In an experim some masses are hung in equilibrium as shown in the figure ab =40 cm and angle cab = 45 and angle cba =30 find the perpendicular distance of ab from c
Answers
Given:
In an experim some masses are hung in equilibrium as shown in the figure ab =40 cm and angle cab = 45 and angle cba =30.
To find:
The perpendicular distance of ab from c
Solution:
Consider the attached figure while going through the following steps.
AB = 40 cm
Now consider,
In Δ ADC,
tan 45° = CD / AD
1 = CD / AD
CD = AD ..........(1)
In Δ BDC,
tan 30° = CD / BD
1 / √3 = CD / BD
CD = 1/√3 BD
√3 CD = BD ..........(2)
adding (1) and (2), we get,
CD + √3 CD = AD + BD
CD (1 + √3) = AB
CD (1 + √3) = 40 (given, AB = 40 cm)
CD = 40 / (1 + √3)
The perpendicular distance of ab from c.
∴ CD = 14.64 cm
SOLUTION---
In triangle ABC, Angle DBC=30* and Angle DAC=45*
Therefore, Angle ACB =105*
Construction-- Draw a line perpendicular to AB from C
TO FIND-- CD , AC , BC
In triangle ADC Angle A = Angle DCA
therefore, CD = AD
AD = CD = 40 - BD -------(1)
In triangle BDC ,
tan30* = CD/BD
1/√3 = (40-BD)/BD (From-- 1)
So, BD= 40√3/√3+1
By rationalizing BD = 20(3-√3)
Therefore, CD = 40 -20(3-√3)
ANS1) CD = 20(√3-1)
To find AC length,
sin45* = CD/AC
1/√2 = 20(√3-1)/AC
Therefore,
ANS2) AC = 20√2(√3-1)
To find BC length,
sin30* = CD/BC
1/2 = 20(√3-1)/BC
Therefore,
ANS3) BC = 40(√3-1)
Hence
CD= 20(√3-1)
AC=20√2(√3-1)
BC=40(√3-1)
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