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

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.

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

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 $$A$$.