WILL GIVE BRAINLIEST ANSWER TO WHOEVER IS ABLE TO FULLY AND CORRECTLY ANSWER THIS QUESTION. Consider and trapezoid in the following image and use them to answer the questions below. is a right triangle and trapezoid is isosceles. Do NOT use the law of cosines, use our knowledge from the content of this course. a. Describe how to find the lengths of segments and . Show your answers for this question. Explain how you can be confident your calculations are correct. b. Describe how to find the angle measures of and . Explain how you can be confident your calculations are correct. c. Are there any segments or angles you can’t calculate based solely on the information in the image? Why? d. If we drew a median into trapezoid and the given measurement of that median was 21.425, would that change your answer for part c of this question?

This is an isosceles trapezoid, because the two angles are supplementary on the right side.

y is 25 based on the Pythagorean theorem

w is sqrt (850)=29.15

z must be 15

Making a right triangle to look at the altitude

cos 59=a/15, so a is 7.73 between the right corner and the edge of x

That means x is 29.15-2(7.73)=13.69

The median should be half way between 13.69 and 29.15

That is half their sum, or half of 42.84 or 21.42

That doesn't change the answer above

Once I know the trapezoid is isosceles, the angles at the base have to be 59d and the angles at the top 121d.

please mark brainliest!

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)