Using the Law of Sines (sina/A=sinb/B=sincC for any triangle)
sinc/28=sin90/53 (sin90=1) multiply both sides by 28
sinc=28/53 take the inverse of sin (arcsin) of both sides
c=arcsin(28/53)°
c≈31.89° (to nearest hundredth of a degree)
So it is obvious that they rounded to nearest tenth of a degree
c≈31.9°
The missing piece of information in the proof is that m∠DOA and m∠BOC are vertical opposite angles.
<h3>How to find vertically opposite angles?</h3>
We want to prove that m∠BOC = 90°.
We are given that m∠AOB is a right angle triangle. Thus;
m∠BOC must also be equal to 90°
Now, since m∠BOC and m∠AOB are right angles, then it means that we must equally say that m∠DOA and m∠BOC are vertical opposite angles and they are as such equal.
Thus, the missing piece of information in the proof is that m∠DOA and m∠BOC are vertical opposite angles.
Read more about vertical opposite angles at; brainly.com/question/24425517
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Answer:
x≤−38
/7
n>2
Step-by-step explanation:
mathpapa.com is v helpful for this kind of stuff :)
dont worry doe, i did da math so u should b good
Answer:
8:12
Step-by-step explanation:
2+3 = 5
20 div 5 = 4
2 times 4 = 8
3 times 4 =12
8:12
Answer:
486 kg
Step-by-step explanation:
(Calculator)
360 + 35%
=486
I hope it is correct
