Question

ABCD is rhombus and AED is an equilateral triangle. E and C lie on opposite sides of AD. If $\angle ABC={78}^{\circ }$, calculate $\angle DCE$ and $\angle ACE$

• A. ${20}^{\circ },{30}^{\circ }$
• B. ${21}^{\circ },{31}^{\circ }$
• C. ${22}^{\circ },{32}^{\circ }$
• D. ${19}^{\circ },{29}^{\circ }$

Answer 1

The correct option is B ${21}^{\circ },{31}^{\circ }$

$\angle BAD=180-78={102}^0$

In triangle ABC, $\angle BAC=\angle BCA={51}^0$ [ABC is an isosceles triangle]

In triangle EDC, angle EDC is $60+78={138}^0$

So, $\angle DCE=\angle DEC={21}^0$

We know angle BCA is ${102}^0$

Out of which angle BCA is 51 deg and angle DCE is 21 deg

Hence the remaining angle, angle ACE is $102-(51+21)={30}^0$

$\angle DCE={21}^0\angle ACE={30}^0$