Question

4.b.6 practice problems
1 practice 1
2 practice 2
consider this triangle.
(right triangle with right angle at j, angle at g is 35°, gh = 9)
write an expression that can be used to find the length of jh and an expression that can be used to find the length of gj.
to find jh:
□ $jh = 9\cos(35°)$ □ $jh = 9\sin(35°)$ □ $jh = \frac{9}{\cos(35°)}$ □ $jh = \frac{9}{\sin(35°)}$
to find gj:
□ $gj = 9\cos(35°)$ □ $gj = 9\sin(35°)$ □ $gj = \frac{9}{\sin(35°)}$ □ $gj = \frac{9}{\cos(35°)}$
how did i do?

Explanation:

To find \( JH \):

We have a right - triangle \( GJH \) with \( \angle G = 35^{\circ} \), \( GH = 9 \) (the hypotenuse of the right - triangle \( \triangle GJH \) where \( \angle J=90^{\circ} \)).
Recall the definition of the sine function in a right - triangle: \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \). For \( \angle G = 35^{\circ} \), the side opposite to \( \angle G \) is \( JH \) and the hypotenuse is \( GH = 9 \).
So, \( \sin(35^{\circ})=\frac{JH}{GH} \). Since \( GH = 9 \), we can solve for \( JH \): \( JH=GH\times\sin(35^{\circ})=9\sin(35^{\circ}) \).

To find \( GJ \):

Recall the definition of the cosine function in a right - triangle: \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \). For \( \angle G = 35^{\circ} \), the side adjacent to \( \angle G \) is \( GJ \) and the hypotenuse is \( GH = 9 \).
So, \( \cos(35^{\circ})=\frac{GJ}{GH} \). Since \( GH = 9 \), we can solve for \( GJ \): \( GJ = 9\cos(35^{\circ}) \).

Answer:

To find \( JH \): \( \boldsymbol{JH = 9\sin(35^{\circ})} \)

To find \( GJ \): \( \boldsymbol{GJ=9\cos(35^{\circ})} \)