QUESTION IMAGE
Question
denis, vera, and yash are rock climbers. yash is connected to vera by a 30 m rope, which is taut, and vera is connected to denis by a 25 m rope, which is also taut.
yash has already completed the climb, and he takes out a camera to take a picture of his friends. denis sees an angle of 41° between vera and yash, as shown.
image of triangle with yash at top, vera and denis at bottom, sides 30m (yash - vera), 25m (vera - denis), angle 41° at denis
what is the minimum camera view angle yash can use to see both denis and vera?
do not round during your calculations. round your final answer to the nearest degree.
Step1: Identify the triangle sides and angle
We have a triangle with sides: \( a = 30 \) m (Yash to Vera), \( b = 25 \) m (Vera to Denis), and the angle \( C = 41^\circ \) (at Denis between Vera and Yash). We need to find the angle \( A \) (at Yash between Vera and Denis) using the Law of Sines. The Law of Sines states that \( \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \), but first we need to find side \( c \) (Yash to Denis) or use the Law of Sines directly for angle \( A \). Wait, actually, let's label the triangle properly: Let’s denote the triangle as \( \triangle YVD \), where \( Y \) is Yash, \( V \) is Vera, \( D \) is Denis. So \( YV = 30 \) m, \( VD = 25 \) m, \( \angle D = 41^\circ \). We need to find \( \angle Y \).
Using the Law of Sines: \( \frac{\sin \angle Y}{VD}=\frac{\sin \angle D}{YV} \)
So \( \sin \angle Y=\frac{VD \times \sin \angle D}{YV} \)
Step2: Substitute the values
Substitute \( VD = 25 \), \( \angle D = 41^\circ \), \( YV = 30 \) into the formula:
\( \sin \angle Y=\frac{25 \times \sin 41^\circ}{30} \)
Calculate \( \sin 41^\circ \approx 0.656059 \)
So \( \sin \angle Y=\frac{25 \times 0.656059}{30}=\frac{16.401475}{30} \approx 0.5467158 \)
Step3: Find the angle
Now, take the inverse sine (arcsin) of \( 0.5467158 \) to find \( \angle Y \):
\( \angle Y=\arcsin(0.5467158) \approx 33.13^\circ \)
Round to the nearest degree: \( 33^\circ \)
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\( 33 \)