- WELCOME TO THE SECOND VIDEO ON THE LAW OF SINES.
THIS VIDEO DEALS WITH THE AMBIGUUS CASE
SO THE GOAL IS TO SOLVE TRIANGLES USING THE LAW OF SINES
GIVEN SIDE-SIDE-ANGLE.
SO THE LAW OF SINES HAS A PROBLEM DEALING
WITH SIDE-SIDE-ANGLE.
IF YOU ARE GIVEN TWO SIDES AND ONE ANGLE,
THE LAW SINES COULD PROVIDE 0, 1, OR 2 SOLUTIONS.
A COUPLE THINGS WE NEED TO REMEMBER BEFORE WE START
IS THE SUM OF THE INTERIOR ANGLES OF A TRIANGLE
IS 180 DEGREES.
THE SINE FUNCTION WILL ONLY RETURN VALUES
IN THE CLOSED INTERVAL FROM -1 TO 1 AND LASTLY,
IF SINE THETA IS BETWEEN 0 AND 1 THETA CAN LIE
IN THE FIRST QUADRANT AS AN ACUTE ANGLE
AND ALSO IN THE SECOND QUADRANT AS AN OBTUSE ANGLE.
LET'S TAKE A LOOK AT THE CONDITIONS
THAT MUST BE PRESENT TO HAVE TWO SOLUTIONS.
IN THIS SKETCH, IF THE LENGTH OF SIDE B IS GIVEN
AND THE LENGTH OF SIDE "A" IS GIVEN AS WELL
AS THE MEASURE OF ANGLE "A" WE'D HAVE SIDE-SIDE-ANGLE.
SO THAT'S THE FIRST CONDITION TO HAVE TWO POSSIBLE SOLUTIONS
SIDE-SIDE-ANGLE.
SECOND, THE GIVEN ANGLE "A" MUST BE ACUTE.
THIRD, THIS SIDE "A" MUST BE LESS THAN SIDE B
AND FOURTH, SIDE "A" IS LONGER THAN THE ALTITUDE THIS SIDE "A"
MUST BE GREATER THAN THIS LENGTH HERE.
REMEMBER YOU CAN THINK OF THIS AS THE HEIGHT OF THE TRIANGLE.
SO THIS LENGTH "A" MUST BE GREATER
THAN THIS HEIGHT HERE.
IF ALL THESE STATEMENTS ARE TRUE THEN ANGLE B,
THIS ONE HERE COULD BE ACUTE AS WE SEE HERE
IN THE DASHED TRIANGLE AND IT COULD ALSO BE OBTUSE
AS WE SEE HERE IN THIS SOLID TRIANGLE.
LET'S TAKE A LOOK AT AN ANIMATION
THAT ILLUSTRATES THIS.
SO HERE WE HAVE A TRIANGLE AND NOTICE AS I CHANGE
THE LENGTH OF B HERE WE SEE IT HAS TWO SOLUTIONS
BECAUSE THIS LENGTH B IS LESS THAN SIDE "A"
BUT GREATER THAN THE HEIGHT OF THIS TRIANGLE
WHICH WOULD BE THE ALTITUDE OF THE TRIANGLE.
SO IF THE LENGTH OF B IS BETWEEN THE HEIGHT
OF THE TRIANGLE BUT LESS THAN THIS OTHER KNOWN SIDE
WE CAN HAVE TWO SOLUTIONS.
IF IT'S TOO SMALL OR LESS THAN THE HEIGHT
WE ACTUALLY HAVE NO SOLUTIONS BECAUSE WE CAN'T CONNECT
THE THREE SIDES TO FORM A TRIANGLE AND IF THE LENGTH
OF SIDE B IS TOO LARGE, AGAIN THERE'S TWO SOLUTIONS
BUT ONCE IT BECOMES LARGER THAN SIDE "A" WE ONLY HAVE ONE
POSSIBLE TRIANGLE.
