Answer:
Step-by-step explanation:
1). m∠AEC = m∠AEB + m∠BEC
= 21° + 37°
= 58°
2). m∠BED = m∠BEC + m∠CED
= 37° + 44°
= 81°
3). m∠IKF = m∠IKH + m∠HKG + m∠GKF
= m∠IKH + m∠HKG + m∠IKH [Since, ∠IKH ≅ ∠GKF]
= 2∠IKH + m∠HKG
103° = 2∠IKH + 41°
2(∠IKH) = 103 - 41
m(∠IKH) = 31°
4). m∠AED = m∠AEB + m∠BEC + m∠CED
= 21° + 37° + 44°
= 102°
5). m∠JKG = 108°
m∠JKG = m∠JKI + m∠IKH + m∠HKG
108° = m∠JKI + 31° + 41°
m∠JKI = 108° - 72°
m∠JKI = 36°
6). m∠HKF = m∠GKF + m∠HKG
= m∠IKH + m∠HKG [Since, m∠GKF = m∠IKH]
= 31° + 41°
= 72°
7). m∠NQO = m∠MQN = 64°
8). m∠JKF = m∠JKI + m∠IKF
= 36° + 103°
= 139°
8). m∠MQO = 2(m∠NQO)
= 2(64)°
= 128°
9). m∠LQO = 156°
m∠LQM = m∠LQO - m∠MQO
= 156° - 128°
= 28°
10. m∠NQP = m∠NQO + m∠OQP
= 64° + m∠LQM [Since ∠OQP ≅ ∠LQM]
= 64° + 28°
= 92°
The complete question in the attached figure
we have that
tan a=7/24 a----> III quadrant
cos b=-12/13 b----> II quadrant
sin (a+b)=?
we know that
sin(a + b) = sin(a)cos(b) + cos(a)sin(b<span>)
</span>
step 1
find sin b
sin²b+cos²b=1------> sin²b=1-cos²b----> 1-(144/169)---> 25/169
sin b=5/13------> is positive because b belong to the II quadrant
step 2
Find sin a and cos a
tan a=7/24
tan a=sin a /cos a-------> sin a=tan a*cos a-----> sin a=(7/24)*cos a
sin a=(7/24)*cos a------> sin²a=(49/576)*cos²a-----> equation 1
sin²a=1-cos²a------> equation 2
equals 1 and 2
(49/576)*cos²a=1-cos²a---> cos²a*[1+(49/576)]=1----> cos²a*[625/576]=1
cos²a=576/625------> cos a=-24/25----> is negative because a belong to III quadrant
cos a=-24/25
sin²a=1-cos²a-----> 1-(576/625)----> sin²a=49/625
sin a=-7/25-----> is negative because a belong to III quadrant
step 3
find sin (a+b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin a=-7/25
cos a=-24/25
sin b=5/13
cos b=-12/13
so
sin (a+b)=[-7/25]*[-12/13]+[-24/25]*[5/13]----> [84/325]+[-120/325]
sin (a+b)=-36/325
the answer issin (a+b)=-36/325
. Correct option C)5.38
<u>Step-by-step explanation:</u>
Here we have , Find the value of x. O is the center of the circle. Round your answer to the nearest hundredth. picture is attached . Let's find out:
In the given figure , Let's draw a line from center to the point where cord of length 8 unit is touching the circle or intersecting or , where this chord finishes . This line is radius of circle and is denoted by x , So now we have a right angle triangle with dimensions as :

By Pythagoras Theorem ,

⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Therefore ,
. Correct option C)5.38
The original area of a face would be a^2. Now that you added b to the edge, the new area of each face would be (a+b)^2. To find how much the are increased, subtract a^2 from (a+b)^2.
So the answer is b(2a+b)