@ 10:49 if we drop a perpendicular from Q on PB we will get two triangles from which one is similar to PNA and another is similar to a triangle buit by another perpendicular from C on PA. Then simple ratios lead to finding all sides.
Let BQ = y, BP = z. Then apply Stewart's Theorem to triangles BPC (cevian is PQ) and ABC (cevian is CP) and Pythagoras to triangle ABC. You get three equations in three unknowns (x, y, z). I'd never try to do them by hand but it presented no problem for Wolfram Alpha. With x = sqrt(41), y = 30sqrt(41) and z = 200 getting the required area is straight forward.
Good demonstration. I proceeded differently: We can easily express PC in terms of X since PQC is a right triangle. We can easily demonstrate angles PQC and ACB are equal, the cosine of which will be 10/2X. From there we can easily determine that the value of X = sqrt(41) from application of the cosine law in triangle ACP. We can then determine the cosine of angle PAC using the cosine law, From there, it is easy to determine the total length of AB and BC, and sub-length BQ. We can then determine the height of triangle BPQ by drawing a line perpendicular to BC and joining point P, the height being 40/sqrt(41).
If cos β = 10/2x (Angle CPQ=90°) then 5/x = cos β but angle APC is not 90°. There's some mistake in the statement of this exercise !!! 5 can't be 5 or 10 can't be 10
If cos β = 10/2x (Angle CPQ=90°) then 5/x = cos β but angle APC is not 90°. There's some mistake in the statement of this exercise !!! 5 can't be 5 or 10 can't be 10 Besides of this, a little mistake in minute 2:08 and 18:00 and others "Area = 5 times segment BM" ???? impossible. Should said "5 cm (or units) multiplied by segment BM" This is the consequence of using "units" instead of the international metric system "cm or meters" "Times" is times, is not "multiplied by", if you don't write the length unit
Nice to add BM to make another pair of similar triangles. Quite an uphill challenge to get k, the ratio between those second two. Thank you for another interesting and clear explanation.
My strategy starts from the observation that the angles QCP and PCA are complementary, therefore the angles PQC and PCA are equal. Let PC = y and the angles PQC=PCA= alpha, we can write three equations with the help of the Pythagorean theorem and the cosine rule. 1) y² = 4x² - 10² 2) 10² + (2x)² - 2*10*2x*cos alpha = y² 3) x² + y² -2xy*cos alpha = 5² Solving the system (finding before all cos alpha) we get: y = PC = 8 x = AC = √ 41 QC = 2√ 41 Now we can find the height PH of triangle PQC doing (h=2A/b): PH = 10*8/2√ 41 = (40√ 41)/41 Finally knowing that triangle ABC and PBH are similar we can write: AC : PH = BC : BH and being QH = PQ²/QC (Euclid's theorem) QH = 50/41√ 41 √ 41 : (40√ 41)/41 = (BQ + 2√ 41) : (BQ + 50/41√ 41) BQ = 30√ 41 Area = 1/2*30√ 41* (40√ 41)/41 = 600
If cos β = 10/2x (Angle CPQ=90°) then cos β = 5/x but angle APC is not 90°. There's some mistake in the statement of this exercise !!! 5 can't be 5 or 10 can't be 10 Or who can explain this??
@@marioalb9726 Angles have sides of triangle for their trigonometric ratio values only applies to right-angled triangles such as triangles CPQ ad ANC but not non-right-angled triangle such as triangle APC though angle beta is one of its angles. Actually sin(APC) = xsin(beta)/5 by sine law for triangle APC. From triangle CPQ, x = sqrt41 and sin(beta) = 8/2sqrt41. Hence sin(APC) = 4/5. Angle APC = 53.13
@@hongningsuen1348 Angle β was labelled by me. I didn't realize that video already had other β I labelled β the angle PQC I wrote "cosβ = 10/(2x) = 5/x" And triangle PQC is right triangle But triangle APC is not right triangle. AP/AC= 5/x = cosβ ??????
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@ 10:49 if we drop a perpendicular from Q on PB we will get two triangles from which one is similar to PNA and another is similar to a triangle buit by another perpendicular from C on PA. Then simple ratios lead to finding all sides.
