Question

find the area of the given triangle. round your answer to the nearest tenth. do not round any intermediate computations. square units

Explanation:

Step1: Identify triangle type and sides

This is a right - triangle with hypotenuse \( c = 12 \) and one angle \( \theta=36^{\circ} \). Let the two legs be \( a \) (opposite to \( 36^{\circ} \)) and \( b \) (adjacent to \( 36^{\circ} \)). We know that \( \sin\theta=\frac{a}{c} \) and \( \cos\theta=\frac{b}{c} \), and the area of a right - triangle is \( A=\frac{1}{2}ab \).

First, find \( a \) and \( b \):
\( a = c\sin\theta=12\times\sin(36^{\circ}) \)
\( b = c\cos\theta = 12\times\cos(36^{\circ}) \)

Step2: Calculate the area

The area formula for a right - triangle is \( A=\frac{1}{2}ab \). Substitute \( a = 12\sin(36^{\circ}) \) and \( b = 12\cos(36^{\circ}) \) into the formula:

\( A=\frac{1}{2}\times(12\sin(36^{\circ}))\times(12\cos(36^{\circ})) \)

We know that \( \sin(2\alpha)=2\sin\alpha\cos\alpha \), so \( \sin\alpha\cos\alpha=\frac{\sin(2\alpha)}{2} \). Here \( \alpha = 36^{\circ} \), so \( 2\alpha=72^{\circ} \)

\( A=\frac{1}{2}\times144\times\sin(36^{\circ})\cos(36^{\circ})=72\times\frac{\sin(72^{\circ})}{2}=36\sin(72^{\circ}) \)

Calculate \( \sin(72^{\circ})\approx0.9511 \)

\( A = 36\times0.9511\approx34.2 \)

Answer:

\( 34.2 \)