3
------------
4(5z+16)
I'm assuming you want to include all of the numbers
#1)
A) b = 10.57
B) a = 22.66; the different methods are shown below.
#2)
A) Let a = the side opposite the 15° angle; a = 1.35.
Let B = the angle opposite the side marked 4; m∠B = 50.07°.
Let C = the angle opposite the side marked 3; m∠C = 114.93°.
B) b = 10.77
m∠A = 83°
a = 15.11
Explanation
#1)
A) We know that the sine ratio is opposite/hypotenuse. The side opposite the 25° angle is b, and the hypotenuse is 25:
sin 25 = b/25
Multiply both sides by 25:
25*sin 25 = (b/25)*25
25*sin 25 = b
10.57 = b
B) The first way we can find a is using the Pythagorean theorem. In Part A above, we found the length of b, the other leg of the triangle, and we know the measure of the hypotenuse:
a²+(10.57)² = 25²
a²+111.7249 = 625
Subtract 111.7249 from both sides:
a²+111.7249 - 111.7249 = 625 - 111.7249
a² = 513.2751
Take the square root of both sides:
√a² = √513.2751
a = 22.66
The second way is using the cosine ratio, adjacent/hypotenuse. Side a is adjacent to the 25° angle, and the hypotenuse is 25:
cos 25 = a/25
Multiply both sides by 25:
25*cos 25 = (a/25)*25
25*cos 25 = a
22.66 = a
The third way is using the other angle. First, find the measure of angle A by subtracting the other two angles from 180:
m∠A = 180-(90+25) = 180-115 = 65°
Side a is opposite ∠A; opposite/hypotenuse is the sine ratio:
a/25 = sin 65
Multiply both sides by 25:
(a/25)*25 = 25*sin 65
a = 25*sin 65
a = 22.66
#2)
A) Let side a be the one across from the 15° angle. This would make the 15° angle ∠A. We will define b as the side marked 4 and c as the side marked 3. We will use the law of cosines:
a² = b²+c²-2bc cos A
a² = 4²+3²-2(4)(3)cos 15
a² = 16+9-24cos 15
a² = 25-24cos 15
a² = 1.82
Take the square root of both sides:
√a² = √1.82
a = 1.35
Use the law of sines to find m∠B:
sin A/a = sin B/b
sin 15/1.35 = sin B/4
Cross multiply:
4*sin 15 = 1.35*sin B
Divide both sides by 1.35:
(4*sin 15)/1.35 = (1.35*sin B)/1.35
(4*sin 15)/1.35 = sin B
Take the inverse sine of both sides:
sin⁻¹((4*sin 15)/1.35) = sin⁻¹(sin B)
50.07 = B
Subtract both known angles from 180 to find m∠C:
180-(15+50.07) = 180-65.07 = 114.93°
B) Use the law of sines to find side b:
sin C/c = sin B/b
sin 52/12 = sin 45/b
Cross multiply:
b*sin 52 = 12*sin 45
Divide both sides by sin 52:
(b*sin 52)/(sin 52) = (12*sin 45)/(sin 52)
b = 10.77
Find m∠A by subtracting both known angles from 180:
180-(52+45) = 180-97 = 83°
Use the law of sines to find side a:
sin C/c = sin A/a
sin 52/12 = sin 83/a
Cross multiply:
a*sin 52 = 12*sin 83
Divide both sides by sin 52:
(a*sin 52)/(sin 52) = (12*sin 83)/(sin 52)
a = 15.11
If you apply the or both
Only 1 of the students would need to know the "or both", therefore maximizing the remaining amount of students you can put in.
Gerald, let's call him, knows French AND German, so there's only one less student that knows french and german. Gerald is 1 student.
MAXIMUM:
There are now 14 monolinguistic French speakers and 16 monolinguistic German's, 30 students + Gerald=31.
Minimum:
As a bonus, the minimum is 15 students knowing french AND German and only 2 monolinguistic German speakers, so 17.
I promise this app called brainily is a life saver for algebra problems like this !! It’s not a scam trust me
and you can expand the numerator if you wish, it won't be simplified further though.
