Question

In the adjoining figure DE is internal bisector of $\angle D$, find the measure of $\angle x$.

• A. ${155}^o$
• B. ${105}^o$
• C. ${115}^o$
• D. ${130}^o$

Answer 1

The correct option is A ${155}^o$

Given $\angle ABC={70}^0$ and $\angle BAC={35}^0$

In $\Delta ABC,$

$\angle ACB=180-\angle ABC-\angle BAC=180-70-35={75}^0$

Given DE is internal bisector of angle D

$\angle EDC={25}^0$

$\therefore FDC=2\times 25={50}^0$

$\angle GCD=180-\angle ACB=180-75=105$

$\angle GFD$ is the external angle of $\Delta FCD$

So $\angle GFD=x^0=\angle GCD+\angle FDC=105+50={155}^0$