5302 Shares

Can Anyone please help me solve these problems for Alg. 2/Trig.? I ahve a test coming up!? Topic: Can Anyone please help me solve these problems for Alg. 2/Trig.? I ahve a test coming up!?
June 20, 2019 / By Delta
Question: I have two question on my study guide that i just can't solve. PLEASE help #1: You are a photographer taking a picutre of a sprinter on a straightaway traveling at 20 ft/second. a) write a trig function to find the angle x you have to turn your camera t seconds later to photograph the runner b) what angle must you be at to take the picture after three seconds? c)If t is twice as long, is the angle twice as big too? This next problem is actually two triangles (SSA case), and i don't know how to solve the 2nd triangle: #2: Given Triangle ABC, find c (the side). Angle A =15, a=8, b=11 Somebody please help me solve these two problems oo sorry about taht. the photographer is standing 50 ft away from the starting position of the runner. hope that helps no the runner is not running towards you. here's kind of a picture. it's like a triangle. the runner is the 8, the photograther it the 9 8---------->> 9 see what i mean? no, there is an answer for finding c on question #2. there is just two possible answers  Caitlyn | 8 days ago
1) According to your information, you must have typed it wrong... You will use the same angle no matter how far away the runner is.... there would be no reason to change the angle unless you were on a hill, and you didn't say that so I must assume you are right in line with him, and if he runs straight you can keep the angle at 0* and he will remain in the picture. Therefore x=0 for A and b=0 and c= no 2) Unless it is told that it is a right triangle there is no way to determine side C.
👍 284 | 👎 8 Originally Answered: PLEASE help me solve this, I have a trig test tomorrow: 4 cos2 x - 4 sin x = 5?
wait, does that say cos 2x or cos^2x? im assuming its cos^2x ok, write out the expression 4 cos^2 x - 4 sin x = 5 now u kno that cos^2x = 1 - sin^2x rite? (u can derive this from the pythagorean identity: sin^2x + cos^2x = 1). plug in 1 - sin^2x in for cos^2x to get the equation be in terms of sine 4 cos^2 x - 4 sin x = 5 4(1 - sin^2 x) - 4 sin x = 5 4 - 4 sin^2 x - 4 sin x = 5 now subtract both sides of the equation by 4 - 4 sin^2 x - 4 sin x = 1 now subtract both sides of the equation by 1 - 4 sin^2 x - 4 sin x - 1 = 0 now multipy the equation by -1 [- 4 sin^2 x - 4 sin x - 1 = 0] * -1 4 sin^2 x + 4 sin x + 1 = 0 now we can make the following substitution. let y = sin x 4y^2 + 4y + 1 = 0 now wat do u realize? its a quadratic equation. u can factor this one into (2y + 1) * (2y + 1) (2y + 1) (2y + 1) = 0 2y + 1 = 0 2y = -1 y = -1/2 remember we let y = sin x. plug in sin x back into y sin x = -1/2 now sine is negative in quadrants III and IV. so there are actually two values of x. u need to take that into consideration. sin x = 1/2 at π/6. that the reference angle we are going to use to find the solutions. to find the solution in the third quadrant, add the reference angle, which is π/6 to π. to find the solution in the fourth quadrant subtract π/6 from 2π x = π/6 + π = 7π/6 x = 2π - π/6 = 11π/6 so x = {7π/6, 11π/6} so theres the answer. hope that helped. good luck on ur test! =] dam, it was cos 2x? then u shouldve put the 2 next to the x in the problem. the double angle formula is denoted by the following cos 2x = cos^2x - sin^2x with now u kno that cos^2x = 1 - sin^2x rite? substitute 1 - sin^2x in for cos^2x. cos 2x = cos^2x - sin^2x cos 2x = (1 - sin^2x) - sin^2x cos 2x = 1 - 2 sin^2x now plug that into the originial problem for cos 2x. it will still turn into a quadratic. follow the same methods i have shown Originally Answered: PLEASE help me solve this, I have a trig test tomorrow: 4 cos2 x - 4 sin x = 5?
Question SHOULD read as :- 4 cos^2x - 4 sin x = 5 and NOT as given. 4 ( 1 - sin^2 x ) - 4 sin x - 5 = 0 - 4 sin^2 x - 4 sin x - 1 = 0 4 sin^2 x + 4 sin x + 1 = 0 ( 2 sin x + 1 ) ( 2 sin x + 1 ) = 0 sin x = - 1/2 x = 210 ° , 330 ° OR x = 7π/6 , 11π/6 Alva
First one: you may integrate "like" words. "like" words listed here are people who've "w". integrate ability to characteristic. then you definately positioned the variable words on one component, and the consistent on the different. Then resolve. integrate: 33 + 15w = 6w Separate: 33 = -9 w resolve -3&2/3= w #2: Multiplication. each little thing in a style of ( ) is prolonged via the consistent in front of the ( ). If there is none, a million is known. then you definately do the property you probably did in concern a million. 60 + 24y= 4y -6 + 9y sixty six = -11y y =-6 3. Slope intercept style is Y= MX+B the place M and B are constants. so which you may circulate each little thing however the y term to the different component, tidy issues up and divide via the y coefficient . circulate: 3y = -8x +3 -8 TIDY 3y = -8x-5 DIVIDE y = - 8/3 x - 5/3 Then -8/3 is the slope and -5/3 is the intercept.
👍 120 | 👎 1 Vance
for the first question, I don't think you have enough information. You need to know where the photographer is standing. For the second, use the law of cosines.
👍 113 | 👎 -6 Originally Answered: How do you solve these trig problems?
csc A > 0 1/ sin A > 0 : sin is positive in Q I and Q II cot B <0 cot B = cos B/ sin B : cos B and sin B are opposite signs in Q II and Q IV sin B >0 in Q I and Q II B must be in Q II

If you have your own answer to the question Can Anyone please help me solve these problems for Alg. 2/Trig.? I ahve a test coming up!?, then you can write your own version, using the form below for an extended answer.