document.write( "Question 1035792: I have to solve a triangle. Given 2 sides and 1 angle from it. I gotta determine if it's based on 1 triangle, 2 triangles, or no triangles.
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Algebra.Com's Answer #650448 by Edwin McCravy(20081) $\"\"$ $\"About$

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I have to solve a triangle. Given 2 sides and 1 angle from
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document.write( "To get a better understanding, let's sketch the geometry of the \r\n" );
document.write( "situation. I start by drawing an arbitrarily long line labeling \r\n" );
document.write( "the endpoint as the vertex of the given angle, in this case B.\r\n" );
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document.write( "At point B we draw the angle B = 10° by drawing side \r\n" );
document.write( "BC = a = 19:\r\n" );
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document.write( "Next we take a compass and open it to the length\r\n" );
document.write( "b = CA = 10.  Put the sharp point on C and swing \r\n" );
document.write( "an arc that cuts the horizontal line that we drew\r\n" );
document.write( "first in two places:\r\n" );
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document.write( "So it appears in this case there are two choices for point A.  \r\n" );
document.write( "Either this way with the green line:\r\n" );
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document.write( "or this way with the blue line:\r\n" );
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document.write( "Now let's do the calculating using the law of sines:\r\n" );
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document.write( "substitute a=19, b=10, B=10°\r\n" );
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document.write( "We use only the first and second parts:\r\n" );
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document.write( "Cross-multiply\r\n" );
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document.write( "Divide both sides by 10\r\n" );
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document.write( "Do the calculation:\r\n" );
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document.write( "[If this had come out greater than 1 there would have \r\n" );
document.write( "been 0 solutions.  If this had come out equal to 1 \r\n" );
document.write( "there would have been 1 solution, and it would have \r\n" );
document.write( "been a right triangle with hypotenuse b.]\r\n" );
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document.write( "Since it came out less than 1, we can tell there is\r\n" );
document.write( "either 1 or 2 solutions.\r\n" );
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document.write( "First solution:\r\n" );
document.write( "Next we calculate the first quadrant (acute) solution\r\n" );
document.write( "for angle A using inverse sine feature on the calculator\r\n" );
document.write( "and get angle A = 19.26462015°.  Then we can calculate the\r\n" );
document.write( "third angle C by adding 19.26462015° and 10° and subtracting\r\n" );
document.write( "from 180°, getting angle C = 150.7353799°.\r\n" );
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document.write( "Now we use the law of sines again:\r\n" );
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document.write( "This time we use only the second and third parts,\r\n" );
document.write( "and substitute 150.7353799° for angle C \r\n" );
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document.write( "Cross-multiply:\r\n" );
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document.write( "Divide both sides by sin(10°)\r\n" );
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document.write( "Next we see if there is a second solution.\r\n" );
document.write( "We calculate the second quadrant (obtuse) solution\r\n" );
document.write( "for angle A by finding the complement of 19.26462015°,\r\n" );
document.write( "by subtracting it from 180°, getting 160.7353798°.\r\n" );
document.write( "Then we can calculate the third angle C by adding \r\n" );
document.write( "160.7353798° and 10° and subtracting\r\n" );
document.write( "from 180°, getting angle C = 9.2646202°.\r\n" );
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document.write( "[If this had come out negative there would have \r\n" );
document.write( "been only the 1 solution.]\r\n" );
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document.write( "So we have a second solution, and we use the law of\r\n" );
document.write( "sines to find c just as we did to find the first\r\n" );
document.write( "solution:\r\n" );
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document.write( "Cross-multiply:\r\n" );
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document.write( "Divide both sides by sin(10°)\r\n" );
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document.write( "Edwin

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