1. ## Trig. Problem

The triangle ABC has sides of length, a,b and c, as shown in the diagram. The point D lies on AB and CD is perpendicular to AB.
a) Show that asinB = bsinA.
b) Show that c = acosB + bcosA.
c) Given that c^2 = 4ab cosA cos B, show that a=b.

Could someone please show me how to do (c) of the question?

I've tried equating (b) squared to (c) but I end up with asinB = b sinA. How do I use this to prove a=b?

2. Originally Posted by xwrathbringerx
The triangle ABC has sides of length, a,b and c, as shown in the diagram. The point D lies on AB and CD is perpendicular to AB.
a) Show that asinB = bsinA.
b) Show that c = acosB + bcosA.
c) Given that c^2 = 4ab cosA cos B, show that a=b.

Could someone please show me how to do (c) of the question?

I've tried equating (b) squared to (c) but I end up with asinB = b sinA. How do I use this to prove a=b?
$(a\cos{B}+b\cos{A})^2 = a^2\cos^2{B} + 2ab\cos{A}\cos{B} + b^2\cos^2{A}$

$a^2\cos^2{B} + 2ab\cos{A}\cos{B} + b^2\cos^2{A} = 4ab\cos{A}\cos{B}$

$a^2\cos^2{B} - 2ab\cos{A}\cos{B} + b^2\cos^2{A} = 0$

$(a\cos{B} - b\cos{A})^2 = 0$

$a\cos{B} = b\cos{A}$

$DB = AD$

what does that say about $\Delta ACB$ ?

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### c=acosB bcosA

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