chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Try sleeping?
anonymous
  • anonymous
try practicing art like painting or drawing
ParthKohli
  • ParthKohli
Reciting a question from my memory: \(ABC\) is a triangle and it's given that \(b>c\). \(AD\) is an altitude and is given to be equal to \(\dfrac{abc}{b^2 - c^2}\). If \(C = 23^{\circ} \) then find angle \(B\).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ParthKohli
  • ParthKohli
|dw:1438169365378:dw|
ganeshie8
  • ganeshie8
|dw:1438169558232:dw|
ParthKohli
  • ParthKohli
\[c \sin B = \frac{abc}{b^2 - c^2}\]\[\sin B = \frac{bc}{b^2 - c^2}\]
ganeshie8
  • ganeshie8
Oh you re looking for numerical value of angle B is it
ParthKohli
  • ParthKohli
Yes.
ganeshie8
  • ganeshie8
|dw:1438169798797:dw|
ganeshie8
  • ganeshie8
thats just sine law anyways
Astrophysics
  • Astrophysics
it's that b>c
ganeshie8
  • ganeshie8
b > c implies B > 23
Astrophysics
  • Astrophysics
Yeah Ik, did you get the numerical value?
ganeshie8
  • ganeshie8
no, still clueless..
ganeshie8
  • ganeshie8
how this \[\frac{abc}{b^2-c^2}=b\sin{c}\] gives this \[\sin{A}=\sin^2{B}-\sin^2{C}\] ?
ganeshie8
  • ganeshie8
I mean how this \[\frac{abc}{b^2-c^2}=b\sin{\color{red}{C}}\] gives this \[\sin{A}=\sin^2{B}-\sin^2{C}\] ?
ganeshie8
  • ganeshie8
amilapsn
  • amilapsn
\[\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=k\]
amilapsn
  • amilapsn
\[\therefore a=k\sin{A},b=k\sin{B},c=k\sin{C}\]
amilapsn
  • amilapsn
\[\therefore \frac{abc}{b^2-c^2}=b\sin{C}\Rightarrow \sin{A}=\sin^2{B}-\sin^2{C}\]
ganeshie8
  • ganeshie8
With you so far !
amilapsn
  • amilapsn
\[\Rightarrow \sin{(B-C)}=1\Rightarrow B=113^0\]
ganeshie8
  • ganeshie8
how \(\sin A = \sin^2B-\sin^2C\) becomes \(\sin(B-C) = 1\) ?
amilapsn
  • amilapsn
\[\sin{A}=\sin^2{B}-\sin^2{C}=(\frac{1-\cos{2B}}{2})-(\frac{1-cos2C}{2})\]
ganeshie8
  • ganeshie8
\[\begin{align}\sin{A}=\sin^2{B}-\sin^2{C}&=(\frac{1-\cos{2B}}{2})-(\frac{1-cos2C}{2})\\~\\ &=\frac{1}{2}(\cos 2C - \cos 2B)\\~\\ &= -\sin(C+B)\sin(C-B)\\~\\ &=-\sin(A)\sin(C-B)\\~\\ \end{align}\] \(\implies \sin(B-C)=1=\sin(90) \implies B = C+90 = 113\) Awesome!
ganeshie8
  • ganeshie8
I did nothing, pls medal amila :)
Astrophysics
  • Astrophysics
Sure thing!
amilapsn
  • amilapsn
\[\Huge\unicode{x1f63a}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.