Chapter 14.4, Problem 19PPSE

Solution

To find:The measured of largest angle of a triangle.

The largest angle of the given triangle is ${66.48}^{\text{ο}}\approx {67}^{\text{ο}}$ .

Given data: The sides of a triangle are 15 in., 17 in. and 16 in. The measure of the smallest angle is ${54}^{\text{ο}}$ .

Method/Formula used:

In a triangle ABC shown in Fig. 1, relations between sides and the angles are:

  $\begin{array}{c}\Delta \left(\text{area}\text{of}\text{triangle}ABC\right)=\frac{1}{2}ab\sin C\\ =\frac{1}{2}bc\sin A\\ =\frac{1}{2}ca\sin B\end{array}\}$ … (1)

  

Calculations:

Let PQR is a given triangle in which $PQ=16\text{in}.$ $QR=17\text{in}.$ , and $PQ=15\text{in}.$ as shown in Fig.2.

  

The smallest angle the angle made by smallest side of 15 in., that is $\angle R={54}^{\text{ο}}$ . The largest angle is the angle made by the side of 17 in. that is angle P , therefore using relations (1), for triangle PQR ,

  $\begin{array}{c}\Delta \left(\text{area}\text{of}\text{triangle}PQR\right)=\frac{1}{2}\left(PQ\cdot PR\right)\text{sin}\angle P\\ =\frac{1}{2}\left(QR\cdot RP\right)\sin \angle R\end{array}$

Therefore,

  $\left(PQ\right)\text{sin}\angle P=\left(QR\right)\sin \angle R$ … (2)

Substitute 15 for PQ , 17 for QR , and ${54}^{\text{ο}}$ for $\angle R$ in result (2),

  $15\text{sin}\angle P=17\sin {54}^{\text{ο}}$

Therefore,

  $\begin{array}{c}\angle P=\text{si}{\text{n}}^{\text{-1}}\left(\frac{17\sin {54}^{\text{ο}}}{15}\right)\\ ={66.48}^{\text{ο}}\end{array}$

Thus, the largest angle of the given triangle is ${66.48}^{\text{ο}}$ .