Question

5 practice 5 (from unit 4, lesson 4)
estimate the values to complete the table. round your answers to the nearest hundredth if necessary.

type your answers in the boxes.
angle adjacent leg ÷ hypotenuse opposite leg ÷ hypotenuse opposite leg ÷ adjacent leg
a 0.31 0.95 3.1
c
how did i do?

Explanation:

Step1: Recall triangle angle relations

In right triangle \(ABC\) (right-angled at \(B\)), \(\angle A+\angle C = 90^\circ\). For complementary angles \(\theta\) and \(90^\circ-\theta\):
- \(\cos(90^\circ - \theta)=\sin\theta\) (adjacent leg/hypotenuse for \(90^\circ - \theta\) is opposite leg/hypotenuse for \(\theta\))
- \(\sin(90^\circ - \theta)=\cos\theta\) (opposite leg/hypotenuse for \(90^\circ - \theta\) is adjacent leg/hypotenuse for \(\theta\))
- \(\tan(90^\circ - \theta)=\frac{1}{\tan\theta}\) (opposite leg/adjacent leg for \(90^\circ - \theta\) is reciprocal of opposite leg/adjacent leg for \(\theta\))

Step2: Find adjacent leg/hypotenuse for \(\angle C\)

For \(\angle A\), adjacent leg/hypotenuse \( = 0.31\) (which is \(\cos A\)), opposite leg/hypotenuse \( = 0.95\) (which is \(\sin A\)), and \(\tan A=\frac{0.95}{0.31}\approx3.1\) (matches the given value).

For \(\angle C\), \(\cos C=\sin A\) (since \(\angle C = 90^\circ-\angle A\)). So adjacent leg/hypotenuse for \(\angle C\) is equal to opposite leg/hypotenuse for \(\angle A\), which is \(0.95\).

Step3: Find opposite leg/hypotenuse for \(\angle C\)

\(\sin C=\cos A\) (because \(\angle C = 90^\circ-\angle A\)). So opposite leg/hypotenuse for \(\angle C\) is equal to adjacent leg/hypotenuse for \(\angle A\), which is \(0.31\).

Step4: Find opposite leg/adjacent leg for \(\angle C\)

\(\tan C=\frac{1}{\tan A}\) (since \(\tan(90^\circ - A)=\cot A=\frac{1}{\tan A}\)). Given \(\tan A = 3.1\), so \(\tan C=\frac{1}{3.1}\approx0.32\) (or more accurately, since \(\tan C=\frac{\sin C}{\cos C}=\frac{0.31}{0.95}\approx0.33\) (wait, correction: \(\sin C=\cos A = 0.31\), \(\cos C=\sin A = 0.95\), so \(\tan C=\frac{\sin C}{\cos C}=\frac{0.31}{0.95}\approx0.33\)? Wait no, earlier relation: \(\tan(90^\circ - A)=\cot A=\frac{\cos A}{\sin A}\). Wait, \(\tan A=\frac{\sin A}{\cos A}=\frac{0.95}{0.31}\approx3.1\), so \(\tan C=\frac{\cos A}{\sin A}=\frac{0.31}{0.95}\approx0.33\)? Wait, let's recast:

For \(\angle A\):
- \(\cos A=\frac{\text{adjacent to }A}{\text{hypotenuse}} = 0.31\) (adjacent leg is \(AB\), hypotenuse \(AC\))
- \(\sin A=\frac{\text{opposite to }A}{\text{hypotenuse}} = 0.95\) (opposite leg is \(BC\), hypotenuse \(AC\))
- \(\tan A=\frac{\text{opposite to }A}{\text{adjacent to }A}=\frac{BC}{AB}=3.1\)

For \(\angle C\):
- Adjacent leg to \(C\) is \(BC\) (opposite to \(A\)), hypotenuse \(AC\), so \(\cos C=\frac{BC}{AC}=\sin A = 0.95\)
- Opposite leg to \(C\) is \(AB\) (adjacent to \(A\)), hypotenuse \(AC\), so \(\sin C=\frac{AB}{AC}=\cos A = 0.31\)
- \(\tan C=\frac{\text{opposite to }C}{\text{adjacent to }C}=\frac{AB}{BC}=\frac{1}{\frac{BC}{AB}}=\frac{1}{\tan A}=\frac{1}{3.1}\approx0.32\) (or \(\frac{0.31}{0.95}\approx0.33\), but let's use the reciprocal of \(\tan A\) since \(\tan C=\cot A\))

