Higher math trigonometry
Angle A is acute and such that tanA= square root of 6 over 3 show clearly that the exact value of sinA can be written in the form 1/5xsquared root of K and state the value of K
This is a very complicated maths question and hopefully you will understand all of the below without a problem! Ok, because we are dealing with trigonometry we can assume the triangle is a right angled triangle. We know that tanA = opp / adj. From the figures you gave in the question therefore the opp is the square root of 6 and the adjacent is 3. Now the first step is to use pythagoras to find the hypotenuse (we need the hypotenuese because sinA = opp / hyp) So 3 squared = 9 and the square root of 6 squared is 6. Using pythagoras 9 add 6 = 15. So the hypotenuese equals the sqaure root of 15. No we can form an equation for sinA. SinA = opp / hyp so SinA = square root 6 / square root 15. The final thing to do is use surds to simplify the answer so that it matches with what the question has asked i.e. in a form 1/5 x square root of K. square root 6 = square root 2 x square root 3 square root 15 = square root 5 x square root 3 When these two are divided then the square 3s will cancel out leaving sinA = square root 2 / square root 5. Now if we manipulate the denominator by multiplying both top and bottom by the square root of 2 we will get square root 2 x square root 5 / 5. square root 2 x square root 5 = square root 10. So the final answer is 1/5 x square root 10. So....K is 10! Phewwww. Hope that made sense
When manipulating the denominator the top and bottom must be multiplied by the square root of 5 not 2!