In the given figure, $ \mathrm{PQ} $, $ \mathrm{RS} $ and $ \mathrm{UT} $ are parallel lines. If $ c=75^{\circ} $ and $ a=(2 / 5) c, $ find $ b+d / 2 $.


A. $ 92^{\circ} $
B. $ 115^{\circ} $
C. $ 112.5^{\circ} $
D. $ 135.5^{\circ} $"

Given:

In the given figure, \( \mathrm{PQ} \), \( \mathrm{RS} \) and \( \mathrm{UT} \) are parallel lines. If \( c=75^{\circ} \) and \( a=(2 / 5) c \).
To do:

We have to find the value of $b+\frac{d}{2}$.
Solution:

$PQ\parallel\ RS\parallel\ UT$

$c=75^o$

$a=\frac{2}{5}c=\frac{2}{5}\times75^o=2\times15^o=30^o$

$PQ\parallel\ UT$ and PT is the transversal.

Therefore,

$c=a+b$   (Alternate angles)

$75^o=30^o+b$

$b=75^o-30^o$

$b=45^o$

$PQ\parallel\ RS$ and PR is the transversal.

Therefore,

$b+d=180^o$    (Angles on the same side of a transversal are supplementary)

$45^o+d=180^o$

$d=180^o-45^o$

$d=135^o$

This implies,

$b+\frac{d}{2}=45^o+\frac{135^o}{2}=45^o+67.5^o$\

$=112.5^o$

Option C is the correct answer.

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Updated on: 10-Oct-2022

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