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How to construct a triangle with $BC=7.5$ cm. $\angle ABC$=$60$° and $AC-AB=1.5$ cm.

At first I constructed $BC$ then $\angle ABC$ ,but I don't know what to do next. Please help me.

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    $\begingroup$ How about utilising the cosine rule, where $AC=AB+1.5$ $\endgroup$ – Mohammad Zuhair Khan Nov 20 '18 at 15:03
  • $\begingroup$ @Raptor please explain more. Give me more hint. Cosine rule over which triangle? $\endgroup$ – Sufaid Saleel Nov 20 '18 at 15:05
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    $\begingroup$ The triangle $\triangle ABC$, so $BC^2+AB^2-2 \cdot AB \cdot BC\cos \angle ABC=AC^2$ $\endgroup$ – Mohammad Zuhair Khan Nov 20 '18 at 15:08
  • $\begingroup$ Thanks! I have done the problem! $\endgroup$ – Sufaid Saleel Nov 20 '18 at 15:11
  • $\begingroup$ @Raptor: The task is to construct, not to calculate the sides of the triangle. You are using a cannot to kill an ant. $\endgroup$ – Oldboy Nov 20 '18 at 15:27
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Using the cosine rule is not the way to solve this problem simply and efficiently. This is a problem about construciton, not trigonometry. You are not supposed to calculate values that are not given.

Suppose that triangle $ABC$ is the solution. Draw a circular arc $l$ with center at point $A$ and radius $AC$ until it meets the ray $AB$ in point $C'$. Obviously $BC'$=$AC-AB$, which is given. So it is possible to construct triangle $BCC'$: we know $BC$, $BC'$ and $\angle CBC'=180^\circ-\angle ABC=120^\circ$.

Triangle $ACC'$ is isosceles so the point $A$ has to be on the median $n$ of segment $CC'$. After the construciton of triangle $CBC'$ just extend $C'B$ until it meets the median of $CC'$. The intersection point is actually your point $A$.

Problems like this one do not need trigonometry at all.

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