A toddler’s playground has two slides on either side. The slide on the left is situated at a 45° angle to the ground and is 52 centimeters away from the base of the playground. The slide on the right is positioned at a 30° angle to the ground and is 98 centimeters away from the playground’s base. What is the total length of both slides on this playground rounded to the nearest tenth of a centimeter?

Question

anonymous · Answer

@ParthKohli Can you help her with this?

parthkohli · Answer

Hmm can you make a diagram for me? I'm not good at understanding verbally :(

anonymous · Answer

attachment

parthkohli · Answer

Oh this is pretty easy. Do you know trigonometric ratios? If not I'll explain this in another way.

parthkohli · Answer

I prefer trigonometric ratios btw.

anonymous · Answer

I don't. I need this question too.

anonymous · Answer

Would the answer happen to be 186.7 or is that completely off?

anonymous · Answer

you can just use cosine to find 2 lengths

anonymous · Answer

well i agree trig ratio is faster but cosine is easier

parthkohli · Answer

Okay, we have to determine the hypotenuse in both triangles.
In the slide towards the left, we'll use cos.

$\Large \color{Black}{\Rightarrow cos45 = {45 \over x}}$

anonymous · Answer

Kk

parthkohli · Answer

xsec45 = 45
We get sqrt2/1 as the sec45, so the hypotenuse would be sqrt2 * 52 = 52sqrt2

anonymous · Answer

Got that.

parthkohli · Answer

Then the right slide.
You use cos again.(cos of 30)

anonymous · Answer

98/cos(30)

anonymous · Answer

What do I add up now? @ParthKohli

parthkohli · Answer

Oops my browser crashed

anonymous · Answer

I'm sorry. That always sucks.

anonymous · Answer

Could you show me how to get the answeR? I have to go to bed soon.

anonymous · Answer

This is my last question.

parthkohli · Answer

$\Large \color{Black}{\Rightarrow cos30 = {98 \over x} }$

Cos30 = sqrt3/2, so

$\Large \color{Black}{\Rightarrow {\sqrt3 \over 2} = {98 \over x} }$

$\Large \color{Black}{\Rightarrow x\sqrt3 = 196 }$

$\Large \color{Black}{\Rightarrow x = {196 \over \sqrt3} }$

$\Large \color{Black}{\Rightarrow x = {196\sqrt{3} \over 3} }$

parthkohli · Answer

Well that's it

parthkohli · Answer

Wait a second. That answer should be 196...

parthkohli · Answer

196 + 52sqrt2

parthkohli · Answer

I did a mistake there.

parthkohli · Answer

$\Large \color{Black}{\Rightarrow  x\sqrt3 = 196}$
Yes, I was wrong

parthkohli · Answer

The answer is 196 + 52sqrt2