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Grade: 12
        
solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
2 years ago

Answers : (4)

mycroft holmes
272 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)
 
2 years ago
mycroft holmes
272 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
2 years ago
mycroft holmes
272 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
2 years ago
mycroft holmes
272 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
2 years ago
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