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Question
determining side lengths and trigonometric ratios
determine the values of sine, cosine, and tangent for angle l.
$a^2 + b^2 = c^2$
$10^2 + \underline{quadquad}^2 = 26^2$
$\underline{quadquad} + ln^2 = 676$
$ln = \sqrt{576}$
$ln = \underline{quadquad}$
label the sides related to $\angle l$ \hypotenuse\, \adjacent\, and \opposite\.
$\sin(l) = \frac{\underline{quadquad}}{26} = \frac{5}{13}$
$\cos(l) = \frac{24}{\underline{quadquad}} = \frac{12}{13}$
$\tan(l) = \frac{\underline{quadquad}}{24} = \frac{5}{12}$
Step1: Solve for LN using Pythagorean theorem
Given the right triangle with hypotenuse \( LM = 26 \) and one leg \( MN = 10 \). The Pythagorean theorem is \( a^{2}+b^{2}=c^{2} \), where \( c \) is the hypotenuse. Here, \( 10^{2}+LN^{2}=26^{2} \). Calculate \( 10^{2}=100 \) and \( 26^{2} = 676 \). So the equation becomes \( 100+LN^{2}=676 \).
Step2: Isolate \( LN^{2} \)
Subtract 100 from both sides: \( LN^{2}=676 - 100=576 \).
Step3: Find LN
Take the square root of 576: \( LN=\sqrt{576} = 24 \).
Step4: Label sides for \( \angle L \)
- Hypotenuse: The side opposite the right angle, so \( LM = 26 \) (hypotenuse).
- Opposite: The side opposite \( \angle L \), which is \( MN = 10 \) (opposite).
- Adjacent: The side adjacent to \( \angle L \) (and part of the right angle), which is \( LN = 24 \) (adjacent).
Step5: Check trigonometric ratios (optional for side length part)
- \( \sin(L)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{10}{26}=\frac{5}{13} \) (matches given).
- \( \cos(L)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{24}{26}=\frac{12}{13} \) (the blank in \( \cos(L) \) denominator should be 26, and simplifying \( \frac{24}{26} \) gives \( \frac{12}{13} \)).
- \( \tan(L)=\frac{\text{opposite}}{\text{adjacent}}=\frac{10}{24}=\frac{5}{12} \) (matches given).
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For the side length of \( LN \), the value is \( 24 \). For the cosine ratio's denominator blank, it is \( 26 \), and the simplified cosine is \( \frac{12}{13} \). (For the side - length part of \( LN \), the key steps show \( LN = 24 \))