Given triangle ABC, which equation could be used to find the measure of ∠B? right triangle ABC with AB measuring 6, AC measuring

10 answers
Question:

Given triangle ABC, which equation could be used to find the measure of ∠B? right triangle ABC with AB measuring 6, AC measuring 3, and BC measuring 3 square root of 5 cos m∠B = square root of 5 over 5 sin m∠B = square root of 5 over 5 cos m∠B = square root of 5 over 2 sin m∠B = 2 square root of 5 all over 5

Answers

 The correct option is (B)

Step-by-step explanation:  We are given a right-angled triangle ABC, where AC is the hypotenuse and

We are to select the equation that could be used to find the measure of ∠B.

With respect to angle B,

perpendicular, AC = 3 units,  base, AB = 6 units  and  hypotenuse, BC = 3√5 units.

So, the sine and cosine of angle B can be written as

and

Thus, the correct equation that could be used to find the measure of ∠B is

Option (B) is CORRECT.

[tex]cos B = \dfrac{2\times \sqrt5}{5}[/tex] is the correct answer.

Step-by-step explanation:

We are given a right angled [tex]\triangle ABC[/tex] with the side measurements as:

AB = 6

AC = 3

BC = 3[tex]\sqrt5[/tex]

In a right angled triangle, the angle of [tex]90^\circ[/tex] is the largest angle and the side opposite to the right angle is the largest.

We are given that the side BC = 3[tex]\sqrt5[/tex] which has a value greater than 6 which means side BC is the largest side.

And angle opposite to BC is [tex]\angle A[/tex] is the largest i.e.

[tex]\angle A=90^\circ[/tex]

Please refer to the attached image for the given dimensions of the triangle.

Now, we can apply trigonometric rules to easily find out the value of [tex]\angle B[/tex]

[tex]cos \theta = \dfrac{Base}{Hypotenuse}\\OR\\cos B = \dfrac{AB}{BC}\\\Rightarrow cos B = \dfrac{6}{3\sqrt5}\\\Rightarrow cos B = \dfrac{2}{\sqrt5}\\\Rightarrow cos B = \dfrac{2\times \sqrt5}{\sqrt5 \times \sqrt5}\\\Rightarrow cos B = \dfrac{2\times \sqrt5}{5}[/tex]


[tex]Given triangle ABC, which equation could be used to find the measure of ∠B? right triangle ABC with[/tex]

A - cos m∠B=2 square root /5

Step-by-step explanation:

A

Step-by-step explanation:

cosB = = =

Rationalise the denominator by multiplying numerator/denominator by

= ( divide 6 and 15 by 3 )

=

Hence

cosB = → A

answer:

solve my question

Step-by-step explanation:

then I will solve this question because it is Easy for me

Check the picture below.

now, let's rationalize the denominator.

Cos m∠B=2²/5

Step-by-step explanation:

sin angle B= square root of 5 / 5

Step-by-step explanation:

took the test and got it right!

Sin m∠B = square root of 5 over 5

Step-by-step explanation:

I took the FLVS quiz and it was right. I guessed so I don't have an explanation.

CosB=6/3sqrt(5). Then just apply arccos to both sides to find the measure of Angle B.

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