Question

IN THE GIVEN FIGURE, IF ABCD IS A RECTANGLE IN WHICH ANGLE APB=100,THEN FIND THE VALUE OF x​

Answer 1

Given:

ABCD is a rectangle

∠APB = 100°

To find:

The value of x

Solution:

From the given figure we can say that APC is a straight line i.e.,

∠APB + ∠CPB = 180° ...... [Linear Pairs]

∵ ∠ APB = 100° (given)

⇒ 100° + ∠CPB = 180°

⇒ ∠CPB = 180° - 100°

⇒ ∠CPB = 80° ....... (i)

We know that → Diagonals of a rectangle are equal in length and bisect each other.

∴ AC = BD & AP = PC = $\frac{1}{2}$AC & BP = PD = $\frac{1}{2}$BD

⇒ AP = PC = BP = PD = $\frac{1}{2}$AC = $\frac{1}{2}$BD

Now, in ΔBPC, we have

BP = PC

∴ ∠PBC = ∠PCB = x ..... [∵ angle opposite to equal sides are equal] ... (ii)

∴ ∠PBC + ∠PCB + ∠CPB = 180°

substituting from (i) & (ii), we get

⇒ x + x + 80° = 180°

⇒ 2x = 180° - 80°

⇒ 2x = 100°

⇒ x = $\frac{100}{2}$

⇒ x = 50°

Thus, the value of x is 50°.

-----------------------------------------------------------------------------------------------------

Also View:

In the adjoining figure, ABCD is a rectangle whose diagonals AC and bd intersect at O. If angle AOB = 110 degree Find the measure of angle ODA and angle OCD????

https://brainly.in/question/6369757

in the given figure ABCD is a rectangle whose diagonals AC and BD intersect each other at O if angle OAB= 20 degree find angle abc

https://brainly.in/question/2510894

In a rectangle ABCD angle aob = 80 find angel ado and angle dca

https://brainly.in/question/12862055

Answer 2

Answer X=50

Explanation: ∠BPC=80°(Linear pair of angles)

PB=PC(The diagonal of a rectangle are equal and bisect each other)

Hence,Triangle PBC is a isosceles triangle.

∠PBC=∠X(The opposite angles of the equal sides are equal)

That is ∠PBC=∠PCB

X+X+80°= 180°(Angle sum property of a triangle)

2X+80°=180°

2X=180°-80°

2X=100°

X=100°/2

X=50°