problems about angles:

Hi. I needed help on some problems for the math section that we are doing in class.

I am trying out some problems in my math book, but they prove quite difficult.

1. In a certain parallelogram, if the measure of one angle is 30 degrees more than the measure of a second angle, find the measure of each of the four angles.

2. In a certain Isosceles triangle, the sum of the measures of the base angles is 88 degrees less than the measure of the remaining angle. Find the measurement of each angle.

3. The degree measures of two angles of a triangle are consecutive even integers. If the measure of the third angle is 50 degrees more than twice the measure of the least angle. Find the measure of each angle.

please guide me on how to start these problems.

1. Ok. Well you know that 2 angles in the parallelogram have the same angle, right? The opposite angles? And the other 2 angles are the same, right? So let's say the small angle is x. That means there are two angles that are x degrees. Now, if the other angles are 30 degrees more, how would you write that? x+30. So, you have 2 x's and 2 (x+30)s. x+x+(x+30)+(x+30)=360 because there are 360 degrees in the parallelogram. Now solve for x. This will be the degrees for the two small angles. Add 30 to get the amount for the larger ones. Make sense?

2. Ok. What are the base angles? Well, in an isoceles triangle, two sides are the same length. There is an angle between them. The base angles are the 2 angles that are NOT between them, and they are EQUAL. So, the two base angles are 88 degrees less that the remaining angle. Let's say the remaining angle is x. What would the sum of the other two angles be? If they are 88 less than the remaining angle, won't the two of them add up to x-88? There are 180 degrees in a triangle, so (x-88)+x=180, solve for x.

ohh.. ^^ thank you :D

now... any1 up for the next 2? ^^

1) start by drawing one,

on the 2 bigger angles put 30+x
on the 2 smaller angles put x
so you have 30+x+30+x+x+x
so now you have 60+4x
in a parallelogram there are 360 degrees
now put it into an equation
60+4x=360
4x=300
x=75
so if the bigger angles are 30+x then they are 30+75=105
so the angles are 105,105,75,75

3, 1/2, 2, -1.5,-2/3,0.75 put in increasing order

Hi please I need help with this question full solution only please. Two angles of a triangle have the same measure and the third one is 9 degrees greater than the measure of each of the other two. Find the measure of the LARGEST angle in the triangle.

Sure, I can help you with these angle problems. Here's the step-by-step guide on how to approach each problem:

1. In a certain parallelogram, if the measure of one angle is 30 degrees more than the measure of a second angle, find the measure of each of the four angles.

Let's call the measure of the second angle x. According to the problem, the measure of one angle is 30 degrees more than x. So, the measure of that angle would be x + 30.

In a parallelogram, opposite angles are congruent. Therefore, the opposite angle to the second angle x is also x. Similarly, the opposite angle to x + 30 would be x + 30.

To find the measure of each angle, you can use the fact that the sum of the measures of the angles in a parallelogram is 360 degrees. So, you can set up an equation:

x + x + (x + 30) + (x + 30) = 360

Solve for x to find the measure of the second angle, and then calculate the measure of the other angles accordingly.

2. In a certain Isosceles triangle, the sum of the measures of the base angles is 88 degrees less than the measure of the remaining angle. Find the measurement of each angle.

Let's call the measure of the base angles in the isosceles triangle x. According to the problem, the sum of the base angles is 88 degrees less than the measure of the remaining angle. So, the equation would be:

2x = (remaining angle) - 88

Since it's an isosceles triangle, the base angles are congruent. So, you can set up another equation using the angle sum property of a triangle:

x + x + (remaining angle) = 180

Now, you have a system of two equations. You can solve for x in the first equation and substitute the value of x in the second equation to find the remaining angle and the base angles.

3. The degree measures of two angles of a triangle are consecutive even integers. If the measure of the third angle is 50 degrees more than twice the measure of the least angle. Find the measure of each angle.

Let's call the measure of the least angle x. According to the problem, the other two angles are consecutive even integers, so their measures would be x + 2 and x + 4.

The measure of the third angle is given as 50 degrees more than twice the measure of the least angle. So, the equation would be:

third angle = 2(x) + 50

Now, you can use the fact that the sum of the measures of the angles in a triangle is 180 degrees. Set up an equation with the three angle measures:

x + (x + 2) + (x + 4) = 180

Solve for x to find the measure of the least angle, then calculate the measures of the other two angles accordingly.

I hope this helps! Let me know if you have any further questions.