Question

Can anybody help me with any of these problems please

Answer 1

Answer:

Part 9)  $x=7$

Part 10)  $m\angle Y=92^o$

Part 11)  $x=5$

Part 12)

$m\angle 1=60^o$

$m\angle 2=41^o$

$m\angle 3=19^o$

$m\angle 4=41^o$

$m\angle 5=19^o$

Part 13) The measure of each side is 56 units

Part 14)

$x=10/3$

$HI=12\ units$

$HJ=12\ units$

$IJ=38/3\ units$

Part 15)

$x=7$

$m\angle C=41^o$

$m\angle D=41^o$

$m\angle A=98^o$

Step-by-step explanation:

Part 9) we know that

An isosceles triangle has two equal sides and two equal interior angles

The triangle of the problem is an isosceles triangle

Because has two equal sides and two equal angles

Remember

The sum of the interior angles in any triangle must be equal to 180 degrees

so

$2(8x-4)^o+76^o=180^o$

solve for x

$16x-8+76=180\\16x=180-68\\16x=112\\x=7$

Part 10) we know that

step 1

Solve for x

An isosceles triangle has two equal sides and two equal interior angles

The triangle XYZ of the problem is an isosceles triangle

Because has two equal sides

so

$m\angle X=m\angle Z$

substitute the given values

$(4x-12)^o=(2x+16)^o$

solve for x

$4x-2x=16+12\\2x=28\\x=14$

Find the measure of angle X

$m\angle X=(4x-12)^o$

substitute the value of x

$m\angle X=(4(14)-12)=44^o$

so

$m\angle Z=44^o$

step 2

Find the measure of angle Y

Remember

The sum of the interior angles in any triangle must be equal to 180 degrees

so

$m\angle Y+2(44^o)=180^o$

$m\angle Y=180^o-88^o=92^o$

Part 11) we know that

we know that

An equilateral triangle has three equal sides and three equal interior angles. The measure of each interior angle is 60 degrees

In this problem we have an equilateral triangle

so

$(14x-10)^o=60^o$

solve for x

$14x=60+10\\14x=70\\x=5$

Part 12)

step 1

Find the measure of angle 1

we know that

An equilateral triangle has three equal sides and three equal interior angles. The measure of each interior angle is 60 degrees

In this problem the triangle WXY is an equilateral triangle

therefore

$m\angle 1=60^o$

step 2

Find the measure of angle 3

we know that

Triangle WZY is an isosceles triangle

so

$m\angle 3=m\angle 5$

Remember that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

$m\angle 3+m\angle 5+142^o=180^o$

$2m\angle 3=180^o-142^o$

$m\angle 3=19^o$

so

$m\angle 5=19^o$

step 3

Find the measure of angle 2

we know that

The measure of each interior angle of an equilateral triangle  is equal to 60 degrees

so

$m\angle 2+m\angle 3=60^o$

substitute the given value

$m\angle 2+19^o=60^o$

$m\angle 2=60^o-19^o$

$m\angle 2=41^o$

step 4

Find the measure of angle 4

we know that

The measure of each interior angle of an equilateral triangle  is equal to 60 degrees

so

$m\angle 4+m\angle 5=60^o$

substitute the given value

$m\angle 4+19^o=60^o$

$m\angle 4=60^o-19^o$

$m\angle 4=41^o$

Part 13) we know that

we know that

An equilateral triangle has three equal sides and three equal interior angles. The measure of each interior angle is 60 degrees

so

AB=BC=AC

solve the equation

AB=BC

substitute the given values

$5x-9=7x-35\\7x-5x=35-9\\2x=26\\x=13$

Find the measure of side AB

$AB=5x-9$

substitute the value of x

$AB=5(13)-9=56\ units$

Verify BC and AC

$BC=7(13)-35=56\ units$ ----> is correct

$AC=4(13)+4=56\ units$ ----> is correct

Part 14) we know that

An isosceles triangle has two equal sides and two equal angles

In this problem Triangle HIJ is an isosceles triangle

because

Angle I is equal to Angle J

so

HI=HJ

substitute the given values

$6x-8=3x+2$

solve for x

$6x-3x=2+8\\3x=10\\x=10/3$

Find the measure of each side

$HI=6(10/3) -8=12\ units$

$HJ=3(10/3)+2=12\ units$

$IJ=5(10/3)-4=38/3\ units$

Part 15) we know that

Triangle ACD is an isosceles triangle

because

has two equal sides

AC=AD

so

$m\angle C=m\angle D$

substitute the given values

$6x-1=7x-8$

solve for x

$7x-6x=8-1\\x=7$

Find the measure of angle C

$m\angle C=6(7)-1=41^o$

Find the measure of angle D

$m\angle D=7(7)-8=41^o$

Find the measure of angle A

we have that

$m\angle C+m\angle D+m\angle A=180^o$

substitute

$41^o+41^o+m\angle A=180^o$

$m\angle A=180^o-82^o$

$m\angle A=98^o$