An isosceles triangle,on Trigonometry, I.S.I Entrance 2016, Solution to Subjective problem no. 6
Understand the problem
[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. If \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\), then prove that the triangle is isosceles.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4" link_option_url="https://cheenta.com/isicmientrance/" link_option_url_new_window="on"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]
Start with hints
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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. Given \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B- \cos A \sin B=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\)
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]\(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B(1-\frac{a}{a+b})=\cos A \sin B(1-\frac{b}{a+b})\) \(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\)
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]\(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\) \(\Rightarrow b\sin A\cos B=a\cos A \sin B\) \(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \)
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]
\(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \) \(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{b}{\sin B}\) [since \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)] \(\Rightarrow \frac{\cos B}{\cos A}=1\) \(\cos B=\cos A\) \(A=B\).
\(\Rightarrow \Delta ABC\) is isosceles (Proved).
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