AwesomeGuy

The length of a shadow of a tree is 130 feet when the angle of elevation of the sun is θ°. a) Write the height h of the tree as a function of θ. b) θ=10° What's height? θ=15° Height? θ=20° How tall is the height? θ=25° What is height?

12 years ago

A passenger in an airplane flying at an altitude of 10 kilometers sees two towns directly to the left of the plane. The angles of depression to the towns are 28° and 55°. How far apart are the towns?

12 years ago

Height of a mountain While traveling across flat land, you notice a mountain directly in front of you. The angle of elevation to the peak is 2.5°. After you drive 18 miles closer to the mountain, the angle of elevation is 10°. Approximate the height of th...

12 years ago

Navigation A ship leaves port at noon and has a bearing of S 29° W. If the ship sails at 20 knots, how many nautical miles south and how many nautical miles west will the ship have traveled by 6:00 P.M.?

12 years ago

Angle of Depression Find the angle of depression from the top of the lighthouse 250 feet above water level to the water line of a ship 2.5 miles offshore.

12 years ago

Height of a Kite A 100-foot line is attached to a kite. When the kite has pulled the line taut, the angle of elevation to the kite is approximately 50°. Now everybody lets approximate the height of the kite!

12 years ago

Wave Motion A buoy oscillates in simple harmonic motion as waves go past. At a given time it is noted that the buoy moves a total of 3.5 feet from its low point to its high point, and that it returns to its high point every 10 seconds. Write an equation t...

12 years ago

Surveying A surveyor wishes to find the distance across a swamp. The bearing from A to B (Segment AB is opposite side of triangle) is N 32° W. The surveyor walks 50 meters from A to C, and at the point C the bearing to B is N 68° W. (Segment AC is adjacen...

12 years ago

A plane is 160 miles north and 85 miles east of an airport. If the pilot wants to fly directly to the airport, what bearing would be taken?

12 years ago

An observer in a lighthouse 350 feet above sea level observes two ships directly offshore. The angles of depression to the ships are 4° and 6.5°. How far apart are the ships?

12 years ago

A ship is 50 miles east and 35 miles south of port. If the captain wants to sail directly to port, what bearing should be taken?

12 years ago

Solve the trigonometric equations. 1.) SIN²x(CSC²x-1) 2.) COT x SEC X 3.) COS²[(π/2)-x]/COS X (Note: These mathematical problems are somewhat tricky, but useful for students as of learning how to their fundamental identities. Let's enjoy solving these pro...

12 years ago

Let's solve this fun trigonometric fun math problem! 1.) SEC²X-1/SIN²X

12 years ago

Google this: 5.1 using fundamental identities Comcast Click and open the first result that appears on your search (Result= PDF Document). Textbook pages will appear and go to page 383 on problem #129. The directions will mention "Solve the right triangle...

12 years ago

Solve the equation. 1.) COS X CSC X

12 years ago

Verify/Solve the identities. 1.) SIN^1/2 X COS X-SIN^5/2 X COS X 2.) Long problem, but it's fun to solve! SEC^6 X(SEC X TAN X)-SEC^4 X(SEC X TAN X)

12 years ago

Verify the identities. 1.) SIN[(π/2)-X]/COS[(π/2)-X]=COT X 2.) SEC(-X)/CSC(-X)= -TAN X 3.) (1 + SIN Y)[1 + SIN(-Y)]= COS²Y 4.) 1 + CSC(-θ)/COS(-θ) + COT(-θ)= SEC θ (Note: Just relax through verifying/solving these nice fun looking math problems! It's heal...

12 years ago

Verify the identity algebraically. TAN X + COT Y/TAN X COT Y= TAN Y + COT X

12 years ago

Verify the identities. 1.) √1-COSθ/1+COSθ= 1+SINθ/SINθ 2.) SEC X SIN(π/2-X)= 1 3.) CSC X(CSC X-SIN X)+SIN X-COS X/SIN X + COT X= CSC²X 4.) CSC^4 X-2 CSC²X+1= COT^4 X 5.) CSC^4 θ-COT^4 θ= 2 CSC²θ-1 6.) TAN^5 X= TAN³X SEC²X-TAN³X 7.) COS³X SIN²X= (SIN²X-SIN...

12 years ago

State the quadrant in which θ lies. 1.) CSC θ is greater than 0 and TAN θ is less than 0. 2.) SEC θ is greater than 0 and SIN θ is less than 0.

12 years ago

Verify the identity algebraically. This problem is very intriguing and awesome at the same time. It's wonderfully amazing! 1.) TAN³α-1/TAN α-1= TAN²α + TAN α + 1

12 years ago

Verify the identities algebraically. 1.) TAN^5 X= TAN³X SEC²X-TAN³X 2.) COS³X SIN²X= (SIN²X-SIN^4X)COS X

12 years ago

Directions: Use a graphing utility to approximate the solutions of the equation in the interval [0,2π) by setting the equation equal to zero, graphing the new equation, and using the ZERO or ROOT feature to approximate the x-intercepts of the graph. (Note...

12 years ago

Verify the identities please. 1.) TAN(X+π)-TAN(π-X)= 2 TAN X 2.) SIN(X+Y)+SIN(X-Y)= 2 SIN X COS Y

12 years ago

Write the trigonometric expression as an algebraic expression. 1.) SIN(ARCSIN X+ARCCOS X) ANSWER: 1 2.) SIN(ARCTAN 2X-ARCCOS X) ANSWER: 2x²-SquareRoot of 1-x²/Square Root of 4x²+1.

12 years ago

Solve the equation for all values of x. -2cos²x-sin x+1=0, on the interval [0,2π).

12 years ago

Solve the equation for all values of x. 2sin(2x)-√3=0, on the interval [0,2π).

12 years ago

Find the inverse of the following quadratic equation. Hint: Complete the square first. y= x²+14x+50

12 years ago

BP= 3091.79 AB/sin 2.5° = 3091.79/sin 4° 3091.785015 sin 2.5° = 134.8617682 134.8617682/sin 4° = 1933.33 Awesome! Thanks for the help!

12 years ago

Since I have also forgot to label the conversion for the solution, it is ultimately measured in FEET.

12 years ago

Reiny thanks a lot for the help! I welcome you with a nice smile! Here is the website where I can show how the problems look/appear like... Google this: 4.8 applications and models Comcast Click the first result that appears on the web search page. Go to...

12 years ago

1.) 1/TAN[(π/2)-X]=COT X BINGO! SOLVED! 2.) SEC X/-CSC X 1/COS X ÷ -1/COS X 1/COS X * -SIN X/1 -TAN X YES BINGO! WOW!

12 years ago

2.) SEC X COS X= 1 1/COS X*COS X/1= 1 BINGO! 3.) CSC²X-CSC X SIN X+SIN X/SINX-TAN X+ COS X/SIN X 1/SIN²X-1/SIN X*SIN X/1-TAN X+COS X/ SIN X 1/SIN²X-1-SIN X/COS X+COS X/SIN X 1/SIN²X-1-TAN X*1/TAN X= CSC²X COOL! Need help on #6 and 7 please. Thank you!

12 years ago

AWESOME! SUPERB! I had it right when you provided support. THANKS!

12 years ago

1.) Apply the following method you used in #2! You should notably understand this already! Goodness grief! 2.) let A = arctan x let B = arccos x then we are looking for sin(A + B) . now, tan A = x , sin A = x/√(x^2+1) cos A = 1/√(x^2+1) ... create a right...

12 years ago