3x- 11 +59 = 90
3x + 48 = 90
3x +48-48 =90-48
3x = 42
3x/3 = 42/3
x= 14
<MNQ = 3x -11
<MNQ = 3(14)-11
<MNQ = 42-11
<MNQ = 31
Answer:
The width is 3 yds
Step-by-step explanation:
Area = length * width
18 = 6*w
Divide each side by 6
18/6 = 6w/6
3=w
The width is 3 yds
Answer:
7.5 ab and cd
4 bd and ac
Step-by-step explanation:
9514 1404 393
Answer:
38.2°
Step-by-step explanation:
The law of sines tells you ...
sin(x)/15 = sin(27°)/11
sin(x) = (15/11)sin(27°) . . . . . multiply by 15
x = arcsin((15/11)sin(27°)) ≈ arcsin(0.619078) ≈ 38.2488°
x ≈ 38.2°
_____
<em>Additional comment</em>
In "law of sines" problems, you need to identify a side and opposite angle that you know both values of. Then, you need to identify whether you're looking for an angle or a side, and whether its opposite side or angle is known. If two angles are known, you can always figure the third from the sum of angles in a triangle.
Here, we have angle 27° opposite side 11. We are looking for an angle, and we know its opposite side. This lets us use the ratio formula directly. Since the angle is the unknown, it is useful to write the equation with sines on top and sides on the bottom.
The given angle is opposite the shorter of the given sides, so this triangle has two solutions. We assume that we want the solution that is an acute angle (141.8° is the other solution). That assumption is based on the drawing. Usually, you're cautioned not to take the drawings at face value.
Answer:
<h3>B = 48.7° , C = 61.3° , b = 12</h3>
Step-by-step explanation:
In order to find B we must first angle C
To find angle C we use the sine rule
That's

From the question
a = 15
A = 70°
c = 14
So we have



C = 61.288
<h3>C = 61.3° to the nearest tenth</h3>
Since we've found C we can use it to find B.
Angles in a triangle add up to 180°
To find B add A and C and subtract it from 180°
That's
A + B + C = 180
B = 180 - A - C
B = 180 - 70 - 61.3
<h3>B = 48.7° to the nearest tenth</h3>
To find b we can use the sine rule
That's



b = 11.9921
<h3>b = 12.0 to the nearest tenth</h3>
Hope this helps you