Question

Sum 6(b) and 7(a). Don't spam.

Answer 1

$\large\underline{\underline{\mathfrak{\pink{1^{st}\:\:question}:-}}}$

$\underline{\sf {In \:\:\triangle ECD,}}$

$\sf{ \measuredangle CEF = \measuredangle ECD + \measuredangle CDE}$

$\longrightarrow \sf \measuredangle CEF = 32^o + 32^o$

$\longrightarrow \underline{\underline{\sf{ \green{\measuredangle CEF = 64^o}}}}$

And,

$\sf \measuredangle FOC = 2\times \measuredangle FDC\quad {\scriptsize{[Centre\:Angle\:=\:2Circum\:Angle]}}$

$\sf \longrightarrow \measuredangle FOC = 2\times \measuredangle EDC$

$\sf \longrightarrow \measuredangle FOC = 2 * 32^o$

$\longrightarrow \underline{\underline{\sf{\green{\measuredangle FOC = 64^o}}}}$

$\sf$

$\sf$

$\large\underline{\underline{\mathfrak{\pink{2^{nd}\:\:question}:-}}}$

As $\sf \measuredangle PRB$ and $\sf \measuredangle PAB$ lie on the same segment, they must be equal.

$\sf \measuredangle PRB = \measuredangle PAB$

$\longrightarrow \underline{\underline{\sf{\green{\measuredangle PRB = 35^o}}}}\quad\quad\dots{\sf{(1)}}$

$\sf$

$\underline{\sf {In \:\:\triangle ABQ,}}$

$\sf \measuredangle ABR = \measuredangle BAQ + \measuredangle AQB$

$\longrightarrow \sf \measuredangle ABR = 35^o + 25^o$

$\longrightarrow \sf \measuredangle ABR = 60^o\quad\quad\dots(2)$

$\sf$

And,

$\underline{\sf {In \:\:\triangle APB,}}$

As $\sf AB$ is the diameter of the circle, and the angle in the circumference of a semicircle is 90°, $\sf \measuredangle APB = 90^o$

So,

$\sf \measuredangle APB + \measuredangle BAP + \measuredangle PBA = 180^o$

$\longrightarrow \sf 90^o + 35^o + \measuredangle PBA + 180^o$

$\longrightarrow \sf \measuredangle PBA = 55^o\quad\quad\dots(3)$

$\sf$

And,

$\sf \measuredangle PBR = \measuredangle ABR + \measuredangle PBA$

From (2) and (3),

$\longrightarrow \sf \measuredangle PBR = 60^o + 55^o$

$\longrightarrow \underline{\underline{\sf{\green{\measuredangle PBR = 115^o}}}}\quad\quad\dots{\sf{(4)}}$

$\sf$

And,

$\underline{\sf {In \:\:\triangle PBR,}}$

$\sf \measuredangle PBR + \measuredangle PRB + \measuredangle BPR = 180^o$

From (1) and (4),

$\sf 115^o + 35^o + \measuredangle BPR = 180^o$

$\longrightarrow \underline{\underline{\sf{\green{\measuredangle BPR = 30 ^o}}}}$