#2 I need help with this math problem!!! (its trigonometry)?

Question

Ok, thanks a bunch fo rthe last problem guys! Umm, this one is harder. The triangle is a right triangle, i need to find the two angles. One side is 12, one is 18 and the hypotenuse is about 21.62 (i used pythagorean thm). How do i do SOHCAHTOA with it inverse? please help and explain! I love you guys. Thanks-

I am sitting here still trying to figure it out you know...

Answer 1

arctan(12/18) is one and arctan(18/12) is the other. Arctan, in case you are not familiar with the term, is just another way to say inverse tan.

Answer 2

at the starting up, tan A = sin A/cos A, and considering the fact that cot A = a million/tan A, therefore cot A = cos A/sin A. also, sec A = a million/cos A, and cosec A = a million/sin A. so, a million. (sin A)(cot A) = (sin A)(cos A/sin A). word that sin A/sin A = a million. so the answer is cos A. 2. (sec A)(cos^2 A) = (a million/cos A)(cos A x cos A) the answer is genuinely cos A between the trigonometry rule is: sin^2 A + cos^2 A = a million. so in case you divide the total equation with cos^2 A, you derive yet another rule that holds a similar relationship: tan^2 A + a million = sec^2 A. on the different hand, in case you divide the equation with sin^2 A instead of cos^2 A, you receives yet another derivation! it turns into a million + cot^2 A = cosec^2 A. for that reason, 3. a million + tan^2 A = sec^2 A. the +/- squareroot is for that reason only +/-(sec A) 4. cot^2 A + a million = cosec^2 A. so, further, the +/- squareroot is +/-(cosec A) 5. cosec A tan A = (a million/sin A)(sin A/cos A) answer is a million/cos A, it really is efectively sec A 6. sin A sec A = (sin A)(a million/cos A) = sin A/cos A = tan A 7. cos A/cot A = cos A x tan A = (cos A)(sin A/cos A) answer is, sure sin A. 8. cot A sec A = (cos A/sin A)(a million/cos A) = a million/sin A = cosec A. TIP: only convert each and every thing to the most ordinary type, ie sin A and cos A only. then it is going to likely be a lot less complicated. together, comprehend what guidelines there are, and word the progression in the equations that is necessary to sparkling up. trigonometry at this degree is extremely relaxing and effortless.

Answer 3

hypotenuse is 6√13, but you don't need it. let A be acute angle between hypotenuse and 12 side, B be other acute angle.

tan A = 18/12, so A = 56.3099°
tan B = 12/18, so B = 33.6901°

these are just calculator problems using inverse tangent.

Answer 4

tan@=12/18=2/3 so @=33.69degrees and the other one
56.31 degrees