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Nishtha Gahlot Grade: 12
solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
one year ago

Answers : (4)

mycroft holmes
271 Points
  1. a cos A = b cos B 
\Rightarrow 2R sin A cos A = 2R sin B cos B \Rightarrow sin 2A = sin 2B
Either A = B (isoceles or equilateral) or 2A = 180o – 2B so that A+B = 90o.(Right-angled)
one year ago
mycroft holmes
271 Points
Let the feet of the altitudes on BC, AC, AB, be D,E,F resp. Let the orthocenter be H.
The following can be proved easily:
​1. HDCE and HFBD are cyclic quadrilaterals. Then chord HE subtends equal angles at C and D, so that \angle HCE = \ang HDE = 90^{\circ} - A.
Similarly \angle HDF = 90^{\circ} - A
​2. Now look at \triangle FBD. You have \angle FDB = \angle ADB - \angle HDF = A
So, \angle BFD = C. In other words \triangle DBF \sim \triangle ABC
3. From right \triangle ABD, we get BD = c cos B.
From \triangle DBF \sim \triangle ABC, we have \frac{DF}{AC} = \frac{DB}{AB}
so that DF = \frac{b c \cos B}{c} = b \cos B
one year ago
mycroft holmes
271 Points
To continue the prev post, we have DF = b \cos B = 2R \sin B \cos B = R \sin 2B
Hence the sides of the orthic triangle are in the ration\sin 2A: \sin 2B : \sin 2C
one year ago
mycroft holmes
271 Points
Draw \triangle OBC which is Isoceles as OB = OC. Now \angle BOC = 2A which means \angle OBC = 90^{\circ} - A
Let D, be the foot of the perp from O on BC ( which is also the midpoint of BC).
Then OD = OC sin (OBC) = R cos A.
Hence the required ratio is cos A: cos B : cos C
one year ago
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