Question

1+cos A+sin A
• 1+cos A-sin A
1+sin A
cos A​

Answer 1

Answer:

The value of x = 3

x + 1 = 4

x + 3 = 6

x + 5 = 8

Step-by-step explanation :

Complete Question :

In triangle ABC ,DE is parallel to BC.Find the value of x ,if AD = x,DB = x + 1,

AE = x + 3,EC = x + 5.

Solution :

Given that :

In triangle ABC ,DE is parallel to BC.

So,

DE ║ BC :

\begin{gathered}\sf \longrightarrow \dfrac{x}{x+1}= \dfrac{x+3}{x+5}\\\\\sf \longrightarrow x^2 = 5x = x^2 +3x +x +3\\\\\sf \longrightarrow x^2 + 5x = x^2+4x=x+3\quad{Here\;x^2\;and\;x^2\;cut\;with\;each\;other}\\\\\sf \longrightarrow 5x-4x = 3\\\\\sf \longrightarrow x = 3.\\\\\\\textbf{\underline{\underline{Therefore,the value of x is 3}}}\end{gathered}

⟶

x+1

x

=

x+5

x+3

⟶x

2

=5x=x

2

+3x+x+3

⟶x

2

+5x=x

2

+4x=x+3Herex

2

andx

2

cutwitheachother

⟶5x−4x=3

⟶x=3.

Therefore,the value of x is 3