Sine, Cosine, and Tangent: Practice

Complementary acute angles exchange opposite and adjacent legs, keeping the same hypotenuse.

Thus the requested cosine is $5/11$. The value $4\sqrt6/11$ is the cosine of $\theta$, not the cosine of its complement.

The other reference angle reverses the two legs in tangent.

Therefore $\tan\phi=9/4$. A triangle with opposite leg $4$ and adjacent leg $9$ relative to $\theta$ has hypotenuse $\sqrt{97}$; the other choices involving it are sine and cosine values.

The adjacent leg for $L$ is $KL$, and the hypotenuse is $JL$.

The $12$-unit side is opposite $L$, so it belongs in sine or tangent rather than in the numerator of this cosine.

Pythagoras gives the hypotenuse.

Thus $c=\sqrt{61}$. Sine divides opposite by hypotenuse.

Rationalizing gives $5\sqrt{61}/61$. The sum of the legs is not the hypotenuse; $5/6$ is tangent.

Use the squared-value identity.

The right side is $576/625$. Cosine is positive for an acute angle, so

Subtracting unsquared values gives $18/25$, the negative root violates the domain, and $7/24$ is tangent.

First find the positive sine from the identity.

Thus $\sin\theta=5/13$. Tangent divides sine by cosine.

The result is $5/12$. The reciprocal reverses the legs; $60/169$ multiplies the ratios instead of dividing.

Divide sine by cosine.

The common denominator cancels, leaving $\sqrt7/3$. The product gives $3\sqrt7/16$, reversing the division gives $3\sqrt7/7$, and copying sine leaves out the adjacent leg.

The two factors are equal complementary-angle values.

Their product is $3/4$. Using the wrong special angle, using one wrong factor, or dropping a denominator factor gives the other choices.

The acute angles are complementary, so their tangent values are reciprocals.

Squaring only the smaller or larger tangent would give $1/3$ or $3$. The value $1/2$ is a sine or cosine value, not this product.

Each squared value is $1/2$.

Adding the two gives $1$. The value $\sqrt2$ adds unsquared ratios, $1/2$ keeps only one term, and $2$ ignores the denominators.

For angle $A$, $CA$ is the adjacent leg and $AB$ is the hypotenuse.

Here the written ratio is $60/61$, whereas tangent is $11/60$ and both listed reciprocal values exceed $1$. Sine would use $BC$ in the numerator.

The sum of the two squares must be $1$. For the repeated five-thirteenths pair,

This differs from $1$, so the pair is impossible. Both $3/5,4/5$ orders come from a right triangle, and the equal values $1/\sqrt2$ come from an equal-leg right triangle.