equal to the sum of the interior opposite angles, we have ACD — A&b = BAC. that is 41° — 23° 45' = 17° 15'. Hence... | |
 | John Farrar - 1833 - 276 páginas
...measure another distance CB = 60 feet in the same direction, and take the angle ABD = 23° 45'. Now, since the exterior angle of a triangle is equal to the sum of the interior opposite angles, we have ACD — ABC = EAC. that is 41° — 23° 45' = 17° 15'. Hence... | |
 | 1836 - 502 páginas
...example, suppose it known that the angles at the base of an isosceles triangle are equal, and also that the exterior angle of a triangle is equal to the sum of the interior and opposite angles. Suppose it also indisputable, that if A and B be respectively double... | |
 | Schoolmaster - 1836 - 926 páginas
...example, suppose it known that the angles at the base ofan isosceles triangle are equal, and also that the exterior angle of a triangle is equal to the sum of the interior and opposite angles. Suppose it also indisputable, that if A and B he respectively double... | |
 | Scottish school-book assoc - 1845 - 444 páginas
...required to find C, the measure of ACB, supposing there to be known APB=P, BPC= p, CP=d, BC=L, AC=R. Since the exterior angle of a triangle is equal to the sum of the two interior opposite angles, (Geo. prop. ] 9), we have, with respect to the triangle IAP, ZAIB=ZP+UAP ; and with... | |
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To prove that the exterior angle of a triangle is equal to the sum of the two interior opposite angles (see fig. 
