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third method should be cosine formula, not sine, you are lucky because at 45 degree it is equal
Yes, it should be cosine instead of sine. Sorry for the mistake, I didn't notice it while solving the problem.
Bhai woh maths ke teacher hai agar algorithm main koi mistake hot tab tum bolte😊
Law of cosine is a lot more reliable anyway. No chance for errors due to pain in the SSA TRIANGLES.
in any triangle a^2= b^2+c^2 -2bc*cos
Yes, there should be cosine instead of sine. Sorry for the mistake.
A simple way to solve this would be to draw a perpendicular CD on AB. BD = CD = 5, AD = 17-BD = 12AC² = CD² + AD²AC = 13
Nice! sin(CDB) = 1 → BD = 5 → AD = 12 → AC = 13or: sin(β) = cos(β) = √2/2 → (AC)^2 = (17)^2 + (5√2)^2 - 2(17)(5√2)cos(β) = 339 - 170 = 169 → AC = 13
by cosine lawAC=13
In 《BCD》two lengths hypotunese BD=CD =5. (1,1,1/sqrt2)1/sqrt2=AD/17 17/sqrt2=AD AD=12 similarth triangle《ADB》~《BCD》BD=12 In 《ADC》 AC²=AD²+CD² AC=13 1/sqrt2*sin(90)=289+50-169/160*sqrt2 sin(90)=289+50-AC²/170 339-AC²/170=1 339-AC²=170 AC²=169AC=13
Without Pythagoras: 5*sqrt 2 is a diagonal of a Square, Hence 5 is BD and likewise DC!
Draw AD perpendicular to BC Hereby AC^2 = AD^2 + ( BD - CD)^2 = AD^2 + BD^2 +CD^2 - 2 BD x CD =AB^2 + CD^2 - 2(BD/AB) AB x CD =17^2 + (5√2)^2 - √2 x 5 √2 x 17 = 17 x 7 + 50 = 169 AC = 13
It is cosine formula. Please note. Cos45 is correct.
Cosine law allowed ?
Yes, i have written sin instead of cosine (in 3rd method). Sorry for the mistake.
Nice
It seems that you like the (5,12,13) triple today.
Por teorema del coseno
a² = b² +c² -2bc cos A child's play
third method should be cosine formula, not sine, you are lucky because at 45 degree it is equal
Yes, it should be cosine instead of sine. Sorry for the mistake, I didn't notice it while solving the problem.
Bhai woh maths ke teacher hai agar algorithm main koi mistake hot tab tum bolte😊
Law of cosine is a lot more reliable anyway. No chance for errors due to pain in the SSA TRIANGLES.
in any triangle a^2= b^2+c^2 -2bc*cos
Yes, there should be cosine instead of sine. Sorry for the mistake.
A simple way to solve this would be to draw a perpendicular CD on AB.
BD = CD = 5, AD = 17-BD = 12
AC² = CD² + AD²
AC = 13
Nice! sin(CDB) = 1 → BD = 5 → AD = 12 → AC = 13
or: sin(β) = cos(β) = √2/2 →
(AC)^2 = (17)^2 + (5√2)^2 - 2(17)(5√2)cos(β) = 339 - 170 = 169 → AC = 13
by cosine law
AC=13
In 《BCD》two lengths hypotunese BD=CD =5.
(1,1,1/sqrt2)
1/sqrt2=AD/17
17/sqrt2=AD
AD=12
similarth triangle《ADB》~《BCD》
BD=12
In 《ADC》 AC²=AD²+CD²
AC=13
1/sqrt2*sin(90)=289+50-169/160*sqrt2
sin(90)=289+50-AC²/170
339-AC²/170=1
339-AC²=170
AC²=169
AC=13
Without Pythagoras: 5*sqrt 2 is a diagonal of a Square, Hence 5 is BD and likewise DC!
Draw AD perpendicular to BC
Hereby
AC^2 = AD^2 + ( BD - CD)^2
= AD^2 + BD^2 +CD^2 - 2 BD x CD
=AB^2 + CD^2 - 2(BD/AB) AB x CD
=17^2 + (5√2)^2 - √2 x 5 √2 x 17
= 17 x 7 + 50 = 169
AC = 13
It is cosine formula. Please note. Cos45 is correct.
Cosine law allowed ?
Yes, i have written sin instead of cosine (in 3rd method). Sorry for the mistake.
Nice
It seems that you like the (5,12,13) triple today.
Por teorema del coseno
a² = b² +c² -2bc cos A child's play