Question

If PAQ is a straight line,find the value of x .
plz tell me this question's answer​

Answer 1

Answer:

∠BAP = 17.5°

∠BAC = 117.5°

∠CAQ = 45°

Step-by-step explanation:

Concept used:

The sum of all angles on a straight line is always equal to 180°

Solution:

Applying the above concept,

   ∠BAP + ∠BAC + ∠CAQ = 180°

$\implies \sf{\bigg(\dfrac{x}{2}\bigg)+\bigg(\dfrac{7x}{2}-5\bigg)+(x+10)=180^{\circ}}$

Rearranging the terms,

$\implies \sf{\bigg(\dfrac{x}{2}+\dfrac{7x}{2}+x\bigg)+(10-5)=180^{\circ}}$

Taking LCM and solving,

$\implies \sf{\dfrac{x+7x+2x}{2}+5=180^{\circ}}$

$\implies \sf{\dfrac{10x}{2}=180^{\circ}-5^{\circ}}$

$\implies \sf{\dfrac{10x}{2}=175^{\circ}}$

$\implies \sf{5x=175^{\circ}}$

$\implies \sf{x=\dfrac{175}{5}}$

$\implies \boxed{\bf{x=35^{\circ}}}$

For finding ∠BAP:

∠BAP

= x/2

= 35/2

= 17.5°

For finding ∠BAC,

∠BAC

= 7x/2 - 5

= [7(35)÷2] - 5

= 117.5°

For finding ∠CAQ,

∠CAQ

= x + 10

= 35 + 10

= 45°