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length of hypotenuse where only part know
Find a rotation where the shape has the least width possible on the x-axisIntegrals using Trig FunctionsUnusual result when comparing trigonometry and Pythagoras in triangles.Using the exponential form of a complex number and De Moivre's theoremUsing Integration Techniques, how do you manipulate specific trig identities?How do I find the length of two legs in a right triangle if I know the length of the hypotenuse and the interior anglesTrigonometric summation involving the squares of cosecant terms.What is the length of the hypotenuse?Trigonometric Equation Solve the equation $(tantheta +1)(sin^2theta - sintheta) = 0$ given that $-pi leq θ leq 2pi$Nested trig functions (incl. inverse trig functions)
$begingroup$
The above links to a question in the 2019 IGCSE maths 0580 specimen paper.
I know the answer is $75$, and have been able to figure this out logically, namely $$frac357 times 8 = 40 + 35 = 75.$$ However the answer paper seems to suggest this can be answered using "correct trig or Pythagoras’ method
leading to value rounding to 40.0". I am a bit stumped by this as I don't see how...any help appreciated.
trigonometry
edited Mar 28 at 15:11
gt6989b
35.2k22557
asked Mar 28 at 15:09
Legal EagleLegal Eagle
61
New contributor
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The above links to a question in the 2019 IGCSE maths 0580 specimen paper.
I know the answer is $75$, and have been able to figure this out logically, namely $$frac357 times 8 = 40 + 35 = 75.$$ However the answer paper seems to suggest this can be answered using "correct trig or Pythagoras’ method
leading to value rounding to 40.0". I am a bit stumped by this as I don't see how...any help appreciated.
trigonometry
edited Mar 28 at 15:11
gt6989b
35.2k22557
asked Mar 28 at 15:09
Legal EagleLegal Eagle
61
New contributor
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Possibly they mean $ell - 35 = 40$?
$endgroup$
– gt6989b
Mar 28 at 15:16
1
$begingroup$
Please don't write $frac357 times 8 = 40 + 35$. I know it's tempting, but $frac357times 8$ is not the same as $40 + 35=75$. The equals sign doesn't mean "Now compute" the way it seems to do on some calculators. It means "Whatever is on the left side is exactly the same as whatever is on the right side".
$endgroup$
– Arthur
Mar 28 at 15:27
add a comment |
$begingroup$
The above links to a question in the 2019 IGCSE maths 0580 specimen paper.
I know the answer is $75$, and have been able to figure this out logically, namely $$frac357 times 8 = 40 + 35 = 75.$$ However the answer paper seems to suggest this can be answered using "correct trig or Pythagoras’ method
leading to value rounding to 40.0". I am a bit stumped by this as I don't see how...any help appreciated.
trigonometry
edited Mar 28 at 15:11
gt6989b
35.2k22557
asked Mar 28 at 15:09
Legal EagleLegal Eagle
61
New contributor
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The above links to a question in the 2019 IGCSE maths 0580 specimen paper.
I know the answer is $75$, and have been able to figure this out logically, namely $$frac357 times 8 = 40 + 35 = 75.$$ However the answer paper seems to suggest this can be answered using "correct trig or Pythagoras’ method
leading to value rounding to 40.0". I am a bit stumped by this as I don't see how...any help appreciated.
trigonometry
trigonometry
edited Mar 28 at 15:11
gt6989b
35.2k22557
asked Mar 28 at 15:09
Legal EagleLegal Eagle
61
New contributor
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 15:11
gt6989b
35.2k22557
asked Mar 28 at 15:09
Legal EagleLegal Eagle
61
New contributor
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 15:11
gt6989b
35.2k22557
edited Mar 28 at 15:11
gt6989b
35.2k22557
edited Mar 28 at 15:11
gt6989b
35.2k22557
35.2k22557
asked Mar 28 at 15:09
Legal EagleLegal Eagle
61
New contributor
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 15:09
Legal EagleLegal Eagle
61
asked Mar 28 at 15:09
Legal EagleLegal Eagle
61
61
New contributor
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Legal Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Possibly they mean $ell - 35 = 40$?
$endgroup$
– gt6989b
Mar 28 at 15:16
1
$begingroup$
Please don't write $frac357 times 8 = 40 + 35$. I know it's tempting, but $frac357times 8$ is not the same as $40 + 35=75$. The equals sign doesn't mean "Now compute" the way it seems to do on some calculators. It means "Whatever is on the left side is exactly the same as whatever is on the right side".
