To solve the problem, we need to determine the measure of \(\angle B\) in \(\triangle ABC\), given that \(\triangle ABC \sim \triangle PQR\), \(\angle A = 32^\circ\), and \(\angle R = 65^\circ\).
1. Understanding Similar Triangles:
Since \(\triangle ABC \sim \triangle PQR\), the corresponding angles of the two triangles are equal. This means:
\[
\angle A = \angle P, \quad \angle B = \angle Q, \quad \angle C = \angle R
\]
Given \(\angle A = 32^\circ\) and \(\angle R = 65^\circ\), we know:
\[
\angle C = \angle R = 65^\circ
\]
2. Sum of Angles in a Triangle:
The sum of the interior angles in any triangle is \(180^\circ\). For \(\triangle ABC\), we have:
\[
\angle A + \angle B + \angle C = 180^\circ
\]
Substituting the known values \(\angle A = 32^\circ\) and \(\angle C = 65^\circ\):
\[
32^\circ + \angle B + 65^\circ = 180^\circ
\]
3. Solving for \(\angle B\):
Combine the known angle measures:
\[
32^\circ + 65^\circ = 97^\circ
\]
Thus:
\[
97^\circ + \angle B = 180^\circ
\]
Solving for \(\angle B\):
\[
\angle B = 180^\circ - 97^\circ = 83^\circ
\]
Final Answer:
The measure of \(\angle B\) is \({83^\circ}\).
In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).
If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).