Chapter 4.8, Problem 13E

To determine

Tofind:the unknown side and angle of the right- angle triangle shown below.

Answer to Problem 13E

$a=91.34$ and $b=420.70$

Explanation of Solution

Given:

  ${\text{A=12}}^{\text{0}}\text{1<msup><mo>5</mo><mo>′</mo></msup>,c=430}\text{.5}$

Concept used:

${\text{1}}^0=60^{\prime }$ and $1^{\prime }=60^{″}$

The sum of three consecutive angles is ${180}^0$ .

Trigonometric formula:

  $\text{sinθ=}\frac{\text{perpendicular}}{\text{hypotenuse}}$

  $\text{cosθ=}\frac{\text{base}}{\text{hypotenuse}}$

  $\tan \text{A}=\frac{\text{opposite}}{\text{adjacent}}$

Calculation:

The right-angled triangle is shown below. For this triangle

${\text{A}}^{\text{0}}=1215^{\prime }$ and $c=430.5$

Now as we know,

  $1^{\prime }=\left(\frac{1}{60}\right)$

Therefore,

  $15^{\prime }=\left(\frac{15}{60}\right)$

  $\Rightarrow 0.25^{\prime }$

So, $1215^{\prime }=12+0.25$

  $\Rightarrow 12.025$

In $\Delta \text{ABC}$ , the sum of three consecutive angles is ${180}^0$ Hence get;

  $\angle \text{A+}\angle \text{B+}\angle {\text{C=180}}^0$

  $\text{12}{\text{.25}}^0\text{+}\angle {\text{B+90}}^0{\text{=180}}^0$

  $\angle {\text{B=180}}^0\text{-12}{\text{.25}}^0{\text{-90}}^0$

Therefore, $\angle \text{B=77}{\text{.75}}^0$

In $\Delta \text{ABC}$ ,

Applying trigonometric identities

  $\sin \text{A=}\frac{a}{c}$

  $\text{a=csinA}$

  $\text{a=430}\text{.5}\times \text{sin12}\text{.25}$

  $a=91.34$

Also,

  $\text{cosA=}\frac{b}{c}$

  $b=c\cos \text{A}$

  $b=430.5\times \cos 12.25$

  $b=420.70$

Therefore, $b=420.70$ .

Hence, $a=91.34$ and $b=420.70$ .