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3 years ago

Find the size of angle AED

Mathematics

1 answer:

kogti [31]3 years ago

8 0

Answer:

The measure of angle AED is 173°

Step-by-step explanation:

Given

The figure above

Required

Calculate <AED

To calculate <AED, we follow the highlighted steps.

1. Calculate angle DEC in triangle DEC.

The angles in ∆DEC are <DEC, <ECD and <CDE where

<ECD = 37° and <CDE = 96°

THE SUM OF ANGLES IN A TRIANGLE IS 180°.

i.e.

<DEC + <ECD + <CDE = 180°

By substituting <ECD = 37° and <CDE = 96°, we have

<DEC + 37° + 96° = 180°

<DEC + 133° = 180°

<DEC = 180° - 133°

<DEC = 47°

2. Calculate angle CEB in triangle CEB.

The angles in ∆CEB are <CEB, <EBC and <BCE where

<EBC = 73° and <BCE = 79°

THE SUM OF ANGLES IN A TRIANGLE IS 180°.

i.e.

<CEB + <EBC + <BCE = 180°

By substituting <EBC = 73° and <BCE = 79°, we have

<CEB + 73° + 79° = 180°

<CEB + 152° = 180°

<CEB = 180° - 152°

<CEB = 28°

3. Calculate <AED

This calculated by taking the sum of angles are point E.

The angles at point E are <AED, <AEB, <CEB and <DEC.

Where <AEB = 112°, <CEB = 28°, <DEC = 47°

Recall that sum of angles at a point is 360°.

So,

<AED + <AEB + <CEB + <DEC = 360

By substituting <AEB = 112°, <CEB = 28°, <DEC = 47°, we have

<AED + 112° + 28° + 47° = 360

<AED + 187° = 360°

<AED = 360° - 187°

<AED = 173°.

Hence, the measure of angle AED is 173°

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Answer:

See explanation

Step-by-step explanation:

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$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

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Therefore,

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Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

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3 years ago

In triangle , side and the perpendicular bisector of meet in point , and bisects . If and , what is the area of triangle

stich3 [128]

In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC。 If AD = 9 and DC = 7, 145–√5 is the area of a triangle.

I supposed here that [ABD] is the perimeter of ▲ ABD.

As BD is a bisector of ∠ABC ,

ABBC=ADDC=97

Let ∠B=2α

Then in isosceles △DBC

∠C=α

BC=2∗DC∗cosα=14cosα

Thus AB=18cosα

The Sum of angles in △ABC is π so

∠A=π−3α

Let's look at AC=AD+DC=16 :

AC=BCcosC+ABcosA

16=14cos2α+18cosαcos(π−3α)

[1]8=7cos2α−9cosαcos(3α)

cos(3α)=cos(α+2α)=cosαcos(2α)−sinαsin(2α)=cosα(2cos2α−1)−2cosαsin2α=cosα(4cos2α−3)

With [1]

8=cos2α(7−9(4cos2α−3))

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cos2α={12,49}

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Thus first root cos2α=12 cannot be chosen and we have to stick with the second root cos2α=49. This gives cosα=23 and sinα=5√3.

The area of a triangle ABD=12h∗AD where h is the distance from B to AC.

h=BCsinC=14cosαsinα

Area of triangle ABD=145–√5

= 145–√5.

Incomplete question please read below for the proper question.

In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC。 If AD = 9 and DC = 7, what is the area of triangle ABD?

Learn more about the Area of the triangle at

brainly.com/question/23945265

#SPJ4

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