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Triangle

XYZ is a right angle triangle.

XY=8 cm
XZ=6 cm
ZY=10 cm
Angle x = 90 degrees

What is the length of the perpendicular distance of x from ZY

How do I work this question out.

Do I use COS SIN TAN or something.
Original post by zed963
XYZ is a right angle triangle.

XY=8 cm
XZ=6 cm
ZY=10 cm
Angle x = 90 degrees

What is the length of the perpendicular distance of x from ZY

How do I work this question out.

Do I use COS SIN TAN or something.


Yeah you can use sohcahtoa

Specifically that sin(theta) = opp/hyp

Figure out what sin(theta) is for either angle. Now draw in the perpindicular and you should have a new right angled triangle containing theta, which will give you sin(theta) = L/something (where L is the perpendicular distance you're looking for.

Alternatively you can use simultaneous equations, area of triangles and probably more methods. You can ask me about those if you'd like, too.
(edited 11 years ago)
Reply 2
Original post by hassi94
Yeah you can use sohcahtoa

Specifically that sin(theta) = opp/hyp

Figure out what sin(theta) is for either angle. Now draw in the perpindicular and you should have a new right angled triangle containing theta, which will give you sin(theta) = L/something (where L is the perpendicular distance you're looking for.

Alternatively you can use simultaneous equations, area of triangles and probably more methods. You can ask me about those if you'd like, too.


I would equate area. http://www.waldomaths.com/TriArea1NL.jsp
(edited 11 years ago)


Yeah as I said there are a fair few methods to tackle this question :smile:

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