Wait, let's compute \(\frac{1}{3.1}\approx0.32\) (more precisely, \(3.1\times0.32 = 0.992\), \(3.1\times0.33 = 1.023\), so \(\frac{1}{3.1}\approx0.32\)). But let's check with the ratios:

\(\tan C=\frac{\sin C}{\cos C}=\frac{0.31}{0.95}\approx0.33\) (since \(0.31\div0.95\approx0.326\approx0.33\)). But maybe the problem expects using the reciprocal of \(\tan A\). Since \(\tan A = 3.1\), \(\tan C=\frac{1}{3.1}\approx0.32\) (rounded to nearest hundredth).

So summarizing:

- Adjacent leg/hypotenuse for \(\angle C\): \(0.95\) (since it's \(\cos C=\sin A\))
- Opposite leg/hypotenuse for \(\angle C\): \(0.31\) (since it's \(\sin C=\cos A\))
- Opposite leg/adjacent leg for \(\angle C\): \(\frac{1}{3.1}\approx0.32\) (or \(\frac{0.31}{0.95}\approx0.33\), but let's use the reciprocal relation)

Wait, let's verify with the triangle:

In right triangle, \(\angle A\) and \(\angle C\) are complementary. So:

- \(\cos C=\sin A = 0.95\) (adjacent over hypotenuse for \(C\) is opposite over hypotenuse for \(A\))
- \(\sin C=\cos A = 0.31\) (opposite over hypotenuse for \(C\) is adjacent over hypotenuse for \(A\))
- \(\tan C=\frac{\sin C}{\cos C}=\frac{0.31}{0.95}\approx0.33\) (wait, \(0.31\div0.95 = 0.3263\approx0.33\) when rounded to nearest hundredth)

But let's check the \(\tan A = 3.1\), so \(\tan C=\frac{1}{3.1}\approx0.32\) (since \(3.1\times0.32 = 0.992\), \(3.1\times0.33 = 1.023\)). There's a slight discrepancy due to rounding. But the key relations are:

- For \(\angle C\):
- adjacent leg/hypotenuse = \(\sin A = 0.95\)
- opposite leg/hypotenuse = \(\cos A = 0.31\)
- opposite leg/adjacent leg = \(\frac{\cos A}{\sin A}=\frac{0.31}{0.95}\approx0.33\) (or \(\frac{1}{\tan A}\approx0.32\))

Given that the problem rounds to nearest hundredth, let's compute:

\(\frac{0.31}{0.95}\approx0.33\) (since \(0.31\div0.95 = 0.3263\approx0.33\))

So the values for \(\angle C\) are:

- adjacent leg/hypotenuse: \(0.95\)
- opposite leg/hypotenuse: \(0.31\)
- opposite leg/adjacent leg: \(\approx0.33\) (or \(0.32\) if using reciprocal of \(3.1\))

But let's confirm with the definitions:

- Adjacent to \(C\): side \(BC\) (opposite to \(A\)), hypotenuse \(AC\). So \(\cos C=\frac{BC}{AC}=\sin A = 0.95\)
- Opposite to \(C\): side \(AB\) (adjacent to \(A\)), hypotenuse \(AC\). So \(\sin C=\frac{AB}{AC}=\cos A = 0.31\)
- \(\tan C=\frac{AB}{BC}=\frac{\cos A}{\sin A}=\frac{0.31}{0.95}\approx0.33\) (since \(AB/BC\) is adjacent to \(A\) over opposite to \(A\), which is \(1/\tan A\))

Yes, so:

- adjacent leg/hypotenuse for \(C\): \(0.95\)
- opposite leg/hypotenuse for \(C\): \(0.31\)
- opposite leg/adjacent leg for \(C\): \(\frac{0.31}{0.95}\approx0.33\) (or \(1/3.1\approx0.32\), but \(0.31/0.95\) is more accurate here)

Answer:

For angle \(C\):

• adjacent leg \(\div\) hypotenuse: \(0.95\)
• opposite leg \(\div\) hypotenuse: \(0.31\)
• opposite leg \(\div\) adjacent leg: \(\approx0.33\) (or \(0.32\) if using \(1/3.1\); precise value from \(\frac{0.31}{0.95}\approx0.33\))

(Note: Depending on rounding, the third value can be \(0.32\) or \(0.33\); the first two are exact due to complementary angle trigonometric identities.)