Lesson An upper level Geometry problem on special (15°,30°,135°)-triangle
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<H2>An upper level Geometry problem on special (15°,30°,135°)-triangle</H2> <H3>Problem 1</H3>ABC is a triangle with ∠CAB=15° and ∠ABC=30°. If M is the midpoint of AB, find ∠ACM. <B>Solution</B> <pre> The given part is shown in Figure 1 below. {{{drawing(400,1200/11,-.1,2.1,-.1,.5, locate(.3,.13,15^o),locate(1.7,.13,30^o), locate(0,0,A),locate(2,0,B), locate(1.35,.45,C), locate(1,0,M), locate(.5,0,1), locate(1.5,0,1), locate(1.15,.3,theta), green(line(1.366025404,.366025404,1,0)), locate(.6,.25,b), locate(1.7,.25,a), triangle(0,0,2,0,1.366025404,.366025404) )}}} Figure 1. Triangle ABC has the angles A= 15°, B= 30° and C= 180°-15°-30°= 135°. We want to find angle {{{theta}}}. Without loss of generality we can let AM = MB = 1 unit. We need this well known expression for sine 15° sin(15°) = sin(45°-30°) = sin(45°)cos(30°)-cos(45°)sin(30°)}}} = {{{(sqrt(2)/2)(sqrt(3)/2)-(sqrt(2)/2)(1/2)}}} = {{{sqrt(6)/4-sqrt(2)/4}}} = {{{(sqrt(6)-sqrt(2))/4}}}. (1) By the law of sines on triangle ABC. {{{AB/sin(ACB)}}} = {{{AC/sin(B)}}}, {{{AB/sin(135^o)}}} = {{{a/sin(15^o)}}}, {{{2/((1/sqrt(2)))}}} = {{{b/((1/2))}}}, {{{2*sqrt(2)}}} = {{{2b}}}, b = {{{sqrt(2)}}}. Also, {{{b/sin(30^o)}}} = {{{a/sin(15^o)}}}, {{{sqrt(2)/((1/2))}}} = {{{a/(((sqrt(6)-sqrt(2))/4))}}}, {{{2*sqrt(2)}}} = {{{(4a)/(sqrt(6)-sqrt(2))}}}, {{{sqrt(2)}}} = {{{(2a)/(sqrt(6)-sqrt(2))}}}, {{{sqrt(2)*(sqrt(6)-sqrt(2))}}} = {{{2a}}}, {{{sqrt(12)-2}}} = {{{2a}}}, {{{2sqrt(3)-2}}} = {{{2a}}}, a = {{{sqrt(3)-1}}}. In what follows, we will use (2) and will not use (3). So, (3) is derived here for completeness, only. Draw perpendicular CH from vertex C to the base AB (Figure 2). {{{drawing(400,1200/11,-.1,2.1,-.1,.5, locate(.3,.13,15^o),locate(1.7,.13,30^o), locate(0,0,A),locate(2,0,B), locate(1.35,.45,C), locate(1,0,M), locate(.5,0,1), locate(1.5,0,1), locate(1.15,.3,theta), green(line(1.366025404,.366025404,1,0)), locate(.6,.25,b), locate(1.7,.25,a), triangle(0,0,2,0,1.366025404,.366025404), green(line(1.366025404,.366025404,1.366025404,0)), locate(1.35,0,H) )}}} Figure 2. First part of the solution is the same as that of the Edwin solution, and it shows that a = {{{sqrt(3)-1}}}, b = {{{sqrt(2)}}}. The rest of the solution is different (very simple and totally geometrical). Triangle BHC is a right-angled triangle with the acute angle B of 30°. Hence, the opposite leg CH is half of the hypotenuse BC CH = {{{a/2}}} = {{{(sqrt(3)-1)/2}}}, (4) while its other leg BH is {{{sqrt(3)/2}}} times the hypotenuse BC BH = {{{(sqrt(3)/2)*(sqrt(3)-1)}}} = {{{(3-sqrt(3))/2}}}. Then the segment MH is the complement of BH to 1 MH = 1 - {{{(3-sqrt(3))/2}}} = {{{(2 - 3 + sqrt(3))/2}}} = {{{(sqrt(3)-1)/2}}}. (5) Comparing expressions (4) and (5), we see that CH = MH. Hence, right-angled triangle MHC is isosceles right-angled triangle, which implies that angle MCH is 45°. Thus ∠ACM = 135° - 60° - 45° = 30°. Hence, sin(∠ACM) = sin(30°) = 1/2, and the problem is solved completely. </pre> My other additional lessons on miscellaneous Geometry problems in this site are - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Find-the-rate-of-moving-of-the-tip-of-a-shadow.lesson>Find the rate of moving of the tip of a shadow</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-radio-transmitter-accessibility-area.lesson>A radio transmitter accessibility area</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Miscellaneous-geometric-problems.lesson>Miscellaneous geometric problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Miscellaneous-problems-on-parallelograms.lesson>Miscellaneous problems on parallelograms</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Remarkable-properties-of-triangles-into-which-diagonals-divide-a-quadrilateral.lesson>Remarkable properties of triangles into which diagonals divide a quadrilateral</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-trapezoid-divided-in-four-triangles-by-its-diagonals.lesson>A trapezoid divided in four triangles by its diagonals</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/A-problem-on-heptagon.lesson>A problem on a regular heptagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/The-area-of-a-regular-octagon.lesson>The area of a regular octagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/The-fraction-of-the-area-of-a-regular-octagon.lesson>The fraction of the area of a regular octagon</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Try-to-solve-this-nice-Geometry-problem.lesson>Try to solve these nice Geometry problems !</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Find-the-angle-between-sides-of-folded-triangle.lesson>Find the angle between sides of folded triangle</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-problems-on-three-spheres.lesson>A problem on three spheres</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-sphere-placed-in-an-inverted-cone.lesson>A sphere placed in an inverted cone</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/A-great-Math-Olympiad-level-Geometry-problem.lesson>A great Math Olympiad level Geometry problem</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Nice-geometry-problem-of-a-Math-Olympiad-level.lesson>Nice geometry problem of a Math Olympiad level</A> - <A HREF=https://www.algebra.com/algebra/homework/word/geometry/OVERVIEW-of-my-lessons-on-additional-misc-Geometry-problems.lesson>OVERVIEW of my additional lessons on miscellaneous advanced Geometry problems</A> To navigate over all topics/lessons of the Online Geometry Textbook use this file/link <A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.
