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Question

In Δ ABC, AD is the median and AD = 12 BC. If BAD = 30°, then the measure of ACB is:

Δ ABC में, AD माध्यिका है और AD = 12 BC। यदि कोण BAD = 30°, तब ACB की माप क्या है?


A
90º
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B
45º
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C
30º
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D
60º
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Solution

The correct option is D 60º
Based on given information, we may draw the diagram which is as follow:


As we know that in ABC, AD is median and AD = 12 BC Therefore, it is evident that all the three sides i.e.
AD, CD and DB are equal.
If BAD = 30º
ABD = 30º [as AD = DB]
ADB = 180º – 30º – 30º = 120º
ADC = 180º – 120º = 60º
Now ADC+ DAC+ ACD=180º
Or 60º + 2 ACD = 180º
Or 2 ACD = 120º
Or ACD = 60º = ACB
So, ACB = 60º
Hence, option (d) is correct.

दी गयी जानकारी के आधार पर, हम नीचे दिए गए अनुसार आरेख बना सकते हैं:

जैसा कि हम जानते हैं कि Δ ABC में AD माध्यिका है और AD = 12 BC
इसलिए, यह स्पष्ट है कि तीनों भुजाएं अर्थात AD, CD और DB बराबर हैं।

यदि BAD = 30º
ABD = 30º [as AD = DB]
ADB = 180º – 30º – 30º = 120º
ADC = 180º – 120º = 60º
त्रिभुज ADC+ DAC+ ACD=180º
या, 60º + 2 ACD = 180º
या, 2 ACD = 120º
या, ACD = 60º = ACB
इसलिए, ACB = 60º
अतः, विकल्प (d) सही है।

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