- OKAY. WELCOME TO A VIDEO
ON APPLICATIONS OF THE LAW OF COSINES.
LET'S GO AHEAD AND REVIEW THE LAW OF COSINES FIRST.
REMEMBER THERE ARE THREE FORMS TO THE LAW OF COSINES.
THE IMPORTANT PART TO REMEMBER
IS WHATEVER ANGLE WE'RE USING HERE ON THE RIGHT SIDE,
WE MUST USE THE OPPOSITE SIDE
ON THE OTHER SIDE OF THE EQUAL SIGN.
AN ENGINEERING FIRM DECIDES TO BID
ON PROPOSED TUNNEL THROUGH A MOUNTAIN.
TO FIND THE TUNNEL LENGTH AND MEASUREMENTS SHOWN ARE TAKEN.
PART "A," HOW LONG IS THE TUNNEL?
PART B, THE COST IS $4,500 PER FOOT.
IF THE COMPANY WANTS A 20% PROFIT, WHAT IS THE BID AMOUNT?
OKAY LET'S GO AHEAD AND START BY FINDING THIS LENGTH HERE.
LET'S GO AHEAD AND CALL THIS SIDE "A."
SO THIS WOULD BE ANGLE "A" THIS WILL BE SIDE B,
AND THIS WILL BE SIDE C.
SO USING THE LAW OF COSINES, "A" SQUARED MUST EQUAL B SQUARED
+ C SQUARED - 2 x B x C x COSINE "A."
NOW WE NEED TO SOLVE FOR "A."
YOU MAY WANT TO CHECK THESE ON YOUR CALCULATOR
BUT NOW WE'RE GOING TO ADD AND SUBTRACT ON THE RIGHT SIDES.
WE HAVE "A" SQUARED = 579,532.05.
NOW WE CAN TAKE THE SQUARE ROOT OF BOTH SIDES.
SO WE HAVE "A" IS APPROXIMATELY 761 YARDS.
SO THIS IS PART "A," HOW LONG IS THE TUNNEL?
FOR PART B, THE COST IS $4,500 PER FOOT
AND THEY WANT TO KNOW WHAT IS THE BID AMOUNT
IF THEY WANT A 20% PROFIT.
LET'S TAKE THIS INFORMATION OVER TO THE NEXT SCREEN.
SO THE FIRST WE HAVE 761 YARDS OF TUNNEL TO DIG.
BUT THE COST IS IN FEET,
SO LET'S GO AHEAD AND CONVERT THIS TO FEET,
ONE YARD IS EQUAL TO THREE FEET.
SO THE UNITS OF YARDS SIMPLIFY OUT
AND WE'RE LEFT WITH 761 x 3
WILL GIVE US THE TOTAL NUMBER OF FEET, THAT'S 2,283 FEET
AND THE COST IS $4,500 PER FOOT.
SO THE COST TO DRILL THAT TUNNEL IS GOING TO BE $10,273,500.
BUT THAT'S JUST THE COST.
IF THEY WANT A PROFIT OF 20%,
WELL 100% PLUS 20% WE WANT 120% OF THIS AMOUNT OR THE COST.
SO WE'LL MULTIPLY THIS BY 1.2
AND THAT WILL GIVE US THE BID AMOUNT.
SO IT LOOKS LIKE WE'LL HAVE $12,328,200.
REMEMBER TO FIND THE PERCENT OF A NUMBER
YOU JUST CONVERT TO A DECIMAL AND MULTIPLY.
OKAY WE HAVE TIME FOR ONE MORE APPLICATION PROBLEM.
THERE IS A TRANSMISSION TOWER NEAR I-10.
THE RANGE OF THE SERVICE FROM THE TOWER
FORMS A 47 DEGREE ANGLE
AND A RANGE OF 28 MILES TO ONE SECTION OF I-10
AND 31 MILES TO ANOTHER POINT ON I-10.
IF A DRIVER IS TRAVELING AT 50 MILES PER HOUR,
HOW LONG WILL SHE HAVE SERVICE?
SO THE DIFFICULT PART ABOUT THIS ONE IS GOING TO BE THE SET UP.
