Question

missing sides
find the measure of each indicated side. round the answer to the nearest tenth
1)
soh cah toa
diagram of triangle with right angle, angle 45°, side and x (hypotenuse) labeled
ac = ———— 111.0
2)
diagram of triangle with right angle, side 75 units, angle labeled
qp = ————
3)
diagram of right triangle def with right angle at e, hypotenuse df = 55 yd, angle at d = 40°, side ef = x
ef = ————
4)
diagram of right triangle imn with right angle at n, side in = 11 ft, angle at i = 60°, side mn = x
mn = ————
5)
soh cah toa
diagram of right triangle uvw with right angle at v, side v w = 61 ft, angle at w = 28°, hypotenuse uw = x
handwritten notes: cos 28 = 61/x, x = 61/cos28
6)
soh cah toa
diagram of right triangle hgf with right angle at g, side gf = 3, angle at h = 50°, side hg = x

Explanation:

Step1: Analyze Triangle 3 (DEF)

Triangle \( DEF \) is right - angled at \( E \). We know that \( \angle D = 40^{\circ} \) and the hypotenuse \( DF=55\) yd. We want to find the length of \( EF \), which is the side opposite to \( \angle D \). In a right - triangle, the sine of an angle \( \theta \) is defined as \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \). So, \( \sin(40^{\circ})=\frac{EF}{DF} \).

Step2: Solve for \( EF \)

We know that \( DF = 55 \) yd and \( \sin(40^{\circ})\approx0.6428 \). Substituting these values into the formula \( EF = DF\times\sin(40^{\circ}) \), we get \( EF=55\times\sin(40^{\circ}) \).
\( EF = 55\times0.6428\approx35.354\approx35.4 \) yd (rounded to the nearest tenth).

Step3: Analyze Triangle 4 (LMN)

Triangle \( LMN \) is right - angled at \( N \). We know that \( \angle L = 60^{\circ} \) and the side \( LN = 11 \) ft. We want to find the length of \( MN \), which is the side opposite to \( \angle L \). In a right - triangle, the tangent of an angle \( \theta \) is defined as \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \). So, \( \tan(60^{\circ})=\frac{MN}{LN} \).

Step4: Solve for \( MN \)

We know that \( \tan(60^{\circ})=\sqrt{3}\approx1.732 \) and \( LN = 11 \) ft. Substituting these values into the formula \( MN=LN\times\tan(60^{\circ}) \), we get \( MN = 11\times\sqrt{3}\approx11\times1.732 = 19.052\approx19.1 \) ft (rounded to the nearest tenth).

Step5: Analyze Triangle 5 (UVW)

Triangle \( UVW \) is right - angled at \( V \). We know that \( \angle W = 28^{\circ} \) and the adjacent side \( VW = 61 \) ft. We want to find the length of the hypotenuse \( UW \) (let's call it \( x \)). In a right - triangle, the cosine of an angle \( \theta \) is defined as \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \). So, \( \cos(28^{\circ})=\frac{VW}{UW} \).

Step6: Solve for \( UW \)

We know that \( \cos(28^{\circ})\approx0.8829 \) and \( VW = 61 \) ft. From \( \cos(28^{\circ})=\frac{61}{x} \), we can solve for \( x \) by cross - multiplying: \( x=\frac{61}{\cos(28^{\circ})} \).
\( x=\frac{61}{0.8829}\approx69.1 \) ft (rounded to the nearest tenth).

Step7: Analyze Triangle 6 (HGF)

Triangle \( HGF \) is right - angled at \( G \). We know that \( \angle H = 50^{\circ} \) and the adjacent side \( GF = 3 \). We want to find the length of \( HG \) (let's call it \( x \)). In a right - triangle, the tangent of an angle \( \theta \) is defined as \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \), and also \( \tan(50^{\circ})=\frac{GF}{HG} \).

Step8: Solve for \( HG \)

We know that \( \tan(50^{\circ})\approx1.1918 \) and \( GF = 3 \). From \( \tan(50^{\circ})=\frac{3}{x} \), we can solve for \( x \) by cross - multiplying: \( x=\frac{3}{\tan(50^{\circ})} \).
\( x=\frac{3}{1.1918}\approx2.5 \) (rounded to the nearest tenth).

(Note: For triangle 2, since the angle measure is not clearly visible, we assume a common angle - side relationship. If we assume the hypotenuse is 75 (from the image) and the angle is, say, \( 30^{\circ} \) (a common angle in such problems), and \( QP \) is the side opposite to the angle. If \( \angle R = 30^{\circ} \) and hypotenuse \( RN = 75 \), then \( \sin(30^{\circ})=\frac{QP}{RN} \), \( QP = 75\times0.5 = 37.5 \). But this is based on an assumption. The above solutions for triangles 3,4,5,6 are based on clear angle - side information from the image.)

Answer:

For problem 3: \( EF\approx35.4 \) yd

For problem 4: \( MN\approx19.1 \) ft

For problem 5: \( UW\approx69.1 \) ft

For problem 6: \( HG\approx2.5 \)

(For problem 2, if we assume the hypotenuse is 75 and angle is \( 30^{\circ} \), \( QP = 37.5 \))