Finding Angles from Side Length in Pre-Calculus
I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.
On my exam my professor wrote the tangent addition identity:
$$tan(A + B) = frac{tan(A) + tan(B)}{1 - tan(A) cdot tan(B)}$$
But I'm not sure what relevance this has.
If C = A + B, find tan(C)
algebra-precalculus trigonometry triangle
asked Nov 21 '18 at 20:43
LuminousNutria
1709
add a comment |
I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.
On my exam my professor wrote the tangent addition identity:
$$tan(A + B) = frac{tan(A) + tan(B)}{1 - tan(A) cdot tan(B)}$$
But I'm not sure what relevance this has.
If C = A + B, find tan(C)
algebra-precalculus trigonometry triangle
asked Nov 21 '18 at 20:43
LuminousNutria
1709
Are the two angles on the top triangles right angles?
– KM101
Nov 21 '18 at 20:45
@KM101 Yes, that's correct.
– LuminousNutria
Nov 21 '18 at 20:46
There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
– KM101
Nov 21 '18 at 20:53
add a comment |
I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.
On my exam my professor wrote the tangent addition identity:
$$tan(A + B) = frac{tan(A) + tan(B)}{1 - tan(A) cdot tan(B)}$$
But I'm not sure what relevance this has.
If C = A + B, find tan(C)
algebra-precalculus trigonometry triangle
asked Nov 21 '18 at 20:43
LuminousNutria
1709
I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.
On my exam my professor wrote the tangent addition identity:
$$tan(A + B) = frac{tan(A) + tan(B)}{1 - tan(A) cdot tan(B)}$$
But I'm not sure what relevance this has.
If C = A + B, find tan(C)
algebra-precalculus trigonometry triangle
algebra-precalculus trigonometry triangle
asked Nov 21 '18 at 20:43
LuminousNutria
1709
asked Nov 21 '18 at 20:43
LuminousNutria
1709
asked Nov 21 '18 at 20:43
LuminousNutria
1709
asked Nov 21 '18 at 20:43
LuminousNutria
1709
asked Nov 21 '18 at 20:43
LuminousNutria
1709
1709
Are the two angles on the top triangles right angles?
– KM101
Nov 21 '18 at 20:45
@KM101 Yes, that's correct.
– LuminousNutria
Nov 21 '18 at 20:46
There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
– KM101
Nov 21 '18 at 20:53
add a comment |
Are the two angles on the top triangles right angles?
– KM101
Nov 21 '18 at 20:45
@KM101 Yes, that's correct.
– LuminousNutria
Nov 21 '18 at 20:46
There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
– KM101
Nov 21 '18 at 20:53
Are the two angles on the top triangles right angles?
– KM101
Nov 21 '18 at 20:45
Are the two angles on the top triangles right angles?
– KM101
Nov 21 '18 at 20:45
@KM101 Yes, that's correct.
– LuminousNutria
Nov 21 '18 at 20:46
@KM101 Yes, that's correct.
– LuminousNutria
Nov 21 '18 at 20:46
There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
– KM101
Nov 21 '18 at 20:53
There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
– KM101
Nov 21 '18 at 20:53
add a comment |
2 Answers
2
active
oldest
votes
Hint: $$tan A = frac{1}{2}$$
$$tan B = frac{3}{4}$$
Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$
answered Nov 21 '18 at 20:52
KM101
5,3141423
How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:54
The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
– KM101
Nov 21 '18 at 20:55
Sorry, I just edited my post. How do you know which side is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:56
2
It seems to be implied by the “squar-ish” angle marks.
– KM101
Nov 21 '18 at 20:58
Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
– LuminousNutria
Nov 21 '18 at 21:00
|
show 1 more comment
$$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$
edited Nov 21 '18 at 22:01
Davide Giraudo
125k16150260
answered Nov 21 '18 at 20:50
WhatToDo
25116
add a comment |
Your Answer
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2 Answers
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2 Answers
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Hint: $$tan A = frac{1}{2}$$
$$tan B = frac{3}{4}$$
Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$
answered Nov 21 '18 at 20:52
KM101
5,3141423
How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:54
The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
– KM101
Nov 21 '18 at 20:55
Sorry, I just edited my post. How do you know which side is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:56
2
It seems to be implied by the “squar-ish” angle marks.
