SOLUTION: Here is the problem I was given and cannot figure out how to start: Draw the locus of P so that AB is the hypotenuse of right triangle APB and A is the vertex of an angle that i

Algebra ->  Triangles -> SOLUTION: Here is the problem I was given and cannot figure out how to start: Draw the locus of P so that AB is the hypotenuse of right triangle APB and A is the vertex of an angle that i      Log On


   

Question 57520: Here is the problem I was given and cannot figure out how to start:
Draw the locus of P so that AB is the hypotenuse of right triangle APB and A is the vertex of an angle that is less than 45 degrees.
Thank you!
Answer by venugopalramana(3286) About Me  (Show Source):
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Here is the problem I was given and cannot figure out how to start:
Draw the locus of P so that AB is the hypotenuse of right triangle APB and A is the vertex of an angle that is less than 45 degrees.
Thank you!
DRAW AB THE HYPOTENUSE = SOME SUITABLE LENGTH SAY 5" OR 15 CM.(SINCE THE LENGTH IS NOT GIVEN)
NOW MARK 'O' THE MID POINT OF AB.FOR THIS PURPOSE YOU CAN DO AS FOLLOWS.
WITH A AS CENTRE AND A SUITABLE RADIUS (MORE THAN AB/2 LENGTH), DRAW ARC COVERING BOTH SIDES OF AB. WITH SAME RADIUS AND B AS CENTRE AGAIN DRAW AN ARC COVERING BOTH SIDES OF AB.LET THE 2 ARCS CUT AT Q AND R ON EITHER SIDE OF AB.JOIN QR.LET IT MEET AB AT 'O'.QOR IS PERPENDICULAR BISECTOR OF AB.
NOW WITH O AS CENTRE AND RADIUS EQUAL TO OA OR OB DRAW A CIRCLE.AB SUBTENDS 90 DEGREES AT ANY POINT ON THIS CIRCLE.BUT WE WANT ANGLE AT A TO BE LESS THAN 45.SO DRAW A 45 DEGREE LINE FRO A MAKING 45 DEREES WITH AB ON EITHER SIDE OF AB.LET IT CUT THE CIRCLE AT U AND V.
P LIES IN THE ARC 'UBV'.BY JOINING ANY POINT P ON THIS ARC WE GET APB TRIANGLE WITH AB AS HYPOTENUSE ...THAT IS ANGLE APB = 90 DEGREES AND ANGLE PAB LESS THAN 45 DEGREES.