In the right triangle ABC shown below what is the value of cos A

10 answers
Question:

In the right triangle ABC shown below what is the value of cos A

Answers

You need to put an image of the triangle or else we can’t help you :)

A

(-2,1)

Step-by-step explanation:

Just did it on Edge

(4,-1)

Step-by-step explanation:

just took the test

(4,-2)

Step-by-step explanation:

because both angle are equal so point E is (4,-2)

-2,1

Step-by-step explanation:

n/a

This question is incomplete because it lacks the diagram of the right angled triangle. Find attached to this answer the diagram of the right angle triangle.

d-50

Step-by-step explanation:

Looking at the attached diagram, the only way to solve for this is the use of the trigonometric function. The trigonometric function to be used is the cosine function.

From the diagram, we are given

Hypotenuse = AB = 14

Adjacent = AC = 9

The measure of angle A to the nearest degree is calculated as:

cos θ = Adjacent / Hypothenuse

cos θ = 9/14

θ = cos -¹ (9/14) or arccos(9/14)

θ = 49.994799115°

To the nearest degree = 50°

Therefore,the measure of angle A to the nearest degree = 50°


[tex]In the diagram of right triangle ABC shown below, AB = 14 and AC = 9. What is the measure of angle A[/tex]

AC = 9

Step-by-step explanation:

[tex]cosine=\dfrac{adjacent}{hypotenuse}\\\\\text{We have:}\\\\\cos C=\dfrac{9\sqrt{130}}{130}\\\\adjacent=AC\\\\hypotenuse=\sqrt{130}\\\\\text{Substitute:}\\\\\dfrac{9\sqrt{130}}{130}=\dfrac{AC}{\sqrt{130}}\qquad\text{cross multiply}\\\\130AC=(9\sqrt{130})(\sqrt{130})\qquad\text{use}\ \sqrt{a}\cdot\sqrt{a}=a\\\\130AC=(9)(130)\qquad\text{divide both sides by 130}\\\\\dfrac{130AC}{130}=\dfrac{(9)(130)}{130}\\\\AC=9[/tex]

Diagram missing.

Question interpreted as:
ABC is a right triangle right angled at A.
BC=17 feet
angle C = 35 degrees.

AC=BCcos(35)=17*0.8192=13.93 feet to the nearest 1/100th of a foot.

First prove that ∆CBD ~ ∆ABC. 
 Each triangle has a right triangle and each includes angle B.   The triangles are similar. 
 You can use similar reasoning to show that ∆ACD ~ ∆ABC. 
 To show that ∆CBD ~ ∆ACD, begin by showing that  angleACD = angle B because they are both complementary to angleDCB.  Then you can use Similarity Postulates.
 I found the image of the triangle that was missing the question! Attached triangle.
[tex]In right triangle abc shown below, altitude cd is drawn to hypotenuse (ab). explain why δabc ∼ δacd.[/tex]

H = -b / 2a = 2 : x coordinate of the vertex of the parabola k = -(2)2 + 4(2) + C = 4 + C : y coordinate of vertex x = (2 + √(4 + C)) , x = (2 - √(4 + C)) : the two x intercepts of the parabola. length of BA = k = 4 + C length of AC = 2 + √(4 + C) - 2 = √(4 + C) area = (1/2)BA * AC = (1/2) (4 + C) * √(4 + C) (1/2) (4 + C) * √(4 + C) = 32 : area is equal to 32 C = 12 : solve above for C.

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