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Please answer correctly !!!!! Will mark Brianliest !!!!!!!!!!!!!!

Please answer correctly !!!!! Will mark Brianliest !!!!!!!!!!!!!!-example-1

2 Answers

6 votes

Answer:

Can't type the Full solution

I'll solve for the triangle with angles 1 and 2

You can go for the rest.

CosA = 0.28

Taking the cos-¹(0.28) to determine its Magnitude

A=73.74°

Now

We can use sine Rule to get the values of each of the other angles in the triangle

For the first triangle

Applying sine rule

What basically happens in the rule is... Each side divided by the angle opposite it.

Look below

The angle opposite 40cm is angle '2' and the angle opposite 41cm is 90°

We have this sine rule when all three sides are given with ONE ANGLE(90° is the given angle in this case)

So

Each side over the angle opposite it

We go!!

Let's call angle "2" an alphabet. Say Q

41/sin90 = 40/SinQ

SinQ = 40sin90/41

Recall sin90=1

SinQ = 40/41

Q= Sin-¹(40/41)

Q= 77.32°

so angle 2 is 77.32°

Now apply this to get others

Angle '1' should be 12.68°

Angle "3" should be 16.26°

Angle "4" should be 73.74°

Therefore

conclusion

Angle "4" is same as A. That's the angle we looking for

Option 4 IS CORRECT!!

User Dimitar Genov
by
3.5k points
4 votes

Answer:

Angle 4 is ∠A

Explanation:

Cos(A) = adjacent / hypotenuse

=> Angle 1: Cos(1) = 40 / 41, or 0.98 (so this is not ∠A)

=> Angle 2: Cos(2) = 9 / 41 or 0.22 (This isn't ∠A either)

=> Angel 3: Cos(3) = 24 / 25 or 0.96 (Not ∠A)

=> Angel 4: Cos(4) = 7 / 25 or 0.28 (This is Angle A, since Cos(A) = 0.28)

Therefore; Angle 4 is ∠A

Hope this helps!

User Computer
by
3.0k points