From a point A on a straight line AB, AC and AC of length 5cm and 7cm are drawing on the same side of AB such that BAC =32 and BAD =55 find the length of angle DC
To find the length of angle DC, we first need to find the length of angle DAC.
In triangle BAD, we have the angle BAD = 55 degrees and the angle ABD = 180 - 55 - 32 = 93 degrees (since the sum of angles in a triangle is 180 degrees).
Using the Law of Sines, we can set up the following equation:
(AB / sin(BAD)) = (AD / sin(ABD))
We know that AB = AC = 5 cm and sin(BAD) = sin(55) = 0.8192. Solving for AD, we get:
(5 / 0.8192) = (AD / sin(ABD))
AD = (5 / 0.8192) * sin(ABD)
AD ≈ 6.10 cm
Now, to find angle DAC, we can use the Law of Cosines:
AC^2 = AD^2 + CD^2 - 2 * AD * CD * cos(DAC)
We know that AC = 5 cm and AD = 6.10 cm. Substituting these values, we get:
25 = 6.10^2 + CD^2 - 2 * 6.10 * CD * cos(DAC)
CD^2 - 12.1 * CD * cos(DAC) + 6.10^2 - 25 = 0
CD^2 - 12.1 * CD * cos(DAC) - 10.39 = 0
This is a quadratic equation in terms of CD. Solving for CD using the quadratic formula, we get:
CD = (12.1 * cos(DAC) ± √(12.1^2 * cos(DAC)^2 + 4 * 10.39)) / 2
Since CD is a positive length, we take the positive square root. Substituting cos(DAC) = cos(180 - 55 - 32) = cos(93) ≈ -0.3917, we get:
CD ≈ (12.1 * (-0.3917) + √(12.1^2 * (-0.3917)^2 + 4 * 10.39)) / 2
CD ≈ (-4.749 * 12.1 + √(63.77 + 169.62)) / 2
CD ≈ (-57.6529 + √(233.39)) / 2
CD ≈ (-57.6529 + 15.278) / 2
CD ≈ -42.3749 / 2
CD ≈ -21.1875
Since length cannot be negative, we discard the negative result. Therefore, the length of angle DC is approximately 21.1875 cm.