OKAY. SO OBVIOUSLY, THAT IS A LOT TO REMEMBER
BUT HERE'S WHAT YOU REALLY NEED TO KNOW.
NUMBER ONE, IF YOU HAVE SIDE- SIDE-ANGLE AND THE GIVEN ANGLE
IS ACUTE YOU COULD HAVE 0, 1, OR 2 POSSIBLE TRIANGLES.
IF WHEN USING THE LAW OF SINES YOU GET AN EQUATION
WHERE SINE THETA IS GREATER THAN ONE
THEN THERE'S NO SOLUTION.
NUMBER THREE, IF WHEN USING THE LAW OF SINES
YOU GET AN EQUATION WHERE SINE THETA
IS LESS THAN 1 WE WILL FIND BOTH THE ACUTE AND OBTUSE ANGLE
FOR THAT EQUATION AND THEN WE'LL JUST USE
THOSE TWO ANGLES TO CHECK TO SEE IF WE CAN FORM
A TRIANGLE WITH BOTH ANGLES, AND THE WAY WE DO THIS IS,
REMEMBER THE SUM OF THE INTERIOR ANGLES CAN'T BE GREATER
THAN 180 DEGREES.
SO IF THE OBTUSE ANGLE IS POSSIBLE
IT WILL HAVE TWO SOLUTIONS.
IF THE OBTUSE ANGLE IS NOT POSSIBLE
THEN WE'LL ONLY USE THE ACUTE ANGLE AND HAVE ONE SOLUTION.
LET'S GO AHEAD AND GIVE IT A TRY.
NOTICE WE HAVE SIDE-SIDE-ANGLE.
LET'S GO AHEAD AND RECORD THE GIVEN INFORMATION.
SO ANGLE C IS 82 DEGREES, SIDE "A" IS 11 METERS,
AND SIDE C IS 7 METERS.
LET'S GO AHEAD AND APPLY THE LAW OF SINES AND SEE WHAT HAPPENS.
LET'S TRY TO DETERMINE THE MEASURE OF ANGLE "A."
SO WE HAVE THE SINE OF 82 DEGREES
IS TO THE OPPOSITE SIDE 7 METERS AS THE SINE OF "A"
DIVIDED BY 11.
WE HAVE A PROPORTION WITH ONE UNKNOWN.
LET'S GO AHEAD AND CROSS MULTIPLY.
7 SINE "A" MUST EQUAL 11 SINE 82 DEGREES.
DIVIDE BOTH SIDES BY 7.
LET'S GO AHEAD AND EVALUATE THIS ON OUR CALCULATOR.
I'M ALREADY IN DEGREE MODE, 11 SINE 82 DEGREES DIVIDED BY 7.
WE HAVE APPROXIMATELY 1.5561.
REMEMBER THE SINE FUNCTION ONLY RETURNS A VALUE
ON THE CLOSED INTERVAL FROM -1 TO +1.
SINE "A" CANNOT EQUAL 1.5561.
THIS IS AN INDICATION THAT THERE IS NO SOLUTION
TO THIS PROBLEM.
LET'S GO AHEAD AND TAKE A LOOK AT ANOTHER PROBLEM.
LET'S SET UP OUR TRIANGLE.
ANGLE "A" IS 88 DEGREES, SIDE "A" IS 110 FEET,
SIDE C IS 54 FEET.
AGAIN, WE HAVE SIDE-SIDE-ANGLE.
THE GIVEN ANGLE IS ACUTE.
LET'S GO AHEAD AND APPLY THE LAW OF SINES AND SEE WHAT HAPPENS.
SO WE HAVE TO USE ANGLE "A" AND SIDE "A"
THE SINE OF 88 DEGREES OVER 110 MUST EQUAL--LET'S TRY TO FIND
ANGLE C SO WE HAVE THE SINE OF C DIVIDED BY 54.
