1.) how are sides rs and rt related to each other? rs are oppostite of t

2.) could 14 inches, 13 inches, and 16 inches be the dimension of a triangle? explain your answer.

Question

1.) how are sides rs and rt related to each other? rs are oppostite of t

2.) could 14 inches, 13 inches, and 16 inches be the dimension of a triangle? explain your answer.

Accepted Answer

### Question 1
# Explanation:
## Step1: Find angle at R
In triangle $ RST $, the sum of angles in a triangle is $ 180^\circ $. Given $ \angle S = 50^\circ $ and $ \angle T = 65^\circ $, so $ \angle R=180^\circ - 50^\circ - 65^\circ = 65^\circ $.
## Step2: Relate sides to angles
In a triangle, sides opposite equal angles are equal. $ \angle T = \angle R = 65^\circ $. Side $ RS $ is opposite $ \angle T $, and side $ RT $ is opposite $ \angle S $? Wait, no: Wait, vertex labels: In triangle $ RST $, side opposite $ \angle S $ is $ RT $, side opposite $ \angle T $ is $ RS $, side opposite $ \angle R $ is $ ST $. Wait, $ \angle R = 65^\circ $, $ \angle T = 65^\circ $, so angles $ \angle R $ and $ \angle T $ are equal. Therefore, sides opposite them: side opposite $ \angle R $ is $ ST $, side opposite $ \angle T $ is $ RS $, wait no, I made a mistake. Wait, vertices: $ R $, $ S $, $ T $. So angle at $ S $: $ \angle S = 50^\circ $, angle at $ T $: $ \angle T = 65^\circ $, so angle at $ R $: $ 180 - 50 - 65 = 65^\circ $. So angles at $ R $ and $ T $ are both $ 65^\circ $. Therefore, sides opposite these angles: side opposite $ \angle R $ is $ ST $, side opposite $ \angle T $ is $ RS $, wait no, angle at $ R $: between $ S $ and $ T $, so side opposite $ \angle R $ is $ ST $. Angle at $ T $: between $ R $ and $ S $, so side opposite $ \angle T $ is $ RS $. Angle at $ S $: between $ R $ and $ T $, so side opposite $ \angle S $ is $ RT $. Wait, so $ \angle R = \angle T = 65^\circ $, so sides opposite them: side opposite $ \angle R $ is $ ST $, side opposite $ \angle T $ is $ RS $. Wait, no, that's not equal. Wait, maybe I mixed up. Let's label the triangle: vertices $ R $, $ S $, $ T $. So side $ RS $ is between $ R $ and $ S $, side $ RT $ is between $ R $ and $ T $, side $ ST $ is between $ S $ and $ T $. So angle at $ S $: between $ R $ and $ T $, so side opposite $ \angle S $ is $ RT $. Angle at $ T $: between $ R $ and $ S $, so side opposite $ \angle T $ is $ RS $. Angle at $ R $: between $ S $ and $ T $, so side opposite $ \angle R $ is $ ST $. Now, $ \angle R = 65^\circ $, $ \angle T = 65^\circ $, so angles $ \angle R $ and $ \angle T $ are equal. Therefore, sides opposite them: side opposite $ \angle R $ is $ ST $, side opposite $ \angle T $ is $ RS $. Wait, that's not equal. Wait, no, $ \angle R $ is at vertex $ R $, between $ S $ and $ T $, so the sides forming $ \angle R $ are $ RS $ and $ RT $, and the side opposite is $ ST $. $ \angle T $ is at vertex $ T $, between $ R $ and $ S $, so sides forming $ \angle T $ are $ RT $ and $ ST $, side opposite is $ RS $. $ \angle S $ is at vertex $ S $, between $ R $ and $ T $, sides forming $ \angle S $ are $ RS $ and $ ST $, side opposite is $ RT $. Now, $ \angle R = 65^\circ $, $ \angle T = 65^\circ $, so angles $ \angle R $ and $ \angle T $ are equal. Therefore, sides opposite them: side opposite $ \angle R $ is $ ST $, side opposite $ \angle T $ is $ RS $. Wait, that's not equal. Wait, I think I messed up the angle - side correspondence. Wait, in a triangle, equal angles have equal opposite sides. So if $ \angle R = \angle T = 65^\circ $, then sides opposite them should be equal. Wait, side opposite $ \angle R $: vertex $ R $, so opposite side is $ ST $. Side opposite $ \angle T $: vertex $ T $, opposite side is $ RS $. Wait, that can't be. Wait, no, let's re - label: let's call the triangle with vertices $ A $, $ B $, $ C $, then side $ a $ is opposite $ \angle A $, side $ b $ opposite $ \angle B $, side $ c $ opposite $ \angle C $. So in triangle $ RST $, let's let $ \angle R = \alpha = 65^\circ $, $ \angle S=\beta = 50^\circ $, $ \angle T=\gamma = 65^\circ $. Then side opposite $ \alpha $ (angle at $ R $) is $ ST $ (side $ a $), side opposite $ \beta $ (angle at $ S $) is $ RT $ (side $ b $), side opposite $ \gamma $ (angle at $ T $) is $ RS $ (side $ c $). Since $ \alpha=\gamma = 65^\circ $, then sides $ a $ ( $ ST $) and $ c $ ( $ RS $) are equal? Wait, no, I'm confused. Wait, angle at $ R $: between $ S $ and $ T $, so the side opposite is $ ST $. Angle at $ T $: between $ R $ and $ S $, so the side opposite is $ RS $. Angle at $ S $: between $ R $ and $ T $, so the side opposite is $ RT $. So if $ \angle R=\angle T = 65^\circ $, then sides $ ST $ (opposite $ \angle R $) and $ RS $ (opposite $ \angle T $) are equal? Wait, no, that's not right. Wait, maybe the problem is about sides $ RS $ and $ RT $. Let's see: side $ RS $ is opposite $ \angle T $, side $ RT $ is opposite $ \angle S $. $ \angle T = 65^\circ $, $ \angle S = 50^\circ $. Since $ 65^\circ>50^\circ $, the side opposite the larger angle is longer. So side $ RS $ (opposite $ \angle T = 65^\circ $) is longer than side $ RT $ (opposite $ \angle S = 50^\circ $)? Wait, no, I think I made a mistake in angle - side correspondence. Let's start over. In triangle $ RST $:

