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Anonymous asked in Science & MathematicsMathematics · 10 years ago

Trig , math problem ?

Hey i'll be askin' alot of trig/ geometry questions , i have a math exam tomorrow so wish me luck , the problem is , AD is a median in triangle ABC, m <ADB= 45 degree, m <CAD=m<B find m <C

THANKS in advance :) !!

2 Answers

  • 10 years ago
    Favorite Answer

    Either my trigonometry is rusty or this is a relatively challenging problem. Good luck on your exam!

    The answer is 15 degrees.

    Here is how I did it, though I suspect there might be a few ways to solve the problem.

    Since m<CAD = m<B and m<C = m<C, Triangle ABC is similar to Triangle DAC.

    We know that m<BAC = 135 because taking advantage of these similar triangles, it corresponds to m<ADC = 180 - m<ADB = 180 - 45 = 135. (we will use this fact later).

    Also, BC/AC = AC/DC because the ratios of any two pairs of corresponding sides of similar triangles should be equal.

    But BC = 2 DC, because D is the midpoint of BC (since AD is a median).

    Substituting, we have, 2DC/AC = AC/DC, or after cross-multiplying,

    AC^2 = 2DC^2,

    AC = DC (sqrt(2)).

    By the Law of Sines (as applied to Triangle DAC),

    AC/ sin (<ADC) = DC / sin (<DAC)

    Substituting 135 for <ADC, and DC (sqrt(2)) for AC, we have

    DC (sqrt(2)) / sin (135) = DC / sin (<DAC)

    DC (sqrt(2)) / (sqrt(2)/2) = DC / sin (<DAC)

    Cancelling DC from both sides and simplifying fractions,

    2 = 1/sin (<DAC)

    sin (<DAC) = 1/2

    m<DAC = 30 or 150.

    However, recall from earlier that we showed that m<BAC = 135. Since m<BAC = 135, m<DAC must be less than this value and so must be 30.

    Since the three angles of Triangle DAC must add up to 180, we have m<C = 180 - m<ADC - m<DAC = 180 - 135 - 30 = 15 degrees.

  • digby
    Lv 4
    5 years ago

    a million)particular, 4sin 21 cos 21 = 2 sin 40 two 2) you do no longer "teach" it, because of the fact that is not an "id" (no longer authentic for ALL x). it is an "Equation" and you remedy it: sin2x+cosx = 0 => 2sinxcosx + cosx = 0 => cosx(2sinx+a million)=0 =>cosx = 0, sinx=-a million/2 => x=ok*PI + PI/2, x=2k*PI - PI/4,x=2k*PI+5PI/4 => (for 0<x<360): PI/2, 3PI/2, 5PI/4, 7PI/4 3) it is likewise an equation (i take advantage of d for delta): cos^2 d = sin^2 d - sin d => a million-sin^2 d = sin^2 d - sin d => 2 sin^2 d - sin d - a million= 0 => sin d = (a million+- sqrt(a million+8))/4 => sin d= a million, sin d= -a million/2 => d = 2k*PI + PI/2, 2k*PI - PI/4, 2k*PI + 5PI/4

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