CROSS MULTIPLY, SOLVE FOR SINE C AND JUST TO SAVE SOME TIME
I'VE ALREADY FOUND THIS VALUE.
IT'S APPROXIMATELY .4906.
LET'S GO AHEAD AND FIND ONE POSSIBLE VALUE FOR C.
WE'LL FIND THE VALUE OF INVERSE SINE .4906.
THIS GIVES US AN ANGLE OF APPROXIMATELY 29 DEGREES.
BUT THERE'S ALSO ANOTHER ANGLE THAT HAS THE SINE FUNCTION
VALUE OF .4906.
REMEMBER, SINE IS POSITIVE IN BOTH THE FIRST
AND THE SECOND QUADRANT.
SO IF WE HAVE A 29 DEGREE ANGLE HERE, IF WE SKETCH
A 29 DEGREE REFERENCE ANGLE IN THE SECOND QUADRANT
THIS ANGLE WILL HAVE THE SAME SINE FUNCTION VALUE.
THIS WOULD BE 180 - 29 DEGREES OR 151 DEGREES.
THE QUESTION BECOMES IS IT POSSIBLE FOR ANGLE C
TO ALSO BE 151 DEGREES.
IT'S ACTUALLY NOT BECAUSE REMEMBER WE ALREADY KNOW
THAT ANGLE "A" IS 88 DEGREES.
WELL 88 DEGREES + 151 DEGREES IS OBVIOUSLY LARGER
THAN 180 DEGREES, WHICH IS THE SUM
OF THE INTERIOR ANGLES OF ANY TRIANGLE.
SO EVEN THOUGH THIS RETURNS THE SAME SINE FUNCTION VALUE
IT'S NOT A POSSIBLE ANGLE FOR THIS TRIANGLE.
SO WE WILL ONLY HAVE ONE SOLUTION TO THIS PROBLEM.
LET'S GO AHEAD AND LABEL ANGLE C AS 29 DEGREES.
WE CAN FIND THE REMAINING ANGLE BECAUSE WE KNOW THE SUM
MUST BE 180 DEGREES.
SO 180 - 29 - 88 WILL GIVE US 63 DEGREES
FOR ANGLE B.
NOW TO FIND THE LENGTH OF SIDE B WE ARE GOING TO HAVE TO USE
THE LAW OF SINES AGAIN.
SO TO GET STARTED LET'S USE ANGLE "A"
AND SIDE "A."
SINE OF 88 DEGREES DIVIDED BY 110
MUST EQUAL THE SINE OF 63 DEGREES DIVIDED BY B.
CROSS MULTIPLY, DIVIDE BY SINE 88.
SO WE HAVE B IS APPROXIMATELY 98 FEET.
SO WE HAD ONE THAT HAD NO SOLUTION,
HERE'S ONE THAT HAS ONE SOLUTION.
LET'S TAKE A LOOK AT THE THIRD POSSIBILITY.
SO ANGLE B IS 40 DEGREES, LENGTH OF B IS 22 CENTIMETERS
AND THE LENGTH OF "A" IS 37.
USING THE LAW OF SINES LET'S TRY TO DETERMINE
THE MEASURE OF ANGLE "A."
SO WE'LL HAVE THE SINE OF B OR THE SINE OF 40 DEGREES
DIVIDED BY 22 MUST EQUAL THE SINE OF "A" DIVIDED BY 30.
CROSS MULTIPLY, DIVIDE BY 22 AND AGAIN,
TO SAVE TIME I'VE ALREADY FOUND THIS VALUE
TO BE APPROXIMATELY .8765.
SO "A" IS APPROXIMATELY EQUAL TO INVERSE SINE .8765.
LET'S GO TO OUR CALCULATOR.
INVERSE SINE .8765 RETURNS AN ANGLE OF 61 DEGREES
BUT REMEMBER THERE'S ALSO ANOTHER ANGLE
IN THE SECOND QUADRANT THAT RETURNS
A SINE FUNCTION VALUE OF .8765.
