Answer:
1) ∠A=84°
2) ∠C=20°
Step-by-step explanation:
1)
First, find ∠C:
<em>(I'm assuming the exterior angle of 126° makes a straight line with ∠C)</em>
The angles on a straight line always add up to 180. Therefore:
∠C+126=180
∠C=180-126
∠C=54
Then find ∠B:
We also know that all the angles in a triangle add up to 180. Therefore:
∠A+∠B+∠C=180
∠A+∠B+54=180
∠A+∠B=126
<em>(we know ∠A=2(∠B))</em>
2(∠B)+∠B=126
3(∠B)=126
∠B=42
Now, find ∠A:
∠A=2(∠B)
∠A=2(42)
∠A=84°
2)
First, find ∠B:
<em>(Again, I'm assuming the exterior angle of 100° makes a straight line with ∠B)</em>
The angles on a straight line always add up to 180. Therefore:
∠B+100=180
∠B=180-100
∠B=80
Then find ∠A:
We also know that all the angles in a triangle add up to 180. Therefore:
∠A+∠B+∠C=180
∠A+80+∠C=180
∠A+∠C=100
<em>(we know ∠A=4(∠C))</em>
4(∠C)+∠C=100
5(∠C)=100
∠C=20°
I believe the answer is either 40 or 250.
I hope this helps you
Area=length ×width
30 1/3=30.3+1/3=91/3
6 1/2=6.2+1/2=13/2
91/3=13/2.width
width =7.13.2/3.13
width =14/3
width = 4 2/3
<span>The problem is to calculate the angles of the triangle. However, it is not clear which angle you have to calculate, so we are going to calculate all of them
</span>
we know that
Applying the law of cosines
c²=a²+b²-2*a*b*cos C------> cos C=[a²+b²-c²]/[2*a*b]
a=12.5
b=15
c=11
so
cos C=[a²+b²-c²]/[2*a*b]---> cos C=[12.5²+15²-11²]/[2*12.5*15]
cos C=0.694------------> C=arc cos (0.694)-----> C=46.05°-----> C=46.1°
applying the law of sines calculate angle B
15 sin B=11/sin 46.1-----> 15*sin 46.1=11*sin B----> sin B=15*sin 46.1/11
sin B=15*sin 46.1/11-----> sin B=0.9826----> B=arc sin (0.9826)
B=79.3°
calculate angle A
A+B+C=180------> A=180-B-C-----> A=180-79.3-46.1----> A=54.6°
the angles of the triangle are
A=54.6°
B=79.3°
C=46.1°
Answer:
D. (7,2)
Step-by-step explanation:
-3 +10 is 7
9-7 is 2.
kinda just put them together, you get (7, 2)
I hope this helps!
pls ❤ and mark brainliest pls!
