Question

In a triangle ABD, right angled at D, AB=100m, and C is a point on BD such that angle ACD=60 degrees and AC=BC. Find AD.
​

Answer 1

$\underline{\underline{\green{\sf Question:}}}$

In a triangle ABD, right angled at D, AB=100m, and C is a point on BD such that angle ACD=60 degrees and AC=BC. Find AD. As perimeter of ΔADC =   perimeter of ΔADB .

$\underline{\underline{\green{\sf Required \; Solution:}}}$

$\sf Let\; length \;of\; AD \;be\; 'l'\; then, \\ \\l+x +80=60+l+(100-x) \\ \\\implies 20=100−2x \\ \\\implies x=40cm \\ \\\therefore length~ of~ BD=(100-x)=60cm$

_______________

Hope it  helps! :)

Answer 2

$\huge\bold{\mathtt{\red{A{\pink{N{\green{S{\blue{W{\purple{E{\orange{R}}}}}}}}}}}}}$

Step-by-step explanation:

In a triangle ABD, right angled at D, AB=100m, and C is a point on BD such that angle ACD=60 degrees and AC=BC. Find AD. As perimeter of ΔADC = perimeter of ΔADB .

\underline{\underline{\red{\sf Required \; Solution:}}}

RequiredSolution:

\begin{gathered}\sf Let\; length \;of\; AD \;be\; 'l'\; then, \\ \\l+x +80=60+l+(100-x) \\ \\\implies 20=100−2x \\ \\\implies x=40cm \\ \\\therefore length~ of~ BD=(100-x)=60cm\end{gathered}

LetlengthofADbe

then,

l+x+80=60+l+(100−x)

⟹20=100−2x

⟹x=40cm

∴length of BD=(100−x)=60cm