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| 1. |
2 circles touch externally at P and AB is a common tangent to the circles . Find angle APB |
| Answer» Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.To find :\xa0{tex}\\angle APB{/tex}Solution : {tex}let\\ \\angle CAP = \\alpha \\ and\\ \\angle CBP = \\beta{/tex}CA = CP [lengths of the tangents from an external point C]In\xa0{tex} \\triangle PAC, \\angle CAP = \\angle APC = \\alpha{/tex}{tex}Similarly \\ CB = CP \\ and \\ \\angle CPB = \\angle PBC = \\beta {/tex}Now in the triangle APB,{tex}\\angle PAB + \\angle PBA + \\angle APB = 180° {/tex} [sum of the interior angles in a triangle]{tex}\\alpha + \\beta + (\\alpha + \\beta ) = 180°{/tex}{tex}=> 2\\alpha + 2\\beta = 180°{/tex}{tex}=> \\alpha + \\beta = 90°{/tex}{tex}\\therefore \\angle APB = \\alpha + \\beta = 90°{/tex} | |