Mathematics, 20.06.2020 06:57 toxsicity

The points A, B, C and D lie on a circle. CDE is a straight line. BA=BD, CB=CD, Angle ABD =32 degrees Work out the size of angle ADE. You must give a reason for each stage of working.

The points A, B, C and D lie on a circle. CDE is a straight line. BA=BD, CB=CD, Angle ABD =32 degre

Answers: 2

Show answers

Answers

Answer from: alannaswitzer

The size of ∠ADE of the cyclic quadrilateral is;

∠ADE = 53°

We are given;

BA = BD

CB = CD

∠ABD = 32°

Now, ΔABD is an isosceles triangle and since sum of angles in a triangle is 180°, then;

∠ABD + ∠BAD + ∠BDA = 180°

∠BAD + ∠BDA = 180 - 32

∠BAD + ∠BDA = 148°

Since isosceles, then;

∠BAD = ∠BDA = 148/2

∠BAD = ∠BDA = 74°

Now, in cyclic quadrilaterals, opposite angles are supplementary and as such add up to 180°. Quadrilateral ABCD is cyclic and Thus;

∠BAD + ∠BCD = 180°

74 + ∠BCD = 180°

∠BCD = 180 - 74

∠BCD = 106°

ΔBCD is an isosceles triangle and as such;

∠CBD = ∠CDB = 106/2

∠CBD = ∠CDB = 53°

Now, sum of angles on a straight line add up to 180°. Thus;

∠CBD + ∠BDA + ∠ADE = 180°

Plugging in the relevant values gives;

53° + 74° + ∠ADE = 180°

∠ADE = 180 - (53 + 74)

∠ADE = 53°

Read more at; link

Answer from: jorgeb12316

$\angle ADE =$ $69^\circ$

Step-by-step explanation:

Given:

$\angle ABD = 32^\circ$

BA = BD

CB = CD

In $\triangle ABD:$

BA = BD

We know that in a triangle, angles opposite to equal sides will also be equal.

$\Rightarrow \angle BAD = \angle BDA$

We know that sum of internal angles of a triangle is equal to $180^\circ$

$\angle BAD + \angle BDA +\angle ABD=180^\circ\\\Rightarrow \angle BAD + \angle BDA +32=180\\\Rightarrow \angle BAD + \angle BDA = 180 -32\\\Rightarrow \angle BAD + \angle BDA = 148\\\Rightarrow \angle BAD = \angle BDA = \dfrac{148}{2} = 74^\circ$

Quadrilateral ABCD is cyclic so, sum of opposite angles is equal to $180^\circ$ .

$\angle BAD + \angle BCD = 180^\circ\\\Rightarrow 74 + \angle BCD = 180\\\Rightarrow \angle BCD = 180-74 = 106^\circ$

In $\triangle BCD:$

BC = CD

We know that in a triangle, angles opposite to equal sides will also be equal.

$\Rightarrow \angle DBC = \angle CDB$

We know that sum of internal angles of a triangle is equal to $180^\circ$

$\angle DBC + \angle BCD +\angle CDB=180^\circ\\\Rightarrow \angle DBC + 106 +\angle CDB=180\\\Rightarrow \angle DBC +\angle CDB=180 - 106\\\Rightarrow \angle DBC =\angle CDB=\dfrac{74}{2} = 37^\circ$

It is given that the line CDE is a straight line, so the sum of angles on one side of point D is $180^\circ$ .

$\angle CDB + \angle BDA +\angle ADE = 180^\circ\\\text{Putting values of } \angle CDB \text{ and }\angle BDA:\\\Rightarrow 37 + 74 + \angle ADE = 180\\\Rightarrow \angle ADE = 180 -111\\\Rightarrow \angle ADE = 69^\circ$

So, the answer is $\angle ADE =$ $69^\circ$

Answer from: Quest

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

Answer from: Quest

im not sure

step-by-step explanation: