Question

Construct a $△PQR$ in which $QR=6.2cm,\angle Q=45°,\angle P=35°$. What type of triangle is this?

Answer 1

Here two angles and their included side is not given so we find the $\angle R$.

$$\begin{gathered} \angle P+\angle Q+\angle R=180° \\ 35°+45°+\angle R=180° \\ 80°+\angle R=180° \\ \angle R=180°-80 \\ \angle R=100° \end{gathered}$$

STEPS OF CONSTRUCTION:

STEP 1:First we draw a line segment $QR$ of length $6.2cm$.

STEP 2:Now we make a ray $RX$ by making an angle $\angle QRX=100°$.

STEP 3:Again we make a ray $QY$ by making an angle $\angle RQY=45°$.

STEP 4: At the point of intersection of these two rays we get the third point$P$. Hence we get the required triangle.

In this triangle one angle $\angle R=100°$ which is an obtuse angle .So it is an obtuse-angled triangle.