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Thread: A geometric problem I am having issues with...

  1. Jun 25th 2009, 07:18 PM #1
    M23
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    A geometric problem I am having issues with...

    This is from an online worksheet. I cannot work this problem out.... Use the following diagram to answer the next two questions... Correct to the nearest tenth, the value of x is A. 5.2 B. 5.3 C. 5.4 D. 5.5 .................................................. .................................................. .................................................. ..... 2. Correct to the nearest half of a degree, the measure of angle ADB is: A. 30.5° B. 32.0° C. 34.5° D. 36.0° Thank you for the help!
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  2. Jun 25th 2009, 07:32 PM #2
    Shyam
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    Quote Originally Posted by M23 View Post
    This is from an online worksheet. I cannot work this problem out.... Use the following diagram to answer the next two questions... Correct to the nearest tenth, the value of x is A. 5.2 B. 5.3 C. 5.4 D. 5.5
    Use Pythagorean theorem in RIGHT TRIANGLE PBC,

    PC^2=PB^2+BC^2<br />

    PC^2=9^2+5^2

    PC=10.3

    x = 10.3 - 5 = 5.3

    Spoiler:
    Pythagorean Theorem in Right Triangle

    a^2 = b^2+c^2

    \Rightarrow(Hypotenuse)^2 = (first\;side)^2+(second \;side)^2
    Last edited by Shyam; Jun 25th 2009 at 08:04 PM.
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  3. Jun 25th 2009, 07:38 PM #3
    Shyam
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    Quote Originally Posted by M23 View Post
    This is from an online worksheet. I cannot work this problem out.... Use the following diagram to answer the next two questions... 2. Correct to the nearest half of a degree, the measure of angle ADB is: A. 30.5° B. 32.0° C. 34.5° D. 36.0° Thank you for the help!
    IN Right triangle PBC,

    \tan P = \frac{5}{9}

    P = 29.05^\circ

    \angle C = 180 - 90 -29.05 = 60.95^\circ

    \angle D = \frac{ 60.95}{2}=30.47=30.5^\circ

    Spoiler:
    \angle D = \frac{\angle C}{2}
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  4. Jun 26th 2009, 06:57 AM #4
    M23
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    Thank you soooo much.
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