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triangleAMT~triangleAHE.In triangleAMT, ...

`triangleAMT~triangleAHE`.In `triangleAMT, AM=6.3 cm,angleTAM=50^(@),AT=5.6cm.(AM)/(AH)=(7)/(5).`
`Construct triangleAHE.`

Text Solution

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The correct Answer is:
The length of side AH will be equal to 5 parts out of 7 equal parts of side AM.
So, if we construct `triangle AMT`, point H will be on side AM, at a distance equal to 5 parts from A.
Now, point E is the point of intersection of ray AT and a line through H, parallel to MT.
`triangle AHE` is the required triangle similar to `triangle AMT`.
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