Solve for d: "Finding Unknowns in Equations with Sin and Cos

[qswdefrg]

Homework Statement

I have this equation, but I'm having trouble solving for the unknown :/ I know that d is 109; I just don't know how to get it.

see below

Homework Equations

-dsin35 = -4.9(dcos35/25)^2

The Attempt at a Solution

-dsin35 = -4.9d^2 cos^2(35)/25^2
-dsin35 = -4.9d^2 cos(35) x cos(35)/25^2

... I get confused after this step. Is it
-dsin35 = -4.9d^2 [cos(35) x cos(35)]/25^2 ?

 

[Mark44]

Homework Statement

I have this equation, but I'm having trouble solving for the unknown :/ I know that d is 109; I just don't know how to get it.

see below

Homework Equations

-dsin35 = -4.9(dcos35/25)^2

The Attempt at a Solution

-dsin35 = -4.9d^2 cos^2(35)/25^2
-dsin35 = -4.9d^2 cos(35) x cos(35)/25^2

Bring everything over to one side so that you have 4.9d^2 cos(35) x cos(35)/25^2
- dsin35 = 0.
This is a quadratic equation in d, so you could use the quadratic formula to solve for d. Or you could simply factor it into d * (the rest) = 0, since there is no constant term.

... I get confused after this step. Is it
-dsin35 = -4.9d^2 [cos(35) x cos(35)]/25^2 ?

 

[qswdefrg]

thanks so much!

 

[HallsofIvy]

Please use parentheses! Is "cos35/25" meant to be "cos(35)/25" or "cos(35/25)"? And you should state whether the arguments for sine and cosine are in degrees or radians. The "default", so to speak, is radians, but it looks to me like you mean degrees.

 

[zgozvrm]

The quadratic equation is not needed here.

-d \times sin35^\circ = -4.9 \left( \frac{d \times cos35^\circ}{25} \right)^2

Since both sides are negative, we can cancel that.
Then expand the squared term on the right hand side:

d \times sind35^\circ = 4.9 \left( \frac{d^2 \times (cos35^\circ)^2}{625} \right)

Multiply the right side out, and then you can divide both sides by d...

 

[Mark44]

The quadratic equation is not needed here.

-d \times sin35^\circ = -4.9 \left( \frac{d \times cos35^\circ}{25} \right)^2

Since both sides are negative, we can cancel that.
Then expand the squared term on the right hand side:

d \times sind35^\circ = 4.9 \left( \frac{d^2 \times (cos35^\circ)^2}{625} \right)

Multiply the right side out, and then you can divide both sides by d...

When you do that, you lose the solution d = 0, which may or may not be meaningful in this problem.

 

[zgozvrm]

When you do that, you lose the solution d = 0, which may or may not be meaningful in this problem.

No. When you do that, you get a d term on the left side and a d^2[/tex] term on the right side. Dividing both sides by d then gives you no d term on the left and a single d term on the right.

 

[Mark44]

Yes, I understand that, and as I said before, you lose the solution d = 0.

Here's a simple example.
2d = d²
If you divide both sides by d, you get
2 = d

However, the original equation has two solutions.
2d = d²
<==> d² - 2d = 0
<==> d(d - 2) = 0
<==> d = 0 or d = 2

 

[zgozvrm]

Pardon me. I read your post to mean that you lose the d-term altogether. You are absolutely correct! (Not enough sleep last night).

But, in my defense, zero is a trivial answer and can be determined by inspection. It is unlikely (although possible, admittedly) that zero is a meaningful answer.

Either way, my point remains valid: you don't need to use the quadratic formula to solve for the value of d.

Even without solving for d=0 by inspection, one only needs to factor d out of the equation to find that either d=0 or set the rest of the equation (the other factor) equal to zero and solve for d (which is what I did by dividing d out of the equation).

 

[Mark44]

Pardon me. I read your post to mean that you lose the d-term altogether. You are absolutely correct! (Not enough sleep last night).

But, in my defense, zero is a trivial answer and can be determined by inspection. It is unlikely (although possible, admittedly) that zero is a meaningful answer.

Maybe zero is physically meaningful in this problem, or maybe it isn't; I don't know enough about the context of this problem to say.

Either way, my point remains valid: you don't need to use the quadratic formula to solve for the value of d.

Nor did I say that you had to use the quadratic formula in this problem. Since there is no constant term, you can find the two roots by factoring.

Even without solving for d=0 by inspection, one only needs to factor d out of the equation to find that either d=0 or set the rest of the equation (the other factor) equal to zero and solve for d (which is what I did by dividing d out of the equation).

And this is what I did: factor the two terms.

My point is that it is dangerous to merely divide through by a variable, since you will be losing a solution. When asked to solve an equation, you should give all solutions, and then decide which to discard because they have no physical meaning.

 

[zgozvrm]

No need to get all defensive. I was agreeing with you!

... You are absolutely correct! ... It is unlikely (although possible, admittedly) that zero is a meaningful answer.

Maybe zero is physically meaningful in this problem, or maybe it isn't;

The OP stated that they knew the answer was 109, therefore we know that d\ne[/tex]0;<br /> I was merely helping the OP to solve for d=109.<br /> <br /> <blockquote data-attributes="" data-quote="Mark44" data-source="post: 2941267" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Mark44 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Since there is no constant term, you can find the two roots by factoring.<br /> And this is what I did: factor the two terms. </div> </div> </blockquote>Again ... I agreed with you!<br /> <br /> <br /> <blockquote data-attributes="" data-quote="Mark44" data-source="post: 2941267" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Mark44 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Nor did I say that you had to use the quadratic formula in this problem. </div> </div> </blockquote>But you stated that you <i>could</i> use it. Why use the quadratic equation if you don&#039;t have to? I merely point out that you don&#039;t have to. (Still, agreeing with you).<br /> <br /> <br /> <br /> <br /> I thought the point of this post was to help others with homework problems. I was trying to address the question the OP asked which was basically, &quot;How do I get d=109?&quot;<br /> My reply helps him/her to get to that result.