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#1 2022-11-04 06:39:05

undertaker
Member
Registered: 2022-11-04
Posts: 6

Geometry - Special Right Trangles Help!

19. And the length of the shortest leg is √(12), what is the length of the longest leg?

A. 29
B. 6
C. 38
D. 56
E. 61
F. 17

For this question I got B, but it doesn't work with question 20.

20. Working from #19, what is the length of the hypotenuse?

A. 8.4197
B. 1.9764
C. 10.5742
D. 6.2414
E. 2.4971
F. 6.9282

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#2 2022-11-04 06:40:08

undertaker
Member
Registered: 2022-11-04
Posts: 6

Re: Geometry - Special Right Trangles Help!

this is a 30-50-90 triangle by the way !

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#3 2022-11-04 07:44:05

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 46,234

Re: Geometry - Special Right Trangles Help!

Hi undertaker,

Welcome to the forum!

How can there be a 30-50-90 degrees triangle? The sum must always be 180°.


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

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#4 2022-11-04 09:40:30

undertaker
Member
Registered: 2022-11-04
Posts: 6

Re: Geometry - Special Right Trangles Help!

Whoops sorry, I meant 30-60-90!

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#5 2022-11-04 22:08:31

Bob
Administrator
Registered: 2010-06-20
Posts: 10,163

Re: Geometry - Special Right Trangles Help!

hi undertaker

Welcome to the forum.

If you have an equilateral triangle, ABC, all sides are equal and so are all three angles.  Suppose the sides are each 2 units long. The angles are 60-60-60.

Draw a line from A to the midpoint, D, of BC.  This line is a line of symmetry, so we have the following:

AB = 2, BD = 1, angle ABD = 60, angle BAD = 30, angle ADB = 90.

So triangle ABD is right angled with angles 30-60-90, and sides 2 and 1.  The third side, AD can be calculated using Pythagoras' theorem.

[math]AD^2 = 2^2 - 1^2 = 3 \implies AD = \sqrt{3}[math]



Thus cos ABD = cos 60 = BD/AB = 1/2 = sin 30

Sin ABD = sin 60 = AD/AB = √3/2 = cos 30

and tan ABD = tan 60 = AD/BD = √3/1 =  √3 so tan 30 = 1/√3

Although I started with a 2-2-2 equilateral, it would also be true for any similar shaped triangle such as the one in question 19. 

Noticing that Q20 wants the hypotenuse I'll assume they mean BD = √12 and they want AD.  You can calculate this using tan 60 = AD/BD.

And question 20 using sin 60 = AD / AB.

Bob


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 smile

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