Question

in the given figure xoy is a straight line find x o p and y o p ​

Answer 1

Step-by-step explanation:

Given:In the given figure xoy is a straight line.

To find: angle x o p and y o p

Solution:

We know that sum of all the angles on one side of straight line is 180°

Here, on line xoy

two angles are formed,

Thus

$\angle \: xop + \angle \: yop = 180° \\ \\ (x + 15)° + (3x + 25)° = 180° \\ \\ x + 3x + 15 °+ 25 °= 180 °\\ \\ 4x + 40° = 180°\\ \\ 4x = 180° - 40° \\ \\ 4x = 140 °\\ \\ x = 35°$

So,

$\angle \: xop = x + 15° \\ \\\angle \: xop = 35 + 15 \\ \\\bold{\pink{\angle \: xop = 50°}} \\ \\\angle \: yop = 3x + 25° \\ \\\angle \: yop = 3 \times 35° + 25° \\ \\ = 105° + 25° \\ \\\bold{\green{\angle yop = 130°}}$

Hope it helps you.

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Answer 2

Step-by-step explanation:

To find: angle x o p and y o p

Solution:

We know that sum of all the angles on one side of straight line is 180°

Here, on line xoy

two angles are formed,

Thus

\begin{gathered} \angle \: xop + \angle \: yop = 180° \\ \\ (x + 15)° + (3x + 25)° = 180° \\ \\ x + 3x + 15 °+ 25 °= 180 °\\ \\ 4x + 40° = 180°\\ \\ 4x = 180° - 40° \\ \\ 4x = 140 °\\ \\ x = 35° \\ \\ \end{gathered}

∠xop+∠yop=180°

(x+15)°+(3x+25)°=180°

x+3x+15°+25°=180°

4x+40°=180°

4x=180°−40°

4x=140°

x=35°

So,

\begin{gathered} \angle \: xop = x + 15° \\ \\\angle \: xop = 35 + 15 \\ \\\bold{\pink{\angle \: xop = 50°}} \\ \\\angle \: yop = 3x + 25° \\ \\\angle \: yop = 3 \times 35° + 25° \\ \\ = 105° + 25° \\ \\\bold{\green{\angle yop = 130°}} \\ \\ \end{gathered}

∠xop=x+15°

∠xop=35+15

∠xop=50°

∠yop=3x+25°

∠yop=3×35°+25°

=105°+25°

∠yop=130°

Hope it helps you