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Triangle Question help

matthew042

New member
Joined
Feb 8, 2009
Messages
15
I have this question that has been driving me nuts. There is given triangle ABC. AB = 20. Angle A is 30 degrees and angle B is 45 degrees. I thought relating the angles to the opposite sides would do but i was wrong. I tried dividing 45/20 and using that number to divide by 30 to get the answer. I was wrong apparently. Please help. Thanks!
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,284
20 / SIN(105) = AC / SIN(45) = BC / SIN(30)
 

matthew042

New member
Joined
Feb 8, 2009
Messages
15
Is there any way to do it without any trig functions like sin and cos? I don't quite understand that. Thanks to everyone helped.
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,284
matthew042 said:
Is there any way to do it without any trig functions like sin and cos? I don't quite understand that.
Well, you did state: "I thought relating the angles to the opposite sides would do but i was wrong."
That means using SIN ! Did you at least understand the "105" in 20 / SIN(105) ?

There is another way.
Draw a representative triangle. Then draw a perpendicular like CD with D on line AB.
You now have 2 right triangles: ACD with angles 30-60-90 and BCD with angles 45-45-90.
With a 30-60-90 triangle, the hypotenuse = twice shorter side, so AC = 2 * CD.
If you let x = BD, then CD = x, AD = 20-x and AC = 2x.
Try that.
 
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