Math Is Fun Forum

apsara123

Member

Registered: 2016-05-14

Posts: 20

Help With Problem

I need help with these problems...can you give detailed solution. Thank you!

1)In right triangle $MNO$, $\tan{M}=\frac{5}{4}$, $OM=8$, and $\angle O = 90^\circ$. Find $MN$. Express your answer in simplest radical form.

2)Points $A$ and $B$ are selected on the graph of $y = -\frac{1}{2}x^2$ so that triangle $ABO$ is equilateral. Find the length of one side of triangle $ABO$.

IMAGE:http://imgur.com/a/gLRIK

3)Find $\cos (-135^\circ)$.

MathWA

Member

From: Somewhere over the rainbow

Registered: 2016-05-22

Posts: 12

Website

Re: Help With Problem

I need help with these problems...can you give detailed solution. Thank you!

1)In right triangle $MNO$, $\tan{M}=\frac{5}{4}$, $OM=8$, and $\angle O = 90^\circ$. Find $MN$. Express your answer in simplest radical form.

2)Points $A$ and $B$ are selected on the graph of $y = -\frac{1}{2}x^2$ so that triangle $ABO$ is equilateral. Find the length of one side of triangle $ABO$.

IMAGE:http://imgur.com/a/gLRIK

3)Find $\cos (-135^\circ)$.

Your latex or bbcode is not working! Can you fix that?

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We cannot solve our problems with the same thinking we used when we created them.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,163

Re: Help With Problem

hi apsara123

Q1.  If you sketch this triangle, you'll find that NO = 10 and MO = 8.  So you can use Pythagoras to calculate MN.

Q2.Let M be the midpoint of AB.  Then angle OAM = 60 so (ignoring negatives) y/x = root 3.  Together with the equation of the curve you can eliminate x and so determine y.  Then x and then AB = 2x.

Q3.  There is an excellent page on trig. ratios above 90 degrees here:

http://www.mathsisfun.com/algebra/trig- … rants.html

Note about your code $ \ etc.  For these problems all that was unnecessary.  I understood all three problems just by ignoring the code.  On this forum, if you want 'fancy' maths to appear you need to use Latex.  See here:

http://www.mathisfunforum.com/viewtopic.php?id=4397

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Help With Problem

Edited with least change.

Last edited by thickhead (2016-07-28 21:28:18)

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
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insert_username

Member

Registered: 2022-02-25

Posts: 1

Re: Help With Problem

Sorry I can only answer your first problem.

We can imagine a right triangle with right angle 'O'. The problem tells us that line 'OM' is 8. It also says that tan M = 5/4. Tangent is opposite/adjacent. The opposite, which is 'NO' is unknown, so we let that be 'X'. Adjacent, which is 'OM' we already know is 8. Now, we see that 5/4 = X/8. 4*2=8, therefore 'X' (or 'NO') = 10. Now that we know the other two sides of the triangle, we can use the Pythagorean Theorem to find that 'MN' = 2* sqrt(41).