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        solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
3 years ago

mycroft holmes
272 Points
								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)

3 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 $DF = \frac{b c \cos B}{c} = b \cos B$

3 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$\sin 2A: \sin 2B : \sin 2C$

3 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

3 years ago
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Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions