Answer:
y = 120°
z = 32°
Step-by-step explanation:
For the sake of simplicity, I labeled one of the interior angles as m < 1 (Please see the attached screenshot). I did this because m < 1 and the exterior angle on the same side of the transversal have the same measure of m < 60°.
Therefore, m < 1 = m < 60°
Next, since m < 1 = 60°, then we can add its measure to < y°, which will sum up to 180° (given that they are supplementary angles).
We can establish the following formula to solve for < y°:
m < 1 + < y ° = 180°
Let's rearrange the formula to isolate < y°:
< y ° = 180° - m < 1
Substitute the value of m < 1 into the revised formula:
< y = 180°- 60°
< y = 120°
Now that we have the value for y°, we can find out what the value of z is by creating another formula:
< (5z - 100)° = 180° - y°
Substitute the value of y° into the formula:
5z - 100° = 180° - 120°
Combine like terms:
5z - 100° = 60°
Add 100° on both sides:
5z - 100° + 100° = 60° + 100°
5z = 160°
Divide both sides by 5:
< z = 32°
Double-check whether we derived the correct answers by plugging in the values of y° and z° into the following formula:
y° + (5z - 100)° = 180° (because they are supplementary angles)
120° + [5(32) - 100]° = 180°
120° + [5(32) - 100]° = 180°
120° + [160 - 100]° = 180°
120° + 60° = 180°
180° = 180° (True statement).
Therefore, the value of y = 120° and z = 32°.
Please mark my answers as the Brainliest if you find my explanations helpful :)
We can see that two sides of that triangle area equal
so, their corresponding angle will also be equal
now, we know that
sum of all angles in any triangle is always 180
so, we can get equation as
now, we can solve for x
so, option-C..............Answer
Answer:
In First method : counting up, counting back on a number line,
If we want the quotient after dividing the number by 5 then we count how many 5 we get from 0 to the dividend.
For example :
Since, from 0 to 30 there are six 5's obtained. ( because 5 × 6 = 30 )
Thus,
In Second Method : dividing by 10, and then doubling the quotient.
First we divide the number by 10 then multiply the quotient by 2.
For Example:
Since,
Thus,
Now, when we compare the above methods then we conclude that for the smaller numbers first method is appropriate because for small numbers we can easily count total 5's from 0. While for large numbers Second method is appropriate because it is hard to count the total 5's for the large number.