– KM101
Nov 21 '18 at 20:58
Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
– LuminousNutria
Nov 21 '18 at 21:00
|
show 1 more comment
Hint: $$tan A = frac{1}{2}$$
$$tan B = frac{3}{4}$$
Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$
answered Nov 21 '18 at 20:52
KM101
5,3141423
How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:54
The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
– KM101
Nov 21 '18 at 20:55
Sorry, I just edited my post. How do you know which side is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:56
2
It seems to be implied by the “squar-ish” angle marks.
– KM101
Nov 21 '18 at 20:58
Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
– LuminousNutria
Nov 21 '18 at 21:00
|
show 1 more comment
Hint: $$tan A = frac{1}{2}$$
$$tan B = frac{3}{4}$$
Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$
answered Nov 21 '18 at 20:52
KM101
5,3141423
Hint: $$tan A = frac{1}{2}$$
$$tan B = frac{3}{4}$$
Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$
answered Nov 21 '18 at 20:52
KM101
5,3141423
answered Nov 21 '18 at 20:52
KM101
5,3141423
answered Nov 21 '18 at 20:52
KM101
5,3141423
answered Nov 21 '18 at 20:52
KM101
5,3141423
5,3141423
How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:54
The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
– KM101
Nov 21 '18 at 20:55
Sorry, I just edited my post. How do you know which side is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:56
2
It seems to be implied by the “squar-ish” angle marks.
– KM101
Nov 21 '18 at 20:58
Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
– LuminousNutria
Nov 21 '18 at 21:00
|
show 1 more comment
How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:54
The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
– KM101
Nov 21 '18 at 20:55
Sorry, I just edited my post. How do you know which side is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:56
2
It seems to be implied by the “squar-ish” angle marks.
– KM101
Nov 21 '18 at 20:58
Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
– LuminousNutria
Nov 21 '18 at 21:00
How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:54
How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:54
The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
– KM101
Nov 21 '18 at 20:55
The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
– KM101
Nov 21 '18 at 20:55
Sorry, I just edited my post. How do you know which side is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:56
Sorry, I just edited my post. How do you know which side is the hypotenuse?
– LuminousNutria
Nov 21 '18 at 20:56
2
2
It seems to be implied by the “squar-ish” angle marks.
– KM101
Nov 21 '18 at 20:58
It seems to be implied by the “squar-ish” angle marks.
– KM101
Nov 21 '18 at 20:58
Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
– LuminousNutria
Nov 21 '18 at 21:00
Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
– LuminousNutria
Nov 21 '18 at 21:00
|
show 1 more comment
$$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$
edited Nov 21 '18 at 22:01
Davide Giraudo
125k16150260
answered Nov 21 '18 at 20:50
WhatToDo
25116
add a comment |
$$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$
edited Nov 21 '18 at 22:01
Davide Giraudo
125k16150260
answered Nov 21 '18 at 20:50
WhatToDo
25116
add a comment |
$$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$
edited Nov 21 '18 at 22:01
Davide Giraudo
125k16150260
answered Nov 21 '18 at 20:50
WhatToDo
25116
$$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$
edited Nov 21 '18 at 22:01
Davide Giraudo
125k16150260
answered Nov 21 '18 at 20:50
WhatToDo
25116
edited Nov 21 '18 at 22:01
Davide Giraudo
125k16150260
edited Nov 21 '18 at 22:01
Davide Giraudo
125k16150260
edited Nov 21 '18 at 22:01
Davide Giraudo
125k16150260
125k16150260
answered Nov 21 '18 at 20:50
WhatToDo
25116
answered Nov 21 '18 at 20:50
WhatToDo
25116
answered Nov 21 '18 at 20:50
WhatToDo
25116
25116
add a comment |
add a comment |
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Are the two angles on the top triangles right angles?
– KM101
Nov 21 '18 at 20:45
@KM101 Yes, that's correct.
– LuminousNutria
Nov 21 '18 at 20:46
There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
– KM101
Nov 21 '18 at 20:53