SO LET'S GO AHEAD AND LABEL THIS AS OUR ROAD, SO HERE'S I-10.
THERE'S THE TRANSMISSION TOWER, LET'S SAY TO THE WEST,
AND IT FORMS A 47 DEGREE ANGLE SO THE TRANSMISSION GOES OUT
FROM THE TOWER LIKE THIS FORMING A 47 DEGREE ANGLE
AND THE RANGE IS 28 MILES TO ONE SECTION OF I-10
AND 31 MILES TO THE OTHER.
SO LET'S JUST LABEL THIS 28 MILES AND THIS 31 MILES.
THIS WOULD BE THE DISTANCE WHERE THE DRIVER WOULD HAVE SERVICE.
SO LET'S GO AHEAD AND LABEL THIS D.
WE FIRST NEED TO FIND THIS LENGTH
AND THEN DETERMINE HOW LONG IT WILL TAKE THAT DRIVER
TO GO THAT DISTANCE DRIVING 50 MILES PER HOUR.
SO AGAIN WE CAN FIND THE VALUE OF D USING THE LAW OF COSINES.
REMEMBER IF WE HAVE THE SIDE D
OR D SQUARED ON THE LEFT SIDE OF THE EQUATION,
WE'LL HAVE COSINE 47 DEGREES ON THE RIGHT.
SO WE'LL HAVE D SQUARED = 28 SQUARED + 31 SQUARED,
- 2 x 28 x 31 x COSINE 47 DEGREES.
REMEMBER IF WE USE THIS ANGLE,
THEN THE SIDE OPPOSITE THAT ANGLE WILL BE ON THE LEFT SIDE
OF OUR EQUATION BY ITSELF.
LET'S GO AHEAD AND START SIMPLIFYING THE RIGHT SIDE
SO D SQUARED IS EQUAL TO 561.0508.
NOW WE'LL TAKE THE SQUARE ROOT OF BOTH SIDES
TO FIND THE VALUE OF D,
REMEMBER D WOULD HAVE TO BE POSITIVE,
SO D IS GOING TO BE APPROXIMATELY EQUAL TO 24 MILES.
SO THE DISTANCE HERE IS APPROXIMATELY 24 MILES.
SO THERE'S A 24 MILE DISTANCE
WHERE THE DRIVER WILL HAVE SERVICE.
NOW SINCE THE DRIVER IS TRAVELING 50 MILES PER HOUR,
THE QUESTION REALLY BECOMES HOW LONG WILL IT TAKE
THE DRIVER TO TRAVEL 24 MILES
BECAUSE THAT'S HOW LONG THE DRIVER WILL HAVE SERVICE
FROM THIS TRANSMISSION TOWER.
SO SINCE DISTANCE = RATE x TIME,
TIME WOULD EQUAL THE DISTANCE DIVIDED BY THE RATE.
DISTANCE IS 24 MILES AND THE RATE IS 50 MILES PER HOUR.
THIS QUOTIENT WILL GIVE US THE TIME AND HOURS
WHICH IS 0.48 HOURS.
LET'S GO TO THE NEXT SCREEN
AND DETERMINE HOW MANY MINUTES THIS WOULD BE 0.48 HOURS,
WELL THERE'S 60 MINUTES IN ONE HOUR.
THE UNITS OF HOURS SIMPLIFY OUT SO 0.48 x 60.
SO THE DRIVER WOULD HAVE SERVICE FOR APPROXIMATELY 29 MINUTES.
NOW THE REASON I'M ALWAYS ROUNDING TO--
AND NOTICE HOW I'M ROUNDING TO 2 SIGNIFICANT DIGITS.
AND THE REASON I AM
IS BECAUSE NOTICE THAT ALL OF THE GIVEN INFORMATION
HAS TWO SIGNIFICANT DIGITS
SO I'M MAINTAINING THAT CONSISTENCY
THROUGHOUT THIS PROBLEM.
SOMETIMES TEXTBOOKS ASK YOU
TO ROUND TO A SPECIFIC PLACE VALUE. OKAY.
I HOPE YOU FOUND THIS VIDEO HELPFUL.
THANK YOU AND HAVE A GOOD DAY.