$endgroup$
– Arthur
Mar 28 at 15:27
add a comment |
$begingroup$
Possibly they mean $ell - 35 = 40$?
$endgroup$
– gt6989b
Mar 28 at 15:16
1
$begingroup$
Please don't write $frac357 times 8 = 40 + 35$. I know it's tempting, but $frac357times 8$ is not the same as $40 + 35=75$. The equals sign doesn't mean "Now compute" the way it seems to do on some calculators. It means "Whatever is on the left side is exactly the same as whatever is on the right side".
$endgroup$
– Arthur
Mar 28 at 15:27
$begingroup$
Possibly they mean $ell - 35 = 40$?
$endgroup$
– gt6989b
Mar 28 at 15:16
$begingroup$
Possibly they mean $ell - 35 = 40$?
$endgroup$
– gt6989b
Mar 28 at 15:16
1
1
$begingroup$
Please don't write $frac357 times 8 = 40 + 35$. I know it's tempting, but $frac357times 8$ is not the same as $40 + 35=75$. The equals sign doesn't mean "Now compute" the way it seems to do on some calculators. It means "Whatever is on the left side is exactly the same as whatever is on the right side".
$endgroup$
– Arthur
Mar 28 at 15:27
$begingroup$
Please don't write $frac357 times 8 = 40 + 35$. I know it's tempting, but $frac357times 8$ is not the same as $40 + 35=75$. The equals sign doesn't mean "Now compute" the way it seems to do on some calculators. It means "Whatever is on the left side is exactly the same as whatever is on the right side".
$endgroup$
– Arthur
Mar 28 at 15:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Obviously either the question meant to ask about just the length of the dotted line segment, or the person supplying the answer thought the question was about that segment rather than the whole line labeled $ell$.
answered Mar 28 at 15:17
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
It can also be solved using similarity of triangles
The triangle having base of 8 cm is similar to the triangle having base of 15 cm . The hypotenuse of the smaller triangle will be
$l-35$ and and that of the larger triangle would be $l$ thus
by using properties of similarity of triangles we get
$$ frac8l-35 = frac15l $$
$$ 8l = 15l -525 $$
$$7l =525 $$
$$ l=75 $$
answered Mar 28 at 15:30
Harsh WasnikHarsh Wasnik
365
$endgroup$
$begingroup$
It would be great if you show how you arrived at 40
$endgroup$
– Harsh Wasnik
Mar 28 at 15:31
$begingroup$
Typographical error: In your second sentence, you meant to write $l - 35$.
$endgroup$
– N. F. Taussig
Mar 29 at 10:40
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Obviously either the question meant to ask about just the length of the dotted line segment, or the person supplying the answer thought the question was about that segment rather than the whole line labeled $ell$.
answered Mar 28 at 15:17
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
Obviously either the question meant to ask about just the length of the dotted line segment, or the person supplying the answer thought the question was about that segment rather than the whole line labeled $ell$.
answered Mar 28 at 15:17
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
Obviously either the question meant to ask about just the length of the dotted line segment, or the person supplying the answer thought the question was about that segment rather than the whole line labeled $ell$.
answered Mar 28 at 15:17
Mark FischlerMark Fischler
33.9k12552
$endgroup$
Obviously either the question meant to ask about just the length of the dotted line segment, or the person supplying the answer thought the question was about that segment rather than the whole line labeled $ell$.
answered Mar 28 at 15:17
Mark FischlerMark Fischler
33.9k12552
answered Mar 28 at 15:17
Mark FischlerMark Fischler
33.9k12552
answered Mar 28 at 15:17
Mark FischlerMark Fischler
33.9k12552
answered Mar 28 at 15:17
Mark FischlerMark Fischler
33.9k12552
33.9k12552
add a comment |
add a comment |
$begingroup$
It can also be solved using similarity of triangles
The triangle having base of 8 cm is similar to the triangle having base of 15 cm . The hypotenuse of the smaller triangle will be
$l-35$ and and that of the larger triangle would be $l$ thus
by using properties of similarity of triangles we get
$$ frac8l-35 = frac15l $$
$$ 8l = 15l -525 $$
$$7l =525 $$
$$ l=75 $$
answered Mar 28 at 15:30
Harsh WasnikHarsh Wasnik
365
$endgroup$
$begingroup$
It would be great if you show how you arrived at 40
$endgroup$
– Harsh Wasnik
Mar 28 at 15:31
$begingroup$
Typographical error: In your second sentence, you meant to write $l - 35$.
