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.

  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.