In the given figure- angle QPB is-45-060-030-015-0

Question

Toppr Admin · Accepted Answer

-x2235-x2220-APB-180-x2218-x2234-x2220-APR-x2220-RPQ-x2220-QPB-x2220-APB-x21D2-2x-x2218-3x-x2218-x-x2218-180-x2218-x21D2-6x-x2218-180-x2218-x21D2-x-1806-30-x2218-x2234-x2220-QPB-30-x2218-

In the given figure, $∠$ QPB is:

$45^{0}$
$60^{0}$
$30^{0}$
$15^{0}$

$∵ ∠ A P B = 180^{\circ}$
$∴ ∠ A P R + ∠ R P Q + ∠ Q P B = ∠ A P B$
$\Rightarrow 2 x^{\circ} + 3 x^{\circ} + x^{\circ} = 180^{\circ}$
$\Rightarrow 6 x^{\circ} = 180^{\circ}$
$\Rightarrow x = \frac{180}{6} = 30^{\circ}$
$∴ ∠ Q P B = 30^{\circ}$

In the given figure, ∠ QPB is:450600300150

∵∠APB=180∘∴∠APR+∠RPQ+∠QPB=∠APB⇒2x∘+3x∘+x∘=180∘⇒6x∘=180∘⇒x=1806=30∘∴∠QPB=30∘

In the given figure, $∠$ QPB is:

$45^{0}$
$60^{0}$
$30^{0}$
$15^{0}$

$∵ ∠ A P B = 180^{\circ}$
$∴ ∠ A P R + ∠ R P Q + ∠ Q P B = ∠ A P B$
$\Rightarrow 2 x^{\circ} + 3 x^{\circ} + x^{\circ} = 180^{\circ}$
$\Rightarrow 6 x^{\circ} = 180^{\circ}$
$\Rightarrow x = \frac{180}{6} = 30^{\circ}$
$∴ ∠ Q P B = 30^{\circ}$