Question

please please solve this

Answer 1

Answer:

sorry I don't know the answer

Step-by-step explanation:

follow me please

Answer 2

PROVE

for the ∆ABC....

2(theta)+2(alpha)+ang(BAC)=180°

2(theta+alpha+(1/2)ang(BAC)=180°

$= > theta + alpha + \frac{1}{2} ang(bac) = 90 \\ = > threta + alpha = 90 - \frac{1}{2} ang(bac)$

now.....

for ∆BPC

theta+alpha+any(BPC)=180°

$= > 90 - \frac{1}{2} ang(bac) + ang(bpc) = 180 \\ = > (bpc) = 180 - 90 + \frac{1}{2} ang(bac) \\ = > ang(bpc) = 90 + \frac{1}{2} ang(bac)$

so proved......