Let BQ = y, BP = z. Then apply Stewart's Theorem to triangles BPC (cevian is PQ) and ABC (cevian is CP) and Pythagoras to triangle ABC. You get three equations in three unknowns (x, y, z). I'd never try to do them by hand but it presented no problem for Wolfram Alpha. With x = sqrt(41), y = 30sqrt(41) and z = 200 getting the required area is straight forward.
Good demonstration. I proceeded differently: We can easily express PC in terms of X since PQC is a right triangle. We can easily demonstrate angles PQC and ACB are equal, the cosine of which will be 10/2X. From there we can easily determine that the value of X = sqrt(41) from application of the cosine law in triangle ACP. We can then determine the cosine of angle PAC using the cosine law, From there, it is easy to determine the total length of AB and BC, and sub-length BQ. We can then determine the height of triangle BPQ by drawing a line perpendicular to BC and joining point P, the height being 40/sqrt(41).
If cos β = 10/2x (Angle CPQ=90°)
then 5/x = cos β
but angle APC is not 90°.
There's some mistake in the statement of this exercise !!!
5 can't be 5 or 10 can't be 10
If cos β = 10/2x (Angle CPQ=90°)
then 5/x = cos β
but angle APC is not 90°.
There's some mistake in the statement of this exercise !!!
5 can't be 5 or 10 can't be 10
Besides of this, a little mistake in minute 2:08 and 18:00 and others
"Area = 5 times segment BM" ???? impossible.
Should said "5 cm (or units) multiplied by segment BM"
This is the consequence of using "units" instead of the international metric system "cm or meters"
"Times" is times, is not "multiplied by", if you don't write the length unit
Nice to add BM to make another pair of similar triangles. Quite an uphill challenge to get k, the ratio between those second two. Thank you for another interesting and clear explanation.
My strategy starts from the observation that the angles QCP and PCA are complementary, therefore the angles PQC and PCA are equal. Let PC = y and the angles PQC=PCA= alpha, we can write three equations with the help of the Pythagorean theorem and the cosine rule.
1) y² = 4x² - 10²
2) 10² + (2x)² - 2*10*2x*cos alpha = y²
3) x² + y² -2xy*cos alpha = 5²
Solving the system (finding before all cos alpha) we get:
y = PC = 8
x = AC = √ 41
QC = 2√ 41
Now we can find the height PH of triangle PQC doing (h=2A/b):
PH = 10*8/2√ 41 = (40√ 41)/41
Finally knowing that triangle ABC and PBH are similar we can write:
AC : PH = BC : BH
and being QH = PQ²/QC (Euclid's theorem)
QH = 50/41√ 41
√ 41 : (40√ 41)/41 = (BQ + 2√ 41) : (BQ + 50/41√ 41)
BQ = 30√ 41
Area = 1/2*30√ 41* (40√ 41)/41 = 600
I did the same way as maisonville. However your method is very elegant.🎉
The picture is too remote from reality. BP = 200 as calculated while AP = 5 as given. A 40x difference!
It will be impossible to make meaningful to-the -scale drawing of ∆ABC.
If cos β = 10/2x (Angle CPQ=90°)
then cos β = 5/x
but angle APC is not 90°.
There's some mistake in the statement of this exercise !!!
5 can't be 5 or 10 can't be 10
Or who can explain this??
@@marioalb9726 Angles have sides of triangle for their trigonometric ratio values only applies to right-angled triangles such as triangles CPQ ad ANC but not non-right-angled triangle such as triangle APC though angle beta is one of its angles. Actually sin(APC) = xsin(beta)/5 by sine law for triangle APC. From triangle CPQ, x = sqrt41 and sin(beta) = 8/2sqrt41. Hence sin(APC) = 4/5. Angle APC = 53.13
@@hongningsuen1348
Angle β was labelled by me.
I didn't realize that video already had other β
I labelled β the angle PQC
I wrote "cosβ = 10/(2x) = 5/x"
And triangle PQC is right triangle
But triangle APC is not right triangle.
AP/AC= 5/x = cosβ ??????