Construction of a triangle

Construction of a triangle ΔABC where BC=7cm, ∠B=75° and AB+AC=13cm.


Explore how a triangle can be constructed with given base-length, one base angle and the sum of other two sides. You can also see how, with a compass, 60°, 120°, 90°, and 75° angles are drawn in the animation.

Construction of a triangle given base-length, base-angle and added value of two sides D C B A P Q Y 60° 120° 90° ∠B = 75° BD = 13cm 7cm 75° Perpendicular bisector to CD BC = 7cm Begin animation Draw BC so that BC = 7cm. Draw BY so that ∠B = 75°. First draw a 60° angle. Then an angle of 120° is drawn. Bisecting 60° and 120° arcs, we get a right angle(90°). Again bisecting 60° and 90° arcs, we finally get the 75° angle. We cut 13cm BD from BY at D, and join CD. On line CD, draw perpendicular bisector PQ. Let PQ intersect BD at point A. Now, ▲ABC is the required triangle, where BC=7cm, ∠B=75°, AB+AC=13cm.

How to Construct a ΔABC where BC=7cm, ∠B=75° and AB+AC=13cm.

* Draw a base BC = 7cm.
* Draw segment BY so that ∠CBY = 75°.

How to draw a 75° angle »

* Put point D on line BY so that BD=13 cm
* Join CD
* Draw a perpendicular bisector PQ on CD
* Let PQ intersect BD at point A.
* Join AC to get the required triangle ABC

--- x ---

Want to leave a message for me?