In right triangle ABC, the measure of angle C is 90^\circ. If \cos(A) = k \sin(A) for some… — SAT

In right triangle $ABC$ , the measure of angle $C$ is $90^\circ$ . If $\cos(A) = k \sin(A)$ for some positive constant $k$ , which of the following expressions represents $\cos(B)$ in terms of $k$ ?

A
$\frac{k}{\sqrt{1 + k^2}}$
B
$\frac{1}{\sqrt{1 - k^2}}$
$\frac{1}{\sqrt{1 + k^2}}$ Cevap
D
$\sqrt{1 + k^2}$

Cevap

The expression that represents

\cos(B)

\frac{1}{\sqrt{1 + k^2}}

In right triangle

ABC

with right angle

C

, the acute angles

A

and

B

are complementary (

A + B = 90^\circ

). By the co-function identity,

\cos(B) = \sin(A)

. Using the given relation

\cos(A) = k \sin(A)

and substituting it into the Pythagorean identity

\sin^2(A) + \cos^2(A) = 1

, we get

\sin^2(A) + (k \sin(A))^2 = 1 \implies \sin^2(A)(1 + k^2) = 1

. Solving for

\sin(A)

gives

\sin(A) = \frac{1}{\sqrt{1 + k^2}}

because

\sin(A)

must be positive for an acute angle. Since

\cos(B) = \sin(A)

, we conclude that the correct expression is

\frac{1}{\sqrt{1 + k^2}}

Adım Adım Çözüm

Use the relationship between the acute angles in right triangle

ABC

Since angle

C

90^\circ

, angles

A

and

B

are complementary. Thus,

\cos(B) = \sin(A)

This allows us to convert the target term

\cos(B)

into

\sin(A)

, which can be related to the given equation.

Substitute the given relation

\cos(A) = k \sin(A)

into the Pythagorean identity.

Using

\sin^2(A) + \cos^2(A) = 1

, we substitute to get

\sin^2(A) + (k \sin(A))^2 = 1

, which simplifies to

\sin^2(A)(1 + k^2) = 1

This sets up a single equation containing only

\sin(A)

and the constant

k

Solve for

\sin(A)

and substitute back to find

\cos(B)

Solving for

\sin(A)

gives

\sin(A) = \frac{1}{\sqrt{1 + k^2}}

(since

\sin(A) > 0

for acute angle

A

). Since

\cos(B) = \sin(A)

, we have

\cos(B) = \frac{1}{\sqrt{1 + k^2}}

This yields the final value of

\cos(B)

in terms of

k

Anahtar Kavram

Applying complementary angle trigonometric identities (co-functions) and the Pythagorean identity in a right triangle.

Alternatif Yöntem

Alternatively, you can model this by setting up a right triangle. Since

\cos(A) = k \sin(A)

, we can divide both sides by

\sin(A)

to get

\cot(A) = k

, which means

\tan(A) = \frac{1}{k}

. In a right triangle

ABC

with right angle

C

\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AC} = \frac{1}{k}

. Let

BC = 1

and

AC = k

. By the Pythagorean theorem, the hypotenuse

AB = \sqrt{1^2 + k^2} = \sqrt{1 + k^2}

. Then,

\cos(B) = \frac{\text{adjacent to } B}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{1}{\sqrt{1 + k^2}}

Tahmini Süre:2m 0s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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Alternatif Yöntem

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