# 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

Answers
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
javaid.aslam
14 January 2011
When manipulating the denominator the top and bottom must be multiplied by the square root of 5 not 2!
javaid.aslam
14 January 2011