Question

Find the values of the unknowns "x" and "y"​

Answer 1

Answer:

we know that a triangle has 180°

than y=70°

and a line has made180°

than x=110

Answer 2

$\sf \pink{Given}\begin{cases}&\sf{\angle NMA=\bf{50\degree.}} \\ \\ &\sf{\angle MAN=\bf{60\degree.}}\end{cases}$

To FinD:-

The values of the unknowns "x" and "y".

SolutioN:-

• All angles in a triangle sum upto 180°.
• Angle MNA = y°

☯ According to the question,

$\\ :\normalsize\implies{\sf{\angle NMA+\angle MAN+\angle MNA=180\degree}}$

$\sf {Here}\begin{cases}&\sf{\angle NMA=\bf{50\degree.}} \\ \\ &\sf{\angle MAN=\bf{60\degree.}} \\ \\ &\sf{\angle MAN=\bf{y\degree.}}\end{cases}$

• Putting the values,

$\\ :\normalsize\implies{\sf{50\degree+60\degree+y\degree=180\degree}}$

$\\ \qquad:\normalsize\implies{\sf{110\degree+y\degree=180\degree}}$

$\\ \qquad \ \ \ \ :\normalsize\implies{\sf{y\degree=180\degree-110\degree}}$

$\\ \qquad\quad \ \ \ \ \normalsize\therefore\boxed{\mathfrak{\pink{y\degree=70\degree.}}}$

VerificatioN:-

$\\ :\normalsize\implies{\sf{\angle NMA+\angle MAN+\angle MNA=180\degree}}$

$\\ \qquad:\normalsize\implies{\sf{50\degree+60\degree+70\degree=180\degree}}$

$\\ \qquad \ \ \ \ :\normalsize\implies{\sf{180\degree=180\degree}}$

$\\ \qquad\quad \ \ \ \ \normalsize\therefore\boxed{\mathfrak{\pink{LHS=RHS.}}}$

• Hence verified.

_________________________________

Now the x° :

• We know that y° = 70°

• We also know that all angles in a straight line sum upto 180°.

☯ According to the question,

$\\ :\normalsize\implies{\sf{\angle MNA+x\degree=180\degree}}$

• Angle MNA = 70°

Putting the values,

$\\ :\normalsize\implies{\sf{70\degree+x\degree=180\degree}}$

$\\ :\normalsize\implies{\sf{x\degree=180\degree-70\degree}}$

$\\ \qquad\quad\normalsize\therefore\boxed{\mathfrak{\pink{x\degree=110\degree.}}}$

Value of y° = 70°.

Value of x° = 110°.