Can Someone Please Help Me With This Trig Problem Step By Step

[lacticacid]

A 15 foot ladder is leaning against the side of a house so that the distance on the ground between the base of the ladder and the house is 7 feet.

What is the angle at the base of the ladder is making with the ground?

Write your answer in a complete sentence in terms of this problem and round your answer to two decimal places.

[Tilly2]

Is this any good to you?


https://www.mathsisfun.com/algebra/trig-finding-angle-right-triangle.html

[Zacs-Master]

Approx 22 deg.

[woofgang]

how far have you got on your own?

[bhg481]

Do you know the definition of cosine?

If you look that up you can easily get the angle from that.

[Prudie]

Imagine you have a right angled triangle with the ladder as the hypoteneuse H at 15 ft and the other side of the angle (x) you need is the distance along the gound - the adjacent A. If you have H and A then you use Cosine
Cos X= Adj/Hyp_ = 7/15

Then you get on a calculator. have you got this far?

[Tilly2]

'Write your answer in a complete sentence in terms of this problem and round your answer to two decimal places.'

Giving lacticacid the answer isn't going to help him/her. This is a homework question and he/she needs to show that he/she can work it out for themselves.

[Peter Pedant]

first of all try to stay awake in class...

draw a diagram - on paper and not in your mind
Label the distances

label the angles as needed - greek ones usually alpha beta or theta

there is a trig function ( which is what you are doing in class )
for which the ratio is 7/15
try to decide what it is ( choices - sin, cos, tan to start with )

yes there is no sin on this thread - the answer is cos - because the cos os an angle is the ratio of adjacent/hypotenuse in a rt angled triangle

write cos theta = 7/15
use calculator to find out what 7/15 is - it is around 0.5
write cos theta = 0.5

on your calculator you have to find the angle that has a cosine of 0.5.
watch out ! angles go round as degrees or radians so make sure how you calculator is toggled

find or press cos-1 (0.5) press go! and see what the answer is
it should be around 40 ( degrees ) or 1.6 radians

select the answer you want - let us say it is 40.5678
round it up to 40.57 ( that is 2 dp - remember the rules on dp and rounding )

write down the angle theta whcih is the angle between the ladder and base is 40.57 deg ro 2 dp.

and pay more attention to class next time
I did this in 1965 or thereabouts - but used things called log tables. on around about page 10 there were cosine tables and we did a true table-look-up. o and we would work out 7/15 either with a slide rule ( you see them in museums now ) or the log tables again.

trip down memory lane

[Xeronema]

Well you've lost me with this one. All I know is that I wouldn't get on a ladder at that angle unless you want to end up with a few broken limbs. I'm considering re-re-taking my maths GCSE.

[Peter Pedant]

Tilly -

you didnt see the fly-on-the-wall prog where in class, a over-made up rusty springboard fourteen year old muses on camera...

wha' IS pie ? Where does Pie come from ? Woss it MEAN ?

I cant imagine the math master allowed them to film without weeks of preparation. I also imagine he shot his brains out the day after transmission....

[Peter Pedant]

No xero

the problem where you have someone standing on the ladder and you are working out whether it slips or not - is a problem in statics

First year A level - and we havent got there yet, altho you are right there is a cos ( or perhpas a sine ) in the answer.

[sandyRoe]

the base of this ladder could be braced against the garden wall. Safe enough then.

[Prudie]

That's odd because I make it 62.18 degrees

[Xeronema]

I'm not amazing at maths however it's obvious it's under 45 degrees. I'm working it out in a practical way though as apposed to a mathematical one. If the ladder was 14 feet then it would be twice the length of its distance from the wall - therefore I'd half 90%. In this case it's longer and must therefore be under a 45% radios. SandyRoe; That tickled me ;)

[Old_Geezer]

This is simply Pythagoras is it not ?

The side of the house is assumed to be 90º to the ground. So you have a right angled triangle with a hypotenuse of 15, and adjacent of 7. And an opposite of sq rt ((15*15) - (7*7)). Just use your sine/cosine/tangent tables, or a calculator. Then write your answer as an English sentence.

[bhg481]

I agree with Prudie.

[Prudie]

Thank you bhg

[Prudie]

And OG not it's not simply Pythagoras unless you particularly want to work it out using sine rather than cosine

[Old_Geezer]

Yes I realised it was not only Pythagoras when I hit the submit, but all this is related. Doesn't one learn it all in the same class lesson ?

[Prudie]

You certainly do although I'd expect Pythagoras to come a few weeks before trig

[Peter Pedant]

No I was 12 when I first met sin .... no er
I was 12 when we first learnt about sine ratios

and we had done pythagoras for a few years before that

He should have started with the second q
it is more interesting
but I think we all have home-work fatigue including the q setter