<p>These are all from the 09-2019 practice test</p>
<ol>
<li>Rectangle ABC has vertices A(4,5) B(0,2) and C(6,-6) find vertex D.. I'm guessing you use the distance formula? I wans't sure how to do this other than going through each of the choices and plugging them in into the distance formula until it matched the distance from B-C. Is there any legit algebra way to do it in one step and get straight to the answer? (10, -3) (9,-2) (8,2) (7, 1) (2, -9)</li>
</ol>
<p>53: A right triangle below, 0<b<a One of the angle measures in the triangle is tan -1(inverse) (a/b). What is the cos[tan-1 (a/b). The triangle is given sides a, b, and squareroot of a²+b² (pythagorean theorem)</p>
<p>a. a/b
b. b/a
c. a/ (squareroot of a²+b²)
d. b/ (squareroot of a²+b²)
e. (squareroot of a²+b²)/a</p>
<p>56: The transmitter site of radio station WGGW and the transmitter site of another radio station, WGWB, are ont he same highway 100 miles apart. The radio signal from the trasnmitter site of WGWB can be received only within a radius of 60miles in all direction from the WGWB transmitter site, WGGW can be received only within a radius of 52miles in all direction. For how many along the highway can the radio signals of both stations be received?</p>
<p>I figure this is an overlap of the radius and area but I just couldn't figure out this one and how to do it.
8, 12, 40 ,44, 48 are the choices</p>
<p>On number ten, its easiest if you start by drawing (or at least imagining) the points they already gave you. Just by looking at the picture, you see that you have one vertex on the y axis and one in the first quadrant and then you have one in the fourth. Just by looking at the picture you can tell that the last point obviously has to go into the fourth quadrant as well- you can now eliminate two of the answer choices. Technically you already have two sides of the rectangle, so I looked at the displacement of the fourth quadrant point (6, -6) from the point it makes a line with in the first quadrant (4, 5)- it moved to the right two and down 11. Therefore, to get the final point, I merely moved to the right two from (0, 2) and down eleven, giving me (0 + 2, 2 - 11) or (2, -9)</p>
<ol>
<li><p>This one is really simple if you think about it. You are given a triangle with lengths A, B, and square root of a squared plus b squared- basically its any right triangle. Then it gives you tan inverse (a/b): all this means is that an angle within the triangle (not the right angle) has a tangent of a/b. So it simply means that the tangent (otherwise known as opposite side over adjacent side) of one of the angles in the triangle is (a/b). Now when it asks you for COSINE (a/h) of the angle whose tangent equals (a/b) you look for adjacent over the hypotenuse. They already gave you the adjacent (b) and the hypotenuse (square root of a squared plus b squared)- so you divide b by the hypotenuse and that is your answer (d)</p></li>
<li><p>Just draw two dots, and say they are 100 miles from each other. One of the dots can only reach sixty miles down the road, which means there are forty miles left in the road in which that dot (or radio station) cannot be heard. At the other end of the road there is another dot which only reaches 52 miles. If point A reaches sixty miles toward point B, and point B reaches 52 miles back, they would overlap by 12 miles. Those 40 miles that point A couldn’t reach to get to point b is met by and taken over by B’s 52 miles by 12.</p></li>
</ol>
<p>Err yes that works the best… doign the slope nuemrical… but it’s down 8 over 6 so it’s (10, -3)not )9, -2( which is what i originally but yes thanks for the explanations to 53 and 56 yeah I can’t believe i didn’t just do 52-(100-60) wow… i should’ve known after drawing so many pictures.</p>
<p>And yeah so 53 is just plug and finding out which angle is being used with the trig functions and then figure out of which angle is it cosined to… ok thanks</p>
<p>Isn’t tan-1 the same thing as cotang…so if the cotang of an angle is a/b that means that the adjacent is a, and the opposite is b, right (since cotang would be a/o). From there I get b on the top of the equation instead of a…so why is a on top? thanks for any help you can provide.</p>
<p>@user83248234: tan-1 is not the same as cotangent. Cotangent is just 1/ tan, and tan-1 helps find the angle of something.</p>
<p>For example, if the tangent of an angle is 1/2, then the cotangent is 1 / (1/2) which equals 2. But, if you put tan-1 (1/2) into your calculator, it gives you the degrees of the ANGLE with a tangent of 1/2: tan-1 (1/2) ~ 26.56 degrees. You can check it by putting it back into your calculator- tangent of 26.56 degrees gives you .499 which is extremely close to .5, or 1/2.</p>
<p>Thanks alot for the reply…So then my last question is how would tan-1 or sin-1 or cos-1 work for soh cah toa (liek in the above problem)…Everything would just stay the same as if it was the normal tan sin or cos function (in terms of o/a o/h a/h)?</p>
<p>so you have a diagram and you know the length of the sides of the diagram
tangent is opposite/hypotenuse. so you find the 1 on the diagram and the 2 on the diagram and figure out from which angle it’s opposite and hypotenuse.</p>