How can I find length of AB and DB? (ABC is a triangle in which m(<B)=90...)

[FireBall]

Can someone please explain how do I solve this? How can I find length of AB and DB?

 

[Dr.Peterson]

View attachment 34778

First, think about how [imath]\sin(A)[/imath] and [imath]\cos(C)[/imath] are related. (That is, first, how are angles A and C related?)

Then think similarly about [imath]\sin(C)[/imath] and [imath]\cos(\angle EDC)[/imath].

 

[The Highlander]

View attachment 34778

Please show us your attempt at proving the required result (after taking into account the good advice provided above). Then we can then confirm your work as correct or offer further advice on how to proceed.

(No one will do it for you in here but we will happily help you along to getting the answer by yourself.)

One final hint: you should be working towards the "Trig Identity": "sin² θ + cos² θ = 1".
(& you don't need to: "find length of AB and DB" )

 

[Dr.Peterson]

Can someone please explain how do I solve this? How can I find length of AB and DB​

View attachment 34778

I hadn't noticed the title.

Interestingly, we not only don't need to find the lengths of AB and DB, we can't! You could slide A and B along lines CE and CD, and it would have no effect on either the consistency of the data, or the answer, as long as AB remains perpendicular.

In fact, we don't even need to know the lengths of DE and EC that are given. There is a lot of extraneous information, which is perhaps put there to distract you. In such a situation, a key idea in problem-solving is to focus on the goal more than the givens (of which there are too many, as in real life).

 

[Steven G]

When I see the sum of the products of two trig functions I immediately think of....?
Pythagoras would be able to help you if he was still alive.

 

[The Highlander]

When I see the sum of the products of two trig functions I immediately think of....?

Escape from the corner?

Pythagoras would be able to help you if he was still alive.

I'm curious to know what "help" he might provide other than finding the length of CD (which doesn't seem to enable any progress in proving what's required); he wasn't big on trig, was he? Care to explain?

 

[pka]

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@FireBall, it seems that you have managed to confuse all here.
In the image you posted it says Prove that : [imath]\sin(A)\cos(C)+\sin(C)\cos\left( \angle D\right)=1[/imath].
[I renamed [imath]\angle EDC~~\angle D[/imath]] Following Pederson's advice, take note that:
[imath]\angle\text{s }A \& C\text{ are complementary}\text{ as are }\angle \text{s }C~ \& ~D [/imath].
Does that lead you to the fact that [imath]m(\angle(A))=m(\angle(D))~?[/imath]
Now can you complete the question?



[imath][/imath][imath][/imath][imath][/imath][imath][/imath]imath][/imath]

 

[Steven G]

Escape from the corner?

I'm curious to know what "help" he might provide other than finding the length of CD (which doesn't seem to enable any progress in proving what's required); he wasn't big on trig, was he? Care to explain?

OK, you got me real good!! Good job!!!
I meant think of the "Trig Identity": "sin² θ + cos² θ = 1"
I'll be in the corner for this one for a long time.

 

[The Highlander]

OK, you got me real good!! Good job!!!
I meant think of the "Trig Identity": "sin² θ + cos² θ = 1"
I'll be in the corner for this one for a long time.

Sometimes it's the best place to be; all the friendliest folks gather there. ???

 

[The Highlander]

@FireBall, it seems that you have managed to confuse all here.

I didn't realise anyone here was "confused"... until I read below. ?

In the image you posted it says Prove that : [imath]\sin(A)\cos(C)+\sin(C)\cos\left( \angle D\right)=1[/imath].
[I renamed [imath]\angle EDC~~\angle D[/imath]] Following Pederson's advice, take note that:
[imath]\angle\text{s }A \& C\text{ are complementary}\text{ as are }\angle \text{s }C~ \& ~D [/imath].

That's all good and the fact that they're complementary is the important thing to recognize.

Does that lead you to the fact that [imath]m(\angle(A))=m(\angle(D))~?[/imath]

Surely, that's another 'redundant' fact; even more confusing (to the OP?) than the suggestion that Pythagoras' theorem is relevant here! The equivalence of those two angles' measures has little bearing on how to prove what's required; it's already clear (from the diagram) that ∠D is the complement of ∠C, there's no need to show it's the same size as ∠A! Sending the OP off down that track is just as much of a wild goose chase as finding side lengths!

Now can you complete the question?

The essence of this problem is just the basic Trigonometric Identities! What the OP needs to do is recognize how the Sine of an angle compares to the Cosine of its complement, ie: (for the OP's benefit) how does sin θ° compare to cos (90° - θ°) and vice versa; as (strongly) hinted at in the first two responses. ?

Final hint: What is sin 30°? And what is cos 60°? IE: they are the ....?

 

[FireBall]

Hello, sorry I'm late, just saw the notifications,

Solution:

Since cos C = 4/5 sin C=3/5
since sin A = cos C
therefore sin A= 4/5

Therefore ((4/5)*(4/5)) + ((3/5)+(3*5))
= (16/28)+(9/25)
=25/25
=1

The answer was right under my nose (sorry?‍), Thank you very much for your help. I thought the functions get affected by the length of the sides, later I found they remain the same as long as the angle is the same size.

 

[The Highlander]

Solution:
Since cos C = 4/5 sin C=3/5
since sin A = cos C
therefore sin A= 4/5

Therefore ((4/5)*(4/5)) + ((3/5)+(3*5)) ((3/5)*(3*5))
= (16/28)+(9/25)
=25/25
=1

Your answer would be acceptable (apart from the small mistake I have highlighted above) but I suspect that the question writer may have intended that you “prove” the issue without resort to actually entering numerical values. ?
There is nothing wrong with that approach and it is always a good idea to enter numbers into your work (where possible) to check that it is 'correct'. Your approach certainly reaches the desired outcome but might be viewed more as “verification” rather than “proof”. ?

A more general 'proof', that relies solely on the trigonometric identities, would (IMNSHO ) be preferable. For example, you might write:-

Replacing “∠EDC” with just “D” the equation becomes:-

sin A cos C + sin C cos D = 1​

but A = 90 – C ∴ sin A = cos C       (∵ sin θ = cos (90 - θ) and vice versa)
and D = 90 – C ∴ cos D = sin C

So it follows that:-

sin A cos C + sin C cos D = cos C cos C + sin C sin C = cos ² C + sin ² C = 1. QED. ??​

The answer was right under my nose (sorry?‍), Thank you very much for your help. I thought the functions get affected by the length of the sides, later I found they remain the same as long as the angle is the same size.

Indeed, that is the whole basis of Trigonometry!​

Hope that helps. ?