Question

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Answer 1

Suppose

AB = X = 40cm

AC = y

BC = Z

Given

<A + <B + <C = 180°

40 °+ 45 °+ <c = 180°

<c = 105°

when a perpendicular line drone from AB to C which bisect angle C .

let the point be D on AB and the line DC be p .

In ∆ADC

tan45° = DC/AD =p/t (where t is the distance between DC) ---(1)

In ∆BDC

tan30 ° = DC/DB = p/(40 - t)

Now

t = p ( form -(1))

so

1/sqrt3 = p/(40 - p)

sqrt3 = 40 /p - 1

(sqrt3 + 1) = 40/p

p = 40/(sqrt3 + 1)

so

t = 40/(sqrt3 + 1)= AD= 14.64cm

BD = 40 - t = {40 - 40/ (sqrt3 + 1)}=35.36cm

Now

sin 45= CD/AC

1/sqrt2 = CD/AC

AC= 40sqrt2/(sqrt3 + 1) = 14.64*sqrt2 =20.7cm

again

in ∆CDB

Sin30 = p/CB

CB = 2( 40 - 40/(sqrt3 + 1) = 2(14.64) = 29.38cm

Answer 2

Checkout Above Provided Attachments

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• Remember Some Important Points :-

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• Equilibrium is only given to Confuse student !

• Trignometric Ratios is must be known by Answerer!

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• Basic Triangular Formulas Should be Known!

• Knowledge of Pythagoras Theorem Required!

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