Question

ANY TEACHER IS HERE. ANSWER NOW.
FIND THE VALUE OF X .
SOLVE ON PAPER​

Answer 1

Answer:

as it is a triangle , we know

∆BAC +∆ ABC+∆ACB =180°

X+46°+90°=180° [∆ABC=90° as we know it's a right angle.]

X=180°-136°

X=44°

Answer 2

$\huge{\underline{\bf\red{QuestiØn} :}}$

• Find the value of x .

$\huge{\underline{\bf\blue{SolutiØn}\ :}}$

We know that,

• in a triangle,
• $\sf ext. \angle = sum \: of \: int. \: opp. \: \angle s$

Thus,

$\sf In \: \triangle CDE \\ \\ \begin{aligned} \sf \angle ECD \: & \sf = \angle AED - \angle EDC \\ \\ &= 132 \degree - 54 \degree \\ \\ & = (132 - 54) \degree \\ \\ & = 78 \degree & & & & & & \end{aligned}$

Now,

$\sf In \: \triangle ABC \\ \\ \begin{aligned} \sf \angle BAC \: & \sf = \angle ACD - \angle ABC \\ \\ & \sf = \angle ECD - \angle ABC \\ \\ &= 78 \degree - 46 \degree \\ \\ & = (78 - 46) \degree \\ \\ & = 32 \degree & & & & & & \end{aligned}$

HENCE ,

$\underline{\boxed{ \bf \quad x = 32 \degree \quad}}$

$\red{\rule{5.5cm}{0.02cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$