IF WE SKETCH THIS IN STANDARD POSITION
THERE'S ANOTHER ANGLE OVER HERE AND THIS ANGLE
WOULD BE 180 - 61 OR 119.
THE QUESTION BECOMES COULD THIS ANGLE
ALSO BE 119 DEGREES AND NOW IT CAN
BECAUSE IF THIS IS 119 DEGREES AND THE OTHER ANGLE
THAT'S KNOWN IS 40 THAT ONLY SUMS TO 159 DEGREES
WHICH IS LESS THAN 180 DEGREES.
SO IT IS POSSIBLE IN THIS PROBLEM
TO HAVE TWO DIFFERENT TRIANGLES.
SO WHAT'S HAPPENING IS THIS SIDE B COULD BE HERE
OR IT COULD ALSO BE ANGLED IN THIS DIRECTION
MAKING THIS ANGLE "A" 119 DEGREES.
LET'S GO AHEAD AND SKETCH THE SECOND POSSIBILITY HERE.
SO ANGLE "A" COULD BE 61 DEGREES FOR OUR FIRST SOLUTION
AND ANGLE "A" COULD ALSO BE 119 DEGREES
FOR THE SECOND SOLUTIONS.
LET'S GO AHEAD AND STICK WITH OUR ORIGINAL SKETCH
AND DETERMINE THE LENGTH OF SIDE C
AND THE MEASURE OF ANGLE C.
SINCE WE KNOW THE SUM OF THESE ANGLES
MUST BE 180 DEGREES THAT WOULD LEAVE 79 DEGREES
FOR ANGLE C.
NOW WE HAVE TO FIND THE LENGTH OF SIDE C
USING THE LAW OF SINES.
SO WE HAVE THE SINE 40 DEGREES DIVIDED BY 22 MUST EQUAL
[ AUDIO SKIP ] THIS COMES OUT TO APPROXIMATELY 34 CENTIMETERS.
HERE'S OUR FIRST POSSIBLE SOLUTION.
NOW LET'S GO AHEAD AND FINISH OFF
THE SECOND POSSIBLE SOLUTION.
AGAIN, IF THIS IS 119 DEGREES AND THIS IS 40 DEGREES
THAT LEAVES 21 DEGREES FOR ANGLE C AND NOW WE NEED
TO FIND THE LENGTH OF SIDE C, FINISH OFF OUR SECOND
[ AUDIO SKIP ] SO THE SET UP WILL BE SIMILAR.
WE'LL HAVE THE SINE OF 40 DEGREES DIVIDED BY 22
MUST EQUAL--NOW THE SINE OF 21 DEGREES DIVIDED BY C,
CROSS MULTIPLY, DIVIDE BY SINE 40.
THE SECOND SOLUTION IS APPROXIMATELY 12 CENTIMETERS
AND AGAIN, THE KEY TO FINDING THIS SECOND SOLUTION
IS YOU'RE GIVEN SIDE-SIDE-ANGLE WHERE THE GIVEN ANGLE IS ACUTE.
WHEN YOU GO TO FIND THAT SECOND ANGLE
MAKE SURE YOU FIND THAT ANGLE THAT IS ACUTE AS WELL AS OBTUSE
THAT HAS THE GIVEN SINE FUNCTION VALUE.
ONCE YOU FIND THIS OBTUSE ANGLE, SEE IF THE SUM OF THE ANGLES
IS MORE THAN 180 DEGREES.
IF IT'S NOT YOU DO HAVE YOUR SECOND [ AUDIO SKIP ]
SO YOU HAVE TO SET UP AND SOLVE TWO DIFFERENT TRIANGLES,
ONE WITH THE ACUTE ANGLE AND ONE WITH THE OBTUSE ANGLE.
I HOPE YOU FOUND THIS VIDEO HELPFUL.
THANK YOU FOR WATCHING.