Question

which of the following expressions is equal to the height, h, of this triangle?
a. 25(cos34°) c. 25(tan34°)
b. 25(cos56°) d. 25(tan56°)
please select the best answer from the choices provided

Explanation:

Step1: Recall SOHCAHTOA

In a right triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.

Step2: Identify sides for angle 34°

Hypotenuse = 25, angle = 34°, height \( h \) is opposite to 34°.

Step3: Use sine for 34° or check other angles

Alternatively, the other non - right angle is \( 90 - 34=56^{\circ} \). For angle 56°, \( h \) is adjacent? No, wait. Wait, for angle 34°, opposite is \( h \), hypotenuse 25. So \( \sin(34^{\circ})=\frac{h}{25}\Rightarrow h = 25\sin(34^{\circ}) \). But also, we can use angle 56°: \( \cos(34^{\circ})=\sin(56^{\circ}) \), and \( \tan(56^{\circ})=\frac{h}{\text{adjacent to }56^{\circ}} \). Wait, let's re - evaluate. Wait, the right triangle has angles 90°, 34°, 56°. Let's take angle 56°: the side opposite to 56° is the adjacent side of 34°, and the side adjacent to 56° is \( h \) (opposite to 34°). Wait, no. Wait, let's list:

For angle 34°:
- Opposite: \( h \)
- Adjacent: let's call it \( a \)
- Hypotenuse: 25

For angle 56°:
- Opposite: \( a \)
- Adjacent: \( h \)
- Hypotenuse: 25

Now, \( \tan(56^{\circ})=\frac{\text{opposite to }56^{\circ}}{\text{adjacent to }56^{\circ}}=\frac{a}{h} \)? No, wait \( \tan(56^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{h} \)? No, opposite to 56° is \( a \) (the side adjacent to 34°), adjacent to 56° is \( h \) (the side opposite to 34°). Wait, no, I made a mistake. Let's start over.

We know that in a right triangle, \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \), \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \), \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \).

Given angle 34°, hypotenuse = 25, opposite side to 34° is \( h \). So \( \sin(34^{\circ})=\frac{h}{25}\Rightarrow h = 25\sin(34^{\circ}) \). But we also know that \( \sin(34^{\circ})=\cos(56^{\circ}) \) and \( \tan(56^{\circ})=\frac{\sin(56^{\circ})}{\cos(56^{\circ})}=\frac{\cos(34^{\circ})}{\sin(34^{\circ})} \)? No, wait, \( \tan(56^{\circ})=\frac{\text{opposite to }56^{\circ}}{\text{adjacent to }56^{\circ}} \). The side opposite to 56° is the adjacent side of 34° (let's call it \( x \)), and the side adjacent to 56° is \( h \) (opposite to 34°). So \( \tan(56^{\circ})=\frac{x}{h} \), and \( \sin(34^{\circ})=\frac{h}{25} \), \( \cos(34^{\circ})=\frac{x}{25} \). Then \( \tan(56^{\circ})=\frac{\cos(34^{\circ})}{\sin(34^{\circ})}=\cot(34^{\circ}) \), which is not helpful. Wait, another approach: Let's check the options.

Option a: \( 25(\cos34^{\circ}) \): \( \cos34^{\circ}=\frac{x}{25} \) (x is adjacent to 34°), so \( 25\cos34^{\circ}=x \), not \( h \).

Option b: \( 25(\cos56^{\circ}) \): \( \cos56^{\circ}=\sin34^{\circ}=\frac{h}{25} \), so \( 25\cos56^{\circ}=h \). Wait, but also, let's check option d: \( 25(\tan56^{\circ}) \). \( \tan56^{\circ}=\frac{h}{x} \), and \( x = 25\cos34^{\circ} \), so \( h=25\cos34^{\circ}\tan56^{\circ} \). But \( \tan56^{\circ}=\cot34^{\circ}=\frac{1}{\tan34^{\circ}} \), so this is not helpful. Wait, maybe I made a mistake in the angle correspondence.

Wait, let's re - express:

The two acute angles are 34° and 56°. For angle 34°:

\( \sin(34^{\circ})=\frac{h}{25}\Rightarrow h = 25\sin(34^{\circ}) \)

\( \cos(56^{\circ})=\sin(34^{\circ})=\frac{h}{25}\Rightarrow h = 25\cos(56^{\circ}) \) (since \( \cos(56^{\circ})=\sin(34^{\circ}) \))

Also, \( \tan(56^{\circ})=\frac{\text{opposite to }56^{\circ}}{\text{adjacent to }56^{\circ}} \). The side opposite to 56° is the adjacent side of 34° (length \( x \)), and the side adjacent to 56° is \( h \). So \( \tan(56^{\circ})=\frac{x}{h} \), and \( x = 25\cos(34^{\circ}) \), so \( h=\frac{25\cos(34^{\circ})}{\tan(56^{\circ})} \), which is not helpful.

Wait, the options are:

a. \( 25(\cos34^{\circ}) \)

b. \( 25(\cos56^{\circ}) \)

c. \( 25(\tan34^{\circ}) \)

d. \( 25(\tan56^{\circ}) \)

Wait, let's calculate the values:

\( \sin(34^{\circ})\approx0.5592 \), \( \cos(56^{\circ})\approx0.5592 \), so \( 25\cos(56^{\circ})\approx25\times0.5592 = 13.98 \)

\( \cos(34^{\circ})\approx0.8290 \), \( 25\cos(34^{\circ})\approx20.725 \)

\( \tan(34^{\circ})\approx0.6745 \), \( 25\tan(34^{\circ})\approx16.86 \)

\( \tan(56^{\circ})\approx1.4826 \), \( 25\tan(56^{\circ})\approx37.065 \)

And \( 25\sin(34^{\circ})\approx25\times0.5592 = 13.98 \), which is equal to \( 25\cos(56^{\circ}) \) (since \( \cos(56^{\circ})=\sin(34^{\circ}) \)). So the correct option is b. \( 25(\cos56^{\circ}) \)

Answer:

B. \( 25(\cos56^{\circ}) \)