Plane Geometry (1 Viewer)

[FDownes]

Lets see if anyone can help me out of this one;

ABC is a triange, in which AQ bisects angle CAB and angle PBA = angle BCA prove that PB = QB.

 

[lolokay]

< QPB = X + @ (exterior of triangle ABP)
< PQB = X + @ (exterior of triangle LAQ)
therefore BPQ is isosceles with PB=QB

 

[-tal-]

In triangle CAQ and triangle APB:
angle CAQ = angle QAB (AQ bisects angle CAB)
angle ACQ = angle PBA (given)
angle CQA = angle APB (third angle in a triangle)
therefore, triangle CAQ is similar to triangle APB (equiangular)

let angle CQA = angle APB = z

therefore, angle QPB = angle PQB = 180 - z (angle sum of a straight line is 180)

therefore, triangle PQB is isosceles (equal base angles)
therefore, PB = QB

 

[FDownes]

Thanks guys, that really clears that up.

 

[tommykins]

回复: Re: Plane Geometry

< QPB = X + @ (exterior of triangle ABP)
< PQB = X + @ (exterior of triangle LAQ)
therefore BPQ is isosceles with PB=QB

Very well done.

 

[Aerath]

< QPB = X + @ (exterior of triangle ABP)
< PQB = X + @ (exterior of triangle LAQ)
therefore BPQ is isosceles with PB=QB

Woooow, that's like....genius. Nice work.

 

[FDownes]

Got another one for you. I'm having trouble with the second part of this question. I'm not sure if I should use the cosine rule or if there's a better method. It asks;

The triangle ABC is isosceles, with angle BCA = ABC = 72^o and sides AB = AC = 1 unit. BD Bisects angle ABC and BC = x.

a) Show that triangle ABC is similar to triangle DCB

b) By using part a), find the value of x, giving you answer in exact form.

 

[lyounamu]

Got another one for you. I'm having trouble with the second part of this question. I'm not sure if I should use the cosine rule or if there's a better method. It asks;

The triangle ABC is isosceles, with angle BCA = ABC = 72^o and sides AB = AC = 1 unit. BD Bisects angle ABC and BC = x.

a) Show that triangle ABC is similar to triangle DCB

b) By using part a), find the value of x, giving you answer in exact form.

x = sin36/sin72

Triangle BCD is an isoceles triangle because its base angles are same. Therefore, BC = BD.

In the triangle ABC, angle BAD is 36 degrees making the triangle ABD an isoceles triangle. Then you can find the angle ADB by using the rule "all angles in the triangle add up to 180 degrees".

Then use the sine rule: 1/sin72 = BD/sin36
BD = sin36/sin72.

But BD = x (side sides of an isoceles triangle are same)
so x = sin36/sin72

If you can solve the question in anyway, use the method that comes out of your head first. In that way, you will not waste time thinking about what to use.

 

[Aerath]

You can't use the cosine rule because the question explicitly states "by using part a", which involves similar triangles.

You do it by using ratio of corresponding sides, which is a property of similar triangles.
If BC = x, therefore BD = x as well (isos triangle), and similarly, AD = x (isos triangle ADB)

We know that AC = 1, and since AD = x (above), therefore, CD must be 1-x.

Therefore, using ratios, we have
x/(1-x) = x/1
x^2 = 1-x
x^2 +x - 1 = 0

Using quadratic formula:
x = (-1 +- r5)/2

But since x is a length, x>0

Therefore x = (-1+r5)/2

(r = squareroot).

 

[lyounamu]

You can't use the cosine rule because the question explicitly states "by using part a", which involves similar triangles.

You do it by using ratio of corresponding sides, which is a property of similar triangles.
If BC = x, therefore BD = x as well (isos triangle), and similarly, AD = x (isos triangle ADB)

We konw that AC = 1, and since AD = x (above), therefore, CD must be 1-x.

Therefore, using ratios, we have
x/(1-x) = x/1
x^2 = 1-x
x^2 +x - 1 = 0

Using quadratic formula:
x = (-1 +- r5)/2

But since x is a length, x>0

Therefore x = (-1+r5)/2

(r = squareroot).

Yeah, true. But that's not necessarily the case. As long as you use the knowledge earnt from the part a), you can solve the question.

 

[FDownes]

Man, I can't believe I forgot about the sine rule. Thanks.

 

[lyounamu]

Man, I can't believe I forgot about the sine rule. Thanks.

Look at Aerath's, it seems more comprehensive.

 

[Aerath]

Well, using sine/cosine rules would get you the answers. But I don't think when that using trigonometric ratios is 'exact'. I think I remember my Yr 8 teacher telling me that when a question asked for 'exact' answer, you couldn't just leave your answer as sin31 or something.

 

[tommykins]

Well, using sine/cosine rules would get you the answers. But I don't think when that using trigonometric ratios is 'exact'. I think I remember my Yr 8 teacher telling me that when a question asked for 'exact' answer, you couldn't just leave your answer as sin31 or something.

When they give you an exact answer question, 95% (probably more) it involves surds of some form, they rarely give any kind fo question that requires you to have sin/cos's in there, and if they do it's from the exact value triangles anyways.

Using part a) means you have to use the fact that they're similar, lyounamu's method will score 1 or so marks, but if it was out of 2/3, only Aerath's method would score full marks.

 

[FDownes]

Okay guys, got time for one more? Here it is;

The rectangles ABCD and DFBE are congruent with BA = FD = 7 and DA = ED = 3.

a) Show that the length AT = 20/7

b) Find the area of the quadrilateral BWDT.

 

[lyounamu]

Okay guys, got time for one more? Here it is;

The rectangles ABCD and DFBE are congruent with BA = FD = 7 and DA = ED = 3.

a) Show that the length AT = 20/7

b) Find the area of the quadrilateral BWDT.

Prove that BFT and TAD are similar.

So BF = 3.

Now, let AT = x so BT = 7-x

Now, use the Pythagoras' theorem to find x (i.e. AT)
x^2 + 3^2 = (7-x)^2
x^2 + 9 = 49 - 14x + x^2
so 14x = 40
x = 20/7

So AT = 20/7

b) A = bh = (7-20/7) . 3
= 29/7 . 3 = 87/7

 

[lolokay]

Got another one for you. I'm having trouble with the second part of this question. I'm not sure if I should use the cosine rule or if there's a better method. It asks;

The triangle ABC is isosceles, with angle BCA = ABC = 72^o and sides AB = AC = 1 unit. BD Bisects angle ABC and BC = x.

a) Show that triangle ABC is similar to triangle DCB

b) By using part a), find the value of x, giving you answer in exact form.

notice how that triangle allows you to calculate the exact value of cos72 (and therefore many other exact trig ratios)? I reckon that's pretty cool

 

[lyounamu]

notice how that triangle allows you to calculate the exact value of cos72 (and therefore many other exact trig ratios)? I reckon that's pretty cool

Yeah. We should learn how to calculate the exact value of those too.

 

[Aerath]

Yeah. We should learn how to calculate the exact value of those too.

I believe you do (somewhat), in 4U Maths (Complex numbers).

 

[lyounamu]

I believe you do (somewhat), in 4U Maths (Complex numbers).

Really, awesome. Looking forward to doing 4 Unit Mathematics.