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2

Answer:

#60^@,50^@" and "70^@#

Explanation:

#"sum the parts of the ratio"#

#rArr6+5+7=18#

#"the sum of the angles in a triangle "=180^@#

#rArr180^@/18=10^@larrcolor(blue)"1 part of the ratio"#

#rArr6" parts "=6xx10^@=60^@#

#rArr5" parts "=5xx10^@=50^@#

#rArr7" parts "=7xx10^@=70^@#

#"the 3 angles are "60^@,50^@" and "70^@#

2

Answer:

See below.

Explanation:

Here we are formulating an equation to solve for #x#.

We know that the interior angles of any triangle adds up #180# degrees.

We have three angles given:

#60#
#x#
#3x#

This means that :

#60+3x+x=180#

Now we collect like terms to simplify.

#60+4x=180#

Now we solve like any linear equation by isolating the variable on one side of the equation with the constant on the other.

Here we must subtract #60# from both sides to isolate the #x#.

#therefore 60 + 4x -60 = 180 -60#

#=>4x=120#
We want one #x#, therefore we divide by the coefficient of #x# on both sides.

Here we divide by #4#

#4x=120#
#=>x=30#

We can check if we are right by putting our value of #x# back into our formulated equation above.

#60 + (4*30) = 60+120 = 180#

2

Answer:

Triangle sum theorem states that all angles in a triangle must add up to #180^@#, a similar theorem applies to quadrilaterals and it states that all angles in a quad. must add up to #360^@#.

Explanation:

You have already applied the triangle sum theorem which states that all 3 angles in a triangle add up to #180# degrees it seems, so now all you have to do is create an algebraic expression that reflects that.

#"Angle 1 + Angle 2 + Angle 3" = 180^@#

#x+3x+60=180#

#4x+60=180#

#4x=120#

So

#x= 30^@#

The angle #x# is 30 degrees and the angle #3x# is #90# degrees

1

Answer:

Longest possible perimeter of the triangle is

#color(blue)(P + a + b + c ~~ 34.7685#

Explanation:

enter image source here

#hatA = (7pi)/12, hatB = pi/4, side = 8#

To find the longest possible perimeter of the triangle.

Third angle #hatC = pi - (7pi)/12 - pi/4 = pi/6#

To get the longest perimeter, smallest angle #hatC = pi/6# should correspond to side length 8#

Using sine law, #a / sin A = b / sin B = c / sin C#

#a = (c * sin A) / sin C = (8 * sin((7pi)/12)) / sin (pi/6) = 15.4548#

#b = (c * sin B) / sin C = (8 * sin(pi/4) )/ sin (pi/6) = 11.3137#

Longest possible perimeter of the triangle is

#color(blue)(P + a + b + c = 15.4548 + 11.3137 + 8 = 34.7685#

1

Answer:

Perimeter of isosceles triangle #color(green)(P = a + 2b = 4.464#

Explanation:

enter image source here

#hatA = (2pi)/3, hatB = pi/6, side = 1#

To find the longest possible perimeter of the triangle.

Third angle #hatC = pi - (2pi)/3 - pi/6 = pi/6#

It’s an isosceles triangle with
#hat B = hat C = pi/6#

Least angle #pi/6# should correspond to the side 1 to get the longest perimeter.

Applying sine law, #a / sin A = c / sin C#

#a = (1 * sin ((2pi)/3)) / sin (pi/6) = sqrt3 = 1.732#

Perimeter of isosceles triangle #color(green)(P = a + 2b = 1 + (2 * 1.732) = 4.464#

1

Answer:

#color(purple)(hatA = 80^@, hatB = 60^@, hatC == 40^@#

Explanation:

enter image source here

Three angles are in the ratio #4 : 3 : 2#

In a triangle sum of the three angles = #180^@#

Let basic unit x be the unit of each angle.

Then, #hatA = 4x, hatB = 3x, hatC = 2x#

#:. 4x + 3 x + 2x = 180^@#

#9x = 180# or #x = 20^@#

#color(purple)(hatA =) 4 x = 4 * 20 = color(purple)(80^@), color(purple)(hatB = )3x = 3* 20=color(purple)( 60^@), color(purple)(hatC = )2 * x = 2 * 20 = color(purple)(40^@#

1

Answer:

No. Of spheres that the cylinder can contain = 6.

Water overflowing volume = 678.54cc

Explanation:

enter image source here

Volume of cylinder #V_c = pi r^2 h#

Given : #r_c= 3.5 cm, h_c= 20cm#

#V_c = pi (3.5)^2 * 20 = 769.69 cc#

enter image source here

Volume of sphere #V_s = (4/3) pi r^3#

Given #r_s= 3 cm #V_s = (4/3) pi 3^3 = 113.09cc#

No. of spheres that can be placed in the cylinder #N = V_c / V_s = 769.69 / 113.09 = 6.8# or only 6 whole spheres can be filled in the cylinder.

Water that will be overflowing after dropping 6 spheres in it = 6 * 113.09 = 678.54# cc

1

Answer:

#80^@,60^@" and "40^@#

Explanation:

#"sum the parts of the ratio "#

#rArr4+3+2=9" parts"#

#• " the sum of the 3 angles in a triangle "=180^@#

#rArr180^@/9=20^@larrcolor(blue)"1 part"#

#rArr4" parts "=4xx20^@=80^@#

#rArr3" parts "=3xx20^@=60^@#

#rArr2" parts "=2xx20^@=40^@#

#"the 3 angles in the triangle are"#

#80^@,60^@" and "40^@#

1

Answer:

9)#" x=13.8"#
10)#" x= 8.1"#
11)#" x= 6.0 "#
12)#" x= 5.5"#

Explanation:

9)

#B=37^@#
#BC=11#(adjacent side)
#BA=x#(hypotenuse)
#cosB=(BC)/(BA)#
#cos37^@=11/(x)#
#x=11/cos37^@#
#x=13.8#

10)

#A=32^@#
#AC=13#(adjacent)
#CB=x#(opposite)
#tanA=(CB)/(AC)#
#tan32^@=x/13#
#x=13tan32^@#
#x=8.1#

11)

#A=50.1^@#
#AC=5#(adjacent)
#CB=x#(opposite)
#tanA=(CB)/(AC)#
#tan50.1^@=x/5#
#x=5tan50.1^@#
#x=6#

12)

#B=60^@#
#BA=11#(hypotenuse)
#BC=x#(adjacent)
#cosB=(BC)/(BA)#
#cos60^@=x/11#
#x=11cos60^@#
#x=5.5#

2

Answer:

#423.9toD#

Explanation:

#"the volume (V) of a cone is calculated using "#

#•color(white)(x)V=1/3pir^2h#

#"where r is the radius of the base and h the height"#

#"we require to find r the radius"#

#"using the formula for the circumference (C) of a circle"#

#•color(white)(x)C=2pir#

#rArr2xx3.14xxr=56.52#

#rArr6.28r=56.52#

#"divide both sides by "6.28#

#(cancel(6.28) r)/cancel(6.28)=56.52/6.28rArrr=9#

#rArrV=(3.14xx81xx5)/3~~423.9toD#