Solved

In the Figure,? and ? Are Positive Angles c=12sin[πβarcsin(0.25β)]sinβc = \frac { 12 \sin [ \pi - \beta - \arcsin ( 0.25 \beta ) ] } { \sin \beta }

Question 22

Multiple Choice

In the figure,? and ? are positive angles.?  In the figure,? and ? are positive angles.?   ? a = 3,b = 12 Use ? = arcsin(0.25sin?) to write c as a function of ?. ? A)   c = \frac { 12 \sin [ \pi - \beta - \arcsin ( 0.25 \beta )  ] } { \sin \beta }  B)   c = \frac { 12 \sin [ \pi + \beta - \arcsin ( 0.25 \beta )  ] } { \sin \beta }  C)   c = \frac { 0.25 \sin [ \pi - \beta - \arcsin ( 12 \beta )  ] } { \sin \beta }  D)   c = \frac { 12 \sin [ \pi + \beta + \arcsin ( 0.25 \beta )  ] } { \sin \beta }  E)   c = \frac { 12 \sin [ \pi - \beta + \arcsin ( 0.25 \beta )  ] } { \sin \beta } ? a = 3,b = 12
Use ? = arcsin(0.25sin?) to write c as a function of ?.
?


A) c=12sin[πβarcsin(0.25β) ]sinβc = \frac { 12 \sin [ \pi - \beta - \arcsin ( 0.25 \beta ) ] } { \sin \beta }
B) c=12sin[π+βarcsin(0.25β) ]sinβc = \frac { 12 \sin [ \pi + \beta - \arcsin ( 0.25 \beta ) ] } { \sin \beta }
C) c=0.25sin[πβarcsin(12β) ]sinβc = \frac { 0.25 \sin [ \pi - \beta - \arcsin ( 12 \beta ) ] } { \sin \beta }
D) c=12sin[π+β+arcsin(0.25β) ]sinβc = \frac { 12 \sin [ \pi + \beta + \arcsin ( 0.25 \beta ) ] } { \sin \beta }
E) c=12sin[πβ+arcsin(0.25β) ]sinβc = \frac { 12 \sin [ \pi - \beta + \arcsin ( 0.25 \beta ) ] } { \sin \beta }

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions