math help thanks?

2014-06-10 4:06 pm
Find the length of the hypotenuse of the right triangle MPQ, if tanM = square root of 6 over 5 and angle P is the right angle.

回答 (4)

2014-06-10 6:30 pm
tan(M) = sqrt(6) / 5 = PQ / MP

Let g = GCD(PQ, MP)

PQ = sqrt(6) * g
MP = 5 * g

MQ^2 = (sqrt(6) * g)^2 + (5 * g)^2 = 6 g^2 + 25 g^2 = 31 g^2
So MQ = sqrt(31) g

MQ = sqrt(31) * GCD(PQ, MP)
2014-06-10 4:48 pm
Tangent is a RATIO. You cannot assume that the legs of the triangle are √6 and 5. They are n√6 and 5n, where n is any positive number.

Since you do not know the lengths of the legs, you cannot determine the length of the hypotenuse.
2014-06-10 4:21 pm
first ur question is not complete..but still i will give u a answer..

tanM = sqrt6/5

means MP=5

PQ=sqrt6

hypotenuse by pyathagoras thm =MQ=sqrt(6+25)
=sqrt31
2014-06-10 4:20 pm
If tanM = sqrt(6)/5 then the two shorter sides of the triangle are sqrt(6) and 5

By Pythagoras Theorem

MQ ^2 = [sqrt(6)]^2 + 5^2

= 6 + 25

= 31

MQ = sqrt(31)

Hypotenuse = sqrt(31)


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