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Given: In quadrilateral ABCD, AB smallest & CD is longest sides.

To Prove: ∠A>∠C & ∠B>∠D

Construction: Join AC. Mark the angles as shown in the figure..

Proof:In △ABC , AB is the shortest side.

BC > AB ∠2>∠4 …(i) [Angle opposite to longer side is greater]

In △ADC , CD is the longest side

CD > AD ∠1>∠3 …(ii) [Angle opposite to longer side is greater]

Adding (i) and (ii), we have

∠2+∠1>∠4+∠3 ⇒∠A>∠C

Similarly, by joining BD, we can prove that ∠B>∠D

find the volume of sphere whose radius is 7

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let the angles of the triangle be (a-d),a,(a+d)

using angle sum property we get

a-d+a+a+d=180

3a=180

a=60 all angles are in degrees

angles are 60,60-d,60+d no. of radian in 60+d /180 *pie hence d=30

so the angles are 30,60 and 90

90 degree

Two Angles are complementary if they sum up to 90°

Let one angle be xThen other angle be 7x/8

Now,x + 7x/8 = 90(8x + 7x) = 90*815x = 720x = 720/15 = 48

Therefore,Angles are 48° and 42°

let one angle be X then48° and 42° other angle be 7x/8 Now,x+7x/8=90(8x+7x)=90*815x=720x= 720/15= 48:• angles are

Complementary angles are two angles whose sum is 90°.

Let one angle be x and then other angle = 7x/8

As per given condition x + 7x/8 = 9015x = 90*8x = 6*8 = 48°

Therefore, Angles are 48°, 48*7/8 = 42°

Given: In quadrilateral ABCD, AB smallest & CD is longest sides.

To Prove: ∠A>∠C & ∠B>∠D

Construction: Join AC. Mark the angles as shown in the figure..

Proof:In △ABC , AB is the shortest side.

BC > AB ∠2>∠4 …(i) [Angle opposite to longer side is greater]

In △ADC , CD is the longest side

CD > AD ∠1>∠3 …(ii) [Angle opposite to longer side is greater]

Adding (i) and (ii), we have

∠2+∠1>∠4+∠3 ⇒∠A>∠C

Similarly, by joining BD, we can prove that ∠B>∠D

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Twocirclesare congruentif they have the same size. The size can be measured as the radius, diameter or circumference.

Here 1st and 2nd circle are congruent because they have same diameter=10cm

corresponding sides are equal and same size and same shape

Temperature dropped in 30 days = 15° CelsiusTemperature dropped in 1 day = 15/30= 0.5° CelsiusTemperature will drop in next 10 days = 0.5× 10= 5°So, there will be a temperature drop of 5° Celsius in next 10 days.

if figure mirror and water image of L or it is L it is right angle .

Let the three angles of a triangle are (a-d),(a),& (a+d).(a-d)+(a)+ (a+d) = 180°3a = 180°a= 180 /3 = 60 ° The 3 angles are 60° -d , 60° and 60°+ d.

ATQ

Greatest angle = sum of two smaller angles

60° + d = 60° -d + 60°60° + d = 120° -dd+d = 120° -60°2d = 60°d = 60° /2= 30°d = 30°

Required angles of a triangle are (60° -30°) , (60°) and (60° + 30°)= 30° , 60°, 90°.

Hence, the required angles of a triangle are 30° , 60°, 90°.

These angles are the multiple of 30°.

Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O' Hence OA = OB = 5 cm O'A = O'B = 3 cm OO' is the perpendicular bisector of chord AB. Therefore, AC = BC Given OO' = 4 cm

Let OC = x Hence O'C = 4 − x

In right angled ΔOAC, by Pythagoras theorem OA^2 = OC^2 + AC^2 ⇒ 5^2 = x^2 + AC^2 ⇒ AC^2 = 25 − x^2.......(1)

In right angled ΔO'AC, by Pythagoras theorem O'A^2 = AC^2 + O'C^2 ⇒ 3^2 = AC^2 + (4 – x)^2 ⇒ 9 = AC^2 + 16 + x^2 − 8x ⇒ AC^2 = 8x − x^2 − 7.......(2)

From (1) and (2), we get 25 − x^2 = 8x − x^2 − 7 8x = 32 Therefore, x = 4

Hence the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle. AC^2 = 25 − x^2 = 25 − 42 = 25 − 16 = 9

Therefore, AC = 3 m Length of the common chord, AB = 2AC = 6 m

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If the radius of two circles is equal then they are called congruent.So in given figureIn triangle ABO & PQOOA=OP(radii) Angle AOB=Angle POQ(given) OB=OQ(radii)By SASTriangle ABO Is congruent to PQOBy CPCTAB=PQHence, proved

Let there be in n sides in the polygon.

Then by geometry, sum of all n interior angles of polygon = (n – 2) * 180°

Also the angles are in A. P. with the smallest angle = 120° , common difference = 5°

∴ Sum of all interior angles of polygon

= n/2[2 * 120 + ( n – 1) * 5

Thus we should have

n/2 [2 * 120 + (n – 1) * 5] = (n – 2) * 180

⇒ n/2 [5n + 235] = (n – 2 ) * 180

⇒ 5n^2+ 235n = 360n – 720

⇒ 5n^2– 125n + 720 = 0 ⇒ n^2– 25n + 144 = 0

⇒ (n – 16 ) (n – 9) = 0 ⇒ n = 16, 9

Also if n = 16 then 16thangle = 120 + 15 * 5 = 195° > 180°

∴ not possible.

Hence n = 9.

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aThe temperature decreases at a rate of 2degrees per hour

It was actually 12 degree above zero

We need to find at what time does the the temperature drops down to 10 degree below zero that is -10

Temperature change = 12-(-10)

=12+10

=22

For one hour it drops by 2degree

So it takes 22/2=11hours to attain -10 degree

It means, at 11'o clock e temperature falls down to -10 degree

24=2*2*2*3nth root(2root(6))=2root(6)so n=1

a=14 as they are given in the ratioso by formula a/30=7/15 you will get the answer

solution of 5)as given that sum of two angles are in the ratio of 2;3 so suppose angle is xthen 2x+3x=905x= 90x=90/5x=18so angles will be 36° and 54°