Question

Use the figure to find x.

Answer 1

Answer:

The value of x is $\frac{7\sqrt{6}}{2}$

Solution given:

AB=7

BD=x

<BAC=60°

<DBC=45°

In right angled triangle ABC

Tan 60°=opposite/adjacent

Tan 60°=BC/AB

Substitute value

$\sqrt{3}$=$\frac{BC}{7}$

BC=$7\sqrt{3}$

again

againIn right angled triangle BCD

againIn right angled triangle BCDUsing Cos angle

Cos 45=adjacent/hypotenuse

Cos45°=BD/BC

Substituting value

$\frac{\sqrt{2}}{2}=\frac{x}{7\sqrt{3}}$

Doing criss cross multiplication

$\frac{\sqrt{2}}{2}*7\sqrt{3}=x$

x=$\frac{7\sqrt{6}}{2}$

Answer 2

Answer:

$x = 8.57$

Step-by-step explanation:

Here two triangles are given to us , which are attached to each other . Here we can use the concept of Trigonometry to find out the value of x. The angles of the triangle are 60° and 45° . Let the common side be p .

Step 1: Use the ratio of tan in upper triangle

$\rm\implies tan60^o = \dfrac{perpendicular}{base}$

Substitute the respective values ,

$\rm\implies \sqrt3=\dfrac{p}{7}$

Cross multiply ,

$\rm\implies p = 7\sqrt3$

Step 2: Use the ratio of cos in lower triangle

$\rm\implies cos45^o = \dfrac{base}{hypontenuse}$

Substitute the respective values ,

$\rm\implies \dfrac{1}{\sqrt2}=\dfrac{x}{7\sqrt3}$

Cross multiply ,

$\rm\implies x= \dfrac{7\sqrt3}{\sqrt2}$

Put the approximate values of √2 and √3

$\rm\implies x= \dfrac{7\times 1.732}{1.414}$

This equals to ,

$\rm\implies \boxed{\blue{\rm \quad x = 8.57\quad}}$

Hence the value of x is 8.57 .