Two circles with centres H and K are 12 cm apart, the radii are 4cm and 8 cm respectively, touching externally. PQT is a tangent to both circles. If angle HTP is A, then what are sin A, cos A and tanA
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@jim_thompson5910 @ganeshie8
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they have a common angle so you know the triangles are similar
so you'll want to set up a ratio and from that you can create an equation to find x
yes but that doesn't exactly help @mathmath333
Oh it helps plenty, actually. You have two similar triangles now. \(\triangle TPK\) and \(\triangle TQH\)
I know and I tried that, does it actually work?
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The radius is the perpendicular bisector of the line TP
\[\triangle THQ \sim \triangle TKP\\So\\ \frac{ TH }{ HQ }=\frac{ TK }{ KP }\\\frac{ TH }{ 4 }=\frac{ TH+HK }{ 8 }\\\frac{ TH }{ 4 }=\frac{ TH+12 }{ 8 }\\\]
\[\frac{x+12}{8}=\frac{x}{4}\]
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