Question

triangle abc is a right triangle and \\(\cos(22.6^\circ)=\frac{b}{13}\\). solve for \\(b\\) and round to the nearest whole number.\
\
which equation correctly uses the value of \\(b\\) to solve for \\(a\\)?\
\\(\tan(22.6^\circ)=\frac{a}{13}\\)\
\\(\tan(22.6^\circ)=\frac{13}{a}\\)\
\\(\tan(22.6^\circ)=\frac{a}{12}\\)\
\\(\tan(22.6^\circ)=\frac{12}{a}\\)

Explanation:

Step1: Identify the right triangle components

In right triangle \( ABC \) with \( \angle C = 90^\circ \), \( \angle A = 22.6^\circ \), hypotenuse \( AB = 13 \) cm, opposite side to \( \angle A \) is \( BC = a \), adjacent side is \( AC = b \).

Step2: Recall trigonometric ratios

For tangent function, \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). Here, \( \theta = 22.6^\circ \), opposite \( = a \), adjacent \( = b \)? Wait, no, wait: Wait, actually, let's re - check the triangle. Wait, the right angle is at \( C \), so sides: \( AC = b \), \( BC = a \), \( AB = 13 \). Wait, maybe I made a mistake. Wait, \( \tan(22.6^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AC}=\frac{a}{b} \)? No, wait, no. Wait, angle at \( A \) is \( 22.6^\circ \), so the sides: opposite to \( A \) is \( BC = a \), adjacent to \( A \) is \( AC = b \), hypotenuse \( AB = 13 \). Wait, but also, we can use Pythagoras? Wait, no, the options are about \( \tan(22.6^\circ) \). Wait, \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). If we consider the sides, maybe \( b \) and \( a \) and the other side. Wait, maybe the triangle has \( AB = 13 \), and we can also find that maybe \( \cos(22.6^\circ)=\frac{b}{13} \) (since \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \), adjacent to \( A \) is \( b \), hypotenuse \( 13 \)), and \( \tan(22.6^\circ)=\frac{a}{b} \) (since \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \), opposite \( a \), adjacent \( b \)). But the options are in terms of \( a \) and 12? Wait, maybe there is a 5 - 12 - 13 triangle? Because \( 5^2+12^2 = 13^2 \). So maybe \( b = 12 \)? Wait, if \( \cos(22.6^\circ)=\frac{b}{13} \), and if \( b = 12 \), then \( \cos(22.6^\circ)=\frac{12}{13}\approx0.923 \), and \( \cos(22.6^\circ)\approx0.923 \), which matches. So then, \( \tan(22.6^\circ)=\frac{a}{b}=\frac{a}{12} \), because \( b = 12 \) (since 5 - 12 - 13 triangle, \( b = 12 \), \( a = 5 \)? Wait, no, wait. Wait, if \( \cos(22.6^\circ)=\frac{b}{13} \), and \( \cos(22.6^\circ)=\frac{12}{13} \), so \( b = 12 \). Then \( \tan(22.6^\circ)=\frac{a}{b}=\frac{a}{12} \), which is one of the options.

Step3: Solve for \( b \) using cosine

We know that \( \cos(22.6^\circ)=\frac{b}{13} \). We know that \( \cos(22.6^\circ)\approx\frac{12}{13}\approx0.923 \). So \( b = 13\times\cos(22.6^\circ) \). Calculating \( 13\times\cos(22.6^\circ) \approx13\times0.923 = 12.0 \) (rounded to nearest whole number).

Step4: Solve for \( a \) using tangent or Pythagoras

Using Pythagoras: \( a=\sqrt{13^{2}-b^{2}} \). Since \( b = 12 \), \( a=\sqrt{169 - 144}=\sqrt{25}=5 \). Or using tangent: \( \tan(22.6^\circ)=\frac{a}{12} \), so \( a = 12\times\tan(22.6^\circ) \). \( \tan(22.6^\circ)\approx0.414 \), so \( 12\times0.414\approx5 \).

For the equation to solve for \( a \): Since \( \tan(22.6^\circ)=\frac{a}{12} \) (because \( b = 12 \), adjacent side is 12, opposite is \( a \)), so the correct equation is \( \tan(22.6^\circ)=\frac{a}{12} \).

Answer:

The equation to solve for \( a \) is \( \tan(22.6^\circ)=\frac{a}{12} \). For \( b \): \( b = 13\times\cos(22.6^\circ)\approx12 \) (rounded to nearest whole number). For \( a \): \( a = 5 \) (using Pythagoras or tangent, rounded to nearest whole number). And the correct equation from the options is the one with \( \tan(22.6^\circ)=\frac{a}{12} \).