In a right triangle , if the hypotenuse is four times as long as the perpendicular from the opposite vertex , one of the acute angles is?

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In a right triangle , if the hypotenuse is four times as long as the perpendicular from the opposite vertex , one of the acute angles is?

Mathematics
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can you draw it please ?
pic us not given..
indifferent just for what is your oppinion from this pic

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Other answers:

|dw:1339328315880:dw| i gues its like dis
|dw:1339328315376:dw|
my ans is theta=cos-1(1/4)
@heena - wrong answer. Read the question again.
i did and i dont find any prob as its saying hypotenuse is four times as long as the perpendicular means we cas asume perpendicular as x and hypotenuse as 4x
|dw:1339328062826:dw| i draw the pic u can find ab=4x^2 ---1 and a^2+b^2=16x^2 ----2 and use ab to represent x^2 at equation 2 we can have a^2+b^2=4ab devide ab two sides, that is a/b+b/a=4 we can let a/b=tanx (x could be the angle of one of two unknown angles) so tanx+1/(tanx)=4 let tanx=n so n+1/n=4, we can rewrite it as n^2-4n+1=0 and \[n=2\pm \sqrt{3}\] \[\tan x= 2\pm \sqrt{3}\] and u can work out the value of x that is the angle
The options 60 45 30 15
The correct answer is \(15^\circ \)
@FoolForMath u r correct how????? PLzz show it!
lol sorry!! \[tan(15^o) = 2 - \sqrt3\] and the other angle will automatically be 90-15 = 75 degrees.
Well, I can think of a radical free beautiful solution :)
i did nt understand the method used!
Don't just think, share it with us. @FoolForMath -_-
@AndrewNJ can u xplain it!
in my answer, the equation 1 ab=4x^2 for area of the triangle equals 0.5ab or 0.5*4x*x
No Idea!
the equation 2, a^2+b^2=4ab means Pythagorean Theorem
Side a is hypot
ok, dont be serious about a and b which is longer, it doesnt matter a is perpendicular to b so 0.5ab is the area of this triangle
and half of hypotenuse times its attitude is also the area of triangle
so these two part, 0.5ab and 0.5*4x*x can be equal
this is the first equation
the second is accroading to Pythagorean Theorem can u got it?
by the way hypotenuse is the longest side in a right triangle
|dw:1339329313013:dw| \[\frac{AM}{MB }= \tan \alpha \] \[ \frac{AM}{MC}=\cot \alpha \] Thus,\[MB+MC =AM(\cot \alpha +\tan \alpha) \] \[\implies BC = 4AM=2AM \frac{1}{ \sin 2 \alpha}\] \[\implies \sin 2\alpha =\frac 12 \implies \alpha = 15^\circ \]
@AndrewNJ i have a doubt in ur 1 st eq a^2=4x^2 is that right?
my first eq is 0.5ab=4x^2 !
@FoolForMath yours looks cool! but i think your method could be a bit difficult
why? I think it's fairly obvious :)
well as far as I'm concerned, i seldom use cot sec :(
high school in my country didnt teach us cot sec,only taught sin cos and tan
which country is that?
guess it!
I am very bad at guessing :(
alright, it's china!
@AndrewNJ, I am from Missouri, the "show me" state lol, and I would like you to please show me how ab=4x^2 (equation 1 in your first post)
i also want it @AndrewNJ
for 0.5ab is the area of triangle and 0.5*4x*x is also the area of triangle so 0.5ab=0.5*4x*x divide 0.5 in two sides is ab=4x^2
Thanks, you had already shown that, just that, I was being to dense to see it. Thanks again
*to too
:)

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