Question

Given $\triangle ABC \sim \triangle PQR$, $\angle A = 30^\circ$ and $\angle Q = 90^\circ$. The value of $(\angle R + \angle B)$ is

• A. $90^\circ$
• B. $180^\circ$
• C. $150^\circ$
• D. $120^\circ$

Hint:

Sum of angles in a triangle is always $180^\circ$. Use this for indirect angle calculations.

Answer 1

The Correct Option is D

Given:
- Triangles \(\triangle ABC \sim \triangle PQR\) (i.e., the triangles are similar)
- \(\angle A = 30^\circ\)
- \(\angle Q = 90^\circ\)
We are to find the value of \(\angle R + \angle B\)

Step 1: Use Triangle Sum Property
In any triangle, the sum of the interior angles is \(180^\circ\)

In \(\triangle PQR\):
\[ \angle P + \angle Q + \angle R = 180^\circ \Rightarrow \angle P + 90^\circ + \angle R = 180^\circ \Rightarrow \angle P + \angle R = 90^\circ \]

In \(\triangle ABC\):
\[ \angle A + \angle B + \angle C = 180^\circ \Rightarrow 30^\circ + \angle B + \angle C = 180^\circ \Rightarrow \angle B + \angle C = 150^\circ \]
Step 2: Use similarity of triangles
Since \(\triangle ABC \sim \triangle PQR\), their corresponding angles are equal:
- \(\angle A = \angle P = 30^\circ\)
- \(\angle B = \angle R\)
- \(\angle C = \angle Q = 90^\circ\)

So from the similarity:
\[ \angle B = \angle R \Rightarrow \angle B + \angle R = 2 \times \angle B \]
From earlier, in \(\triangle ABC\):
\(\angle B + \angle C = 150^\circ\), and since \(\angle C = 90^\circ\),
\[ \angle B = 60^\circ \Rightarrow \angle R = 60^\circ \Rightarrow \angle B + \angle R = 60^\circ + 60^\circ = 120^\circ \]

Final Answer:
\(\angle R + \angle B = \boxed{120^\circ}\)