Question

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Answer 1

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Answer 2

$\sf\huge\bold{Answer:}$

$\fbox{Solution 1}$

Given :

PC intersects QB

PQ || BC

To find :

value of u and y

Solution :

Since,PC intersects QB

$\angle \: PAQ = \angle \: BAC = 60 \degree$

$\triangle \: ABC$ is an isosceles triangles

where AC=BC

Since, sides are equal then,angles facing them will also be equal.

$\angle \: ABC = \angle \: BAC$

$\angle \: ABC = \angle \: BAC =60 \: \degree$

Sum of all angles of triangle =180°

In $\triangle \: ABC$

$⟶\angle \: ABC + \angle \: ACB + \angle \: BAC = 180 \degree$

$⟶60 \degree + u + 60 \degree = 180 \degree$

$⟶u = 180 \degree - 120 \degree$

$⟶u = 60 \degree$

Sum of opposite interior angle is equal to Exterior angle.

$⟶\angle \: ABC + \angle \: BAC = \angle \: ACD$

$⟶60 \degree+ 60\degree = y$

$⟶y = 120\degree$

Hence,

value of u=60° and y =120°

$\fbox{Solution 2}$

Given :

$\triangle\: TRS \: and \: \triangle \: PQR$

To find :

value of u and y

Solution :

Sum of all angles of triangle =180°

In $\triangle \: PQR$

$⟶\angle \: PRQ + \angle RQP + \angle \: QPR = 180 \degree$

⟶100°+60°+u = 180°

⟶160°+u = 180°

⟶u = 180°-160°

⟶u = 20°

$⟶\angle \: PRS + \angle \: PRQ = 180 \degree \: \: \: (linear \: pair)$

$⟶\angle \: PRS + 100 \degree= 180 \degree \: \: \:$

$⟶\angle \: PRS = 180 \degree - 100 \degree$

$⟶\angle \: PRS = 80 \degree$

Sum of all angles of triangle =180°

In$\triangle\: TRS$

$⟶\angle \: TSR + \angle SRT + \angle RTS = 180 \degree$

⟶y+80°+65°=180°

⟶y+145°=180°

⟶y=180°-145°

⟶y=35°

Hence,

value of u=20° and y= 35°