$endgroup$
– N. F. Taussig
Mar 29 at 10:40
add a comment |
$begingroup$
It can also be solved using similarity of triangles
The triangle having base of 8 cm is similar to the triangle having base of 15 cm . The hypotenuse of the smaller triangle will be
$l-35$ and and that of the larger triangle would be $l$ thus
by using properties of similarity of triangles we get
$$ frac8l-35 = frac15l $$
$$ 8l = 15l -525 $$
$$7l =525 $$
$$ l=75 $$
answered Mar 28 at 15:30
Harsh WasnikHarsh Wasnik
365
$endgroup$
$begingroup$
It would be great if you show how you arrived at 40
$endgroup$
– Harsh Wasnik
Mar 28 at 15:31
$begingroup$
Typographical error: In your second sentence, you meant to write $l - 35$.
$endgroup$
– N. F. Taussig
Mar 29 at 10:40
add a comment |
$begingroup$
It can also be solved using similarity of triangles
The triangle having base of 8 cm is similar to the triangle having base of 15 cm . The hypotenuse of the smaller triangle will be
$l-35$ and and that of the larger triangle would be $l$ thus
by using properties of similarity of triangles we get
$$ frac8l-35 = frac15l $$
$$ 8l = 15l -525 $$
$$7l =525 $$
$$ l=75 $$
answered Mar 28 at 15:30
Harsh WasnikHarsh Wasnik
365
$endgroup$
It can also be solved using similarity of triangles
The triangle having base of 8 cm is similar to the triangle having base of 15 cm . The hypotenuse of the smaller triangle will be
$l-35$ and and that of the larger triangle would be $l$ thus
by using properties of similarity of triangles we get
$$ frac8l-35 = frac15l $$
$$ 8l = 15l -525 $$
$$7l =525 $$
$$ l=75 $$
answered Mar 28 at 15:30
Harsh WasnikHarsh Wasnik
365
edited Mar 29 at 17:47
answered Mar 28 at 15:30
Harsh WasnikHarsh Wasnik
365
answered Mar 28 at 15:30
Harsh WasnikHarsh Wasnik
365
answered Mar 28 at 15:30
Harsh WasnikHarsh Wasnik
365
365
$begingroup$
It would be great if you show how you arrived at 40
$endgroup$
– Harsh Wasnik
Mar 28 at 15:31
$begingroup$
Typographical error: In your second sentence, you meant to write $l - 35$.
$endgroup$
– N. F. Taussig
Mar 29 at 10:40
add a comment |
$begingroup$
It would be great if you show how you arrived at 40
$endgroup$
– Harsh Wasnik
Mar 28 at 15:31
$begingroup$
Typographical error: In your second sentence, you meant to write $l - 35$.
$endgroup$
– N. F. Taussig
Mar 29 at 10:40
$begingroup$
It would be great if you show how you arrived at 40
$endgroup$
– Harsh Wasnik
Mar 28 at 15:31
$begingroup$
It would be great if you show how you arrived at 40
$endgroup$
– Harsh Wasnik
Mar 28 at 15:31
$begingroup$
Typographical error: In your second sentence, you meant to write $l - 35$.
$endgroup$
– N. F. Taussig
Mar 29 at 10:40
$begingroup$
Typographical error: In your second sentence, you meant to write $l - 35$.
$endgroup$
– N. F. Taussig
Mar 29 at 10:40
add a comment |
Legal Eagle is a new contributor. Be nice, and check out our Code of Conduct.
Legal Eagle is a new contributor. Be nice, and check out our Code of Conduct.
Legal Eagle is a new contributor. Be nice, and check out our Code of Conduct.
Legal Eagle is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Possibly they mean $ell - 35 = 40$?
$endgroup$
– gt6989b
Mar 28 at 15:16
1
$begingroup$
Please don't write $frac357 times 8 = 40 + 35$. I know it's tempting, but $frac357times 8$ is not the same as $40 + 35=75$. The equals sign doesn't mean "Now compute" the way it seems to do on some calculators. It means "Whatever is on the left side is exactly the same as whatever is on the right side".
$endgroup$
– Arthur
Mar 28 at 15:27