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 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°

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Step-by-step explanation:

Given:

BA = BD

CB = CD

In

BA = BD

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

We know that sum of internal angles of a triangle is equal to

Quadrilateral ABCD is cyclic so, sum of opposite angles is equal to .

In

BC = CD

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

We know that sum of internal angles of a triangle is equal to

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

So, the answer is

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50