The problem could have been solved without trigonometry 3:28. However, it is nice to learn new methods, so thanks for that method. Since angles A and B are the same ( 90 -theta), then AC =15 and since triangle ACD is a 3-4-5, scaled up by 3; its height is 12. And since its base = 15, ( 9 + 6) then its area = 1/2 base * height = 1/2 (15 *12) = 1/2 ( 180) = 90 Answer
EXTENDED LAW OF SINE: 2*R = a/sin(A) (easy to show by drawing a diameter from one of the vertex) therefore: R = a / (2*sin(arccos((b^2+c^2-a^2)/(2bc))) and simplify the sin(arccos()) I enjoy the problems you're posting. Thanks for sharing!
It is worth making note that, if tan(2Θ) = 3/4, then tan(Θ) = 1/3, and, if tan(2Θ) = 4/3, tan(Θ) = 1/2. (These 2 values for tan(2Θ) just happen to appear in the 3-4-5 right triangle.) In this problem, we can take an educated guess that tan(2Θ) is either 3/4 or 4/3. If tan(2Θ) = 3/4, tan(Θ) = 1/3 and AD has length 18. tan(2Θ) = 18/9 = 2, which is not 3/4. If tan(2Θ) = 4/3, tan(Θ) = 1/2 and AD has length 12. tan(2Θ) = 12/9 = 4/3, which is the desired value. So, length AD = h = 12 and length BC = b = 6 + 9 = 15. Area = (1/2)bh = (1/2)(15)(12) = 90, as Math Booster also found.
Reflecting ∆ABD about BD, leads to a ∆A'BD, with chord AB subtending θ at A' and 2θ at C. Hence, a circle passes through ABA' with C as the center, with Radius = BC= 15, implying AC = Radius= 15 ∆ADC is a (3,4,5) triangle, hence AD = 12 Area=½*15*9= 90
90 Let's label the triangle to the right CDP, then angle P = 90 - 2 thetas ]180-( 90 + 2 thetas) ] Hence, angle A= 90 - theta ( 90 -2 thetas + 1 theta) For triangle ABD, angle B= 90 - theta [ 180 - ( 90 + theta)] Hence, triangle ABC is an isosceles triangle Hence, line BC = line AC = 6+ 9 = 6+ 9 = 15=15 Since AC = 15 and DC = 9, if we divide both by 3, we get 5 and 3. Hence, triangle ACD is a 3-4-5 right triangle scaled up by 3. Hence, the other side is 12 (4*3) Therefore, triangle ABC has a height of 12 and a base of 15 Hence, the area = 12 * 15 * 1/2 = 6 * 15 = 90 Answer
The answer is 90 unita square. Also it *really* similar to another problem on your channel. I shall include that problem as well as the last problem as familiar problems to practice on. Also the last problem, I mis-commented. I meant x=3 for yesterday's problem. And now that that I followed every step and am reminded of similar problem, I hope that that means that I definitely can do these problems easily!!!
Bro, my brain is turning soft and mushy like an egg yolk with self referential ruminations and I am slowly but surely descending into a swamp only to sink deep into it to never see the daylight again.
3:26 As far as we need 'h' but not theta it is easier to eliminate tan(theta) (substitute tg(theta)=6/h into tg(2*theta) and solve for h furthermore. We get h/9=(2-6/h)/(1-(6/h)^2) => h= +/-12 => our h=12.
