 | Charles Hutton - 1811 - 424 pages
...instrument, or the summit of the observed angle APB : it is required to find c, the measure of ACS, supposing there to be known APB = p, BPC = p, CP ==...we have, with respect to the triangle IAP, AIB = P + IAF-, and with regard to the triangle BIC, AIB = c + CBP. Making these two values of AIB. equal,... | |
 | Charles Hutton - 1811 - 406 pages
...: -it is required to find c, the measure of ACE, supposing there to be • known АРБ = p, UPC = p, CP =: d, BC = L, AC = R. Since the exterior angle...we have, with respect to the triangle IAP, AIB = p -f- IAP ; and with regard to the triangle BIC, AIB = с + СВР. Making thest two values of AIB equal,... | |
 | John Gummere - 1814 - 400 pages
...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... | |
 | Olinthus Gregory - 1816 - 276 pages
...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 BIC, AIB... | |
 | John Farrar - 1822 - 270 pages
...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 - 278 pages
...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 pages
...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 pages
...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 pages
...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 pages
...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 regard to the triangle... | |
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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. 
