Question
✓
Answer
In $\triangle$ABC and $\triangle$PQR,
$\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{\mathrm{BC}}{\mathrm{QP}}=\frac{6}{12}=\frac{1}{2} $ and $\frac{\mathrm{CA}}{\mathrm{PR}}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
That is, $\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{BC}}{\mathrm{QP}}=\frac{\mathrm{CA}}{\mathrm{PR}}$
So,$\triangle \mathrm{ABC} \sim \triangle \mathrm{RQP}$ (SSS similarity criterion)
Therefore, $\angle$C = $\angle$P (Corresponding angles of similar triangles)
But $\angle$C = 180° – $\angle$A – $\angle$B (Angle sum property of triangle)
$\angle$C = 180° – 80° – 60°
$\angle$C = 40°
So, $\angle$P = 40°
$\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{\mathrm{BC}}{\mathrm{QP}}=\frac{6}{12}=\frac{1}{2} $ and $\frac{\mathrm{CA}}{\mathrm{PR}}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
That is, $\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{BC}}{\mathrm{QP}}=\frac{\mathrm{CA}}{\mathrm{PR}}$
So,$\triangle \mathrm{ABC} \sim \triangle \mathrm{RQP}$ (SSS similarity criterion)
Therefore, $\angle$C = $\angle$P (Corresponding angles of similar triangles)
But $\angle$C = 180° – $\angle$A – $\angle$B (Angle sum property of triangle)
$\angle$C = 180° – 80° – 60°
$\angle$C = 40°
So, $\angle$P = 40°
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Prove that:
$\sin(50^\circ+\theta)-\cos(40^\circ-\theta)\\+\tan1^\circ+\tan10^\circ\tan80^\circ\tan89^\circ=1$
→$\sin(50^\circ+\theta)-\cos(40^\circ-\theta)\\+\tan1^\circ+\tan10^\circ\tan80^\circ\tan89^\circ=1$
Find the sum:
$25 + 28 + 31 + ..... + 100.$
→$25 + 28 + 31 + ..... + 100.$
What is a composite number?
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$1.0, 1.7, 2.4, 3.1, .....$
→$1.0, 1.7, 2.4, 3.1, .....$
Are the given numbers form an $AP?$ If they form an $AP$, write the next two terms: $1, -1, -3, -5, ...$
→Prove the following trigonometric identities.
$\frac{\text{cosec A}}{\text{cosec A}-1}+\frac{\text{cosec A}}{\text{cosec A}+1}=2\sec^2\text{A}$
→$\frac{\text{cosec A}}{\text{cosec A}-1}+\frac{\text{cosec A}}{\text{cosec A}+1}=2\sec^2\text{A}$
Write the discriminant of the following quadratic equation.
$x^2- 2x + k = 0$, $\text{k}\in\text{R}$
→$x^2- 2x + k = 0$, $\text{k}\in\text{R}$
Solve the following system of equations graphically:
$2x - 3y = 1, 4x - 3y + 1 = 0$
→$2x - 3y = 1, 4x - 3y + 1 = 0$
Find the $30^{th}$ term of an $A P\ 10,7,4,....$
→In Figure, $OA$ $\cdot$ $OB = OC$ $\cdot$ $OD$. Show that $\angle$ $A =$ $\angle$$C$ and $\angle$$B =$ $\angle$$D$
→