Help with trigonometry homework

[NeatSquidYT]

'Sup LTT,

I'm having issues with my trig. homework.

https://answers.yahoo.com/question/index?qid=20150109133003AAobcH3

 

Answers on the page or this thread are appreciated. Thanks.

[werto165]

20*tan(19) AFAIK 

 

if it's a bi-sector it the angle below should be 19o thus tan(19) = x/20 from soh cah toa.

 

[NeatSquidYT]

20*tan(19) AFAIK

No, that wouldn't work afaik. You need to work out the bottom triangle's angle

[Ciccioo]

No, that wouldn't work afaik. You need to work out the bottom triangle's angle

with it being a bisector, the bottom angle is 19°

so you can apply the formula that is 20*tan(19)

[Tacitus]

EDIT: Nvm, just read what you added.

[NeatSquidYT]

with it being a bisector, the bottom angle is 19°

so you can apply the formula that is 20*tan(19)

 

20*tan(19) AFAIK 

 

if it's a bi-sector it the angle below should be 19o thus tan(19) = x/20 from soh cah toa.

 

I can try it.

[werto165]

No, that wouldn't work afaik. You need to work out the bottom triangle's angle

a bisector is a line that divides something into two equal parts.

[mansoor_]

The middle line can not be an angle bisector according to the other dimensions. 

i.e 2*(20tan(19)) != 20tan(19*2)

I have the two equations but i cannot be asked to solve them, far too tedious.

tan(y) = 10/x 

sin(19)*sqrt(400+4x^2) = x*sin(19+y)

{edit - I have defined the angle at the top corner as y}

solution :http://www.wolframalpha.com/input/?i=solve+simultaneous+equation+tan%28y%29+%3D+10%2Fx+%2C+sin%2819%29*sqrt%28400%2B4x%5E2%29+%3D+xsin%2819%2By%29

[Ciccioo]

The middle line can not be an angle bisector according to the other dimensions. 

I have the two equation but i cannot be asked to solve them, far too tedious.

tan(y) = 20/x 

sin(19)*sqrt(400-4x^2) = x*sin(19+y)

he's pretty damn right

that is a median, not a bisector

[colonel_mortis]

I might be wrong, and I'm sure there's a faster way to do it, but this is my proof:

Given that a is the corner with the angle, b is the right angle, c is the other corner of the triangle and m is the centre of bc:

 

• Split the top triangle so that the new line is perpendicular to ab and meets bc at m. Label the point where that line intersects ab as n.
• Using trig, prove that nm = xcos19
• Using trig on the right angled triangle anm, prove that am is equal to xcos19/sin19, which also equals xcot19 if you have done cosecs.
• Using pythag on triangle abm, 20²+x²=(xcot19)²
• therefore, 20²+x²=x²cot²19=8.43x²
• subtract x² from both sides (and swap the sides) to give 7.43x²=400
• x²=58.84
• x=7.34 (m).

I have probably made a mistake somewhere in here (especially rounding errors), but hopefully this should give you a starting point for you to then get it right.

Hope I'm not too late.

[NeatSquidYT]

Could someone solve this: (it's the same but with different measurements)

 

[colonel_mortis]

Could someone solve this: (it's the same but with different measurements)

• It's an angle bisector this time, so BAD is also 20 degrees
• BC = 24tan(40) = 20.14
• BD = 24tan(20) = 8.74
• x = 20.14-8.74 = 11.40 cm

[NeatSquidYT]

 

• It's an angle bisector this time, so BAD is also 20 degrees
• BC = 24tan(40) = 20.14
• BD = 24tan(20) = 8.74
• x = 20.14-8.74 = 11.40 cm

 

I'll be back in a min with the verdict

[NeatSquidYT]

 

• It's an angle bisector this time, so BAD is also 20 degrees
• BC = 24tan(40) = 20.14
• BD = 24tan(20) = 8.74
• x = 20.14-8.74 = 11.40 cm

 

You're a lifesaver. Thanks so much. I got my whole family involved in this, including my sister's boyfriend