What is the measure of angle 1?
1 answer:
Answer:
27 degrees
Step-by-step explanation:
The easy way is by remembering the formula (a-b)/2=c, where a is the larger angle, and b is the smallest angle. (90-36)/2=27.
The longer, more drawn out answer goes as follows. See the image to understand the notation I use:
- AOE + BOD + BOA + DOE = 360
- AOE + BOD = 90 + 36 = 126
- BOA + DOE = 360 - AOE - BOD = 234
- Since the sum of a triangle's angles is 180, ODE = (180 - DOE) / 2
- Likewise, OBA = (180 - BOA) / 2
- Since CDE is 180, CDO = 180 - ODE = 180 - (180 - DOE) / 2
- Likewise, CBA is 180, so CBO = 180 - OBA = 180 - (180 - BOA) / 2
- The interior angles of the irregular polygon CBOD add up to 360, so CBO + CDO + BOD + BCD = 360.
- Substituting what we already found, 180 - (180 - BOA)/2 + 180 - (180 - DOE)/2 + 36 + BCD = 360
- Cleaning it all up, we get 180 + (BOA + DOE)/2 + 36 + BCD = 360
- As we found in line 3, BOA + DOE = 234, so substituting that in, 180 + 117 + 36 + BCD = 360
- Finally, solving for BCD (360 - 36 - 117 - 180) we get our answer, 27
Note: The long drawn out method shown above is a way to derive the formula for the secant theorem. You do not need to use this method every time. Just remember, large angle minus small angle, all divided by 2. That is it.
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THIS IS THE COMPLETE QUESTION
basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modeled by the equation h = 6 + 30t – 16t2. Two-tenths of a second after the shot is launched, an opposing player leaps up to block the shot. The height of the shot blocker’s outstretched hands t seconds after he leaps is modeled by the equation h = 9 + 25t – 16t2. If the ball reaches the net 1.7 seconds after the shooter launches it, does the leaping player block the shot?
Yes, exactly 0.6 seconds after the shot is launched.
Yes, between 0.64 seconds and 0.65 seconds after the shot is launched
Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.
No, the shot is not blocked.
Answer:
Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.
Given that the
equation for shot block height h = 9 + 25t – 16t2.
equation for ball height h = 6 + 30t – 16t2.
From the question ,it can be deducted that the shot is made before two tenths of a second or 0.2seconds
CHECK THE ATTACHMENT TO COMPLETE THE DETAILED SOLUTION
