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Post by John Marston on Feb 9, 2021 15:55:43 GMT
Haha. This subject tests your limit. If anybody ready, I have some problems.
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Post by Gerd von Rundstedt on Feb 9, 2021 16:01:01 GMT
OK, I haven't taken trig yet, but hit me.
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Post by John Marston on Feb 9, 2021 16:04:45 GMT
OK, I haven't taken trig yet, but hit me. Muahahaha!! John Marston defeats Gerd von Rundstedt again! Simple one - In a triangle ABC, AB = 24cm, BC = 7 cm. Find - 1. Sin A, Cos A 2. Sin C, Cos C |
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Post by John Marston on Feb 9, 2021 16:08:37 GMT
OK, I haven't taken trig yet, but hit me. Toughest one -( Cosec A - Sin A)(Sec A - Cos A) = 1/(tan A + cot A) |
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Post by Gerd von Rundstedt on Feb 9, 2021 16:12:50 GMT
OK, I haven't taken trig yet, but hit me. Muahahaha!! John Marston defeats Gerd von Rundstedt again! Simple one - In a triangle ABC, AB = 24cm, BC = 7 cm. Find - 1. Sin A, Cos A 2. Sin C, Cos C I swear I did not use the internet, nor will I unless you want me to. Sin (A) = 7/25, Cos (A) = 24/25, Sin (C) = 24/25, Cos (C) = 7/25 |
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Post by Gerd von Rundstedt on Feb 9, 2021 16:13:29 GMT
OK, I haven't taken trig yet, but hit me. Toughest one -( Cosec A - Sin A)(Sec A - Cos A) = 1/(tan A + cot A) What am I supposed to solve for? |
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Post by John Marston on Feb 9, 2021 16:14:24 GMT
Toughest one -( Cosec A - Sin A)(Sec A - Cos A) = 1/(tan A + cot A) What am I supposed to solve for? Prove L.H.S = R.H.S |
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Post by John Marston on Feb 9, 2021 16:15:08 GMT
I swear I did not use the internet, nor will I unless you want me to. Sin (A) = 7/25, Cos (A) = 24/25, Sin (C) = 24/25, Cos (C) = 7/25 Right!! I bet you studied trigonometry right? |
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Post by Gerd von Rundstedt on Feb 9, 2021 16:16:43 GMT
Nope, I am only at Alg 2 as a 9/10th grader, and I specialized in geometry, especially abstract.
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Post by Gerd von Rundstedt on Feb 9, 2021 16:17:20 GMT
What am I supposed to solve for? Prove L.H.S = R.H.S Pardon me , but can you explain those acronyms? |
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Post by Warlord247 on Feb 9, 2021 17:18:03 GMT
I believe those are the two triangles (LHS RHS) and you must prove there equal. H acts as the median between the two. So if you use law of medians, it should start you on the right path. Or I could be completely wrong right now, I have no idea.
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Post by Gerd von Rundstedt on Feb 9, 2021 18:20:04 GMT
I believe those are the two triangles (LHS RHS) and you must prove there equal. H acts as the median between the two. So if you use law of medians, it should start you on the right path. Or I could be completely wrong right now, I have no idea. No, there is angle A |
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Post by John Marston on Feb 10, 2021 2:34:03 GMT
I believe those are the two triangles (LHS RHS) and you must prove there equal. H acts as the median between the two. So if you use law of medians, it should start you on the right path. Or I could be completely wrong right now, I have no idea. No, there is angle A No. No such law is applied here. |
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Post by Nobunaga Oda on Feb 10, 2021 3:58:36 GMT
No. No such law is applied here. It's just the equation on the left being equivalent to the equation on the right by forming either counterpart using the other or obtaining an equivalent expression on both sides, correct? Is there a negative sign on the LHS? If not, I just obtain the equivalent expression of sin A cos A on both sides. |
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Post by John Marston on Feb 10, 2021 10:40:10 GMT
No. No such law is applied here. It's just the equation on the left being equivalent to the equation on the right by forming either counterpart using the other or obtaining an equivalent expression on both sides, correct? Is there a negative sign on the LHS? If not, I just obtain the equivalent expression of sin A cos A on both sides. Yes. And then we must express them in terms of the given values accordingly. That is not a minus Nobunaga Oda. |
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