## Yahoo! Group Title 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? 2 years ago 2 years ago

1. jhonyy9 Group Title

can you draw it please ?

2. Yahoo! Group Title

pic us not given..

3. jhonyy9 Group Title

indifferent just for what is your oppinion from this pic

4. heena Group Title

|dw:1339328315880:dw| i gues its like dis

5. apoorvk Group Title

|dw:1339328315376:dw|

6. heena Group Title

my ans is theta=cos-1(1/4)

7. apoorvk Group Title

@heena - wrong answer. Read the question again.

8. heena Group Title

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

9. AndrewNJ Group Title

|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

10. Yahoo! Group Title

The options 60 45 30 15

11. FoolForMath Group Title

The correct answer is $$15^\circ$$

12. Yahoo! Group Title

@FoolForMath u r correct how????? PLzz show it!

13. apoorvk Group Title

lol sorry!! $tan(15^o) = 2 - \sqrt3$ and the other angle will automatically be 90-15 = 75 degrees.

14. FoolForMath Group Title

Well, I can think of a radical free beautiful solution :)

15. Yahoo! Group Title

i did nt understand the method used!

16. apoorvk Group Title

Don't just think, share it with us. @FoolForMath -_-

17. Yahoo! Group Title

@AndrewNJ can u xplain it!

18. AndrewNJ Group Title

in my answer, the equation 1 ab=4x^2 for area of the triangle equals 0.5ab or 0.5*4x*x

19. Yahoo! Group Title

No Idea!

20. AndrewNJ Group Title

the equation 2, a^2+b^2=4ab means Pythagorean Theorem

21. Yahoo! Group Title

Side a is hypot

22. AndrewNJ Group Title

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

23. AndrewNJ Group Title

and half of hypotenuse times its attitude is also the area of triangle

24. AndrewNJ Group Title

so these two part, 0.5ab and 0.5*4x*x can be equal

25. AndrewNJ Group Title

this is the first equation

26. AndrewNJ Group Title

the second is accroading to Pythagorean Theorem can u got it?

27. AndrewNJ Group Title

by the way hypotenuse is the longest side in a right triangle

28. FoolForMath Group Title

|dw:1339329313013:dw| $\frac{AM}{MB }= \tan \alpha$ $\frac{AM}{MC}=\cot \alpha$ Thus,$MB+MC =AM(\cot \alpha +\tan \alpha)$ $\implies BC = 4AM=2AM \frac{1}{ \sin 2 \alpha}$ $\implies \sin 2\alpha =\frac 12 \implies \alpha = 15^\circ$

29. Yahoo! Group Title

@AndrewNJ i have a doubt in ur 1 st eq a^2=4x^2 is that right?

30. AndrewNJ Group Title

my first eq is 0.5ab=4x^2 !

31. AndrewNJ Group Title

@FoolForMath yours looks cool! but i think your method could be a bit difficult

32. FoolForMath Group Title

why? I think it's fairly obvious :)

33. AndrewNJ Group Title

well as far as I'm concerned, i seldom use cot sec :(

34. AndrewNJ Group Title

high school in my country didnt teach us cot sec,only taught sin cos and tan

35. FoolForMath Group Title

which country is that?

36. AndrewNJ Group Title

guess it!

37. FoolForMath Group Title

I am very bad at guessing :(

38. AndrewNJ Group Title

alright, it's china!

39. radar Group Title

@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)

40. Yahoo! Group Title

i also want it @AndrewNJ

41. AndrewNJ Group Title

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

42. radar Group Title

Thanks, you had already shown that, just that, I was being to dense to see it. Thanks again

43. radar Group Title

*to too

44. AndrewNJ Group Title

:)