- Vertex $ S $: angle $ \angle S = 50^\circ $, the side opposite to $ \angle S $ is $ RT $.
- Vertex $ T $: angle $ \angle T = 65^\circ $, the side opposite to $ \angle T $ is $ RS $.
- Vertex $ R $: angle $ \angle R=180 - 50 - 65 = 65^\circ $, the side opposite to $ \angle R $ is $ ST $.

Since $ \angle T=\angle R = 65^\circ $, the sides opposite to them ($ RS $ and $ ST $) are equal? Wait, no, the question is about $ RS $ and $ RT $. Let's compare $ \angle S $ and $ \angle T $. $ \angle S = 50^\circ $, $ \angle T = 65^\circ $. The side opposite $ \angle S $ is $ RT $, the side opposite $ \angle T $ is $ RS $. In a triangle, the larger the angle, the longer the opposite side. Since $ \angle T>\angle S $ ( $ 65^\circ>50^\circ $ ), then $ RS>RT $ (because $ RS $ is opposite $ \angle T $, $ RT $ is opposite $ \angle S $). Wait, but also, angle at $ R $ is $ 65^\circ $, same as angle at $ T $, so side $ RS $ (opposite $ \angle T $) and side $ ST $ (opposite $ \angle R $) are equal? Maybe the initial approach was wrong. Let's use the basic triangle angle - side relationship: in a triangle, if two angles are equal, their opposite sides are equal. Here, $ \angle R=\angle T = 65^\circ $, so sides opposite them: side opposite $ \angle R $ is $ ST $, side opposite $ \angle T $ is $ RS $, so $ RS = ST $. But the question is about $ RS $ and $ RT $. Let's look at angles opposite $ RS $ and $ RT $. Angle opposite $ RS $ is $ \angle T = 65^\circ $, angle opposite $ RT $ is $ \angle S = 50^\circ $. Since $ 65^\circ>50^\circ $, by the angle - side relationship (larger angle opposite longer side), $ RS>RT $.

# Answer:
$ RS $ is longer than $ RT $ (because $ \angle T = 65^\circ $ (opposite $ RS $) is greater than $ \angle S = 50^\circ $ (opposite $ RT $), and in a triangle, the larger the angle, the longer the opposite side).

### Question 2
# Explanation:
## Step1: Recall triangle inequality theorem
The triangle inequality theorem states that for three side lengths $ a $, $ b $, $ c $ to form a triangle, the sum of any two sides must be greater than the third side. Mathematically, $ a + b>c $, $ a + c>b $, and $ b + c>a $.
## Step2: Check the given lengths
Let $ a = 14 $ inches, $ b = 13 $ inches, $ c = 16 $ inches.

- Check $ a + b>c $: $ 14+13 = 27 $, and $ 27>16 $.
- Check $ a + c>b $: $ 14 + 16=30 $, and $ 30>13 $.
- Check $ b + c>a $: $ 13+16 = 29 $, and $ 29>14 $.

Since all three conditions of the triangle inequality theorem are satisfied, the lengths 14 inches, 13 inches, and 16 inches can form a triangle.

# Answer:
Yes, 14 inches, 13 inches, and 16 inches can be the dimensions of a triangle because they satisfy the triangle inequality theorem (the sum of any two sides is greater than the third side: $ 14 + 13>16 $, $ 14+16>13 $, and $ 13 + 16>14 $).

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Find angle at R

Step2: Relate sides to angles

Step1: Recall triangle inequality theorem

Step2: Check the given lengths

Answer:

Question 2