Câu 5: Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
=>\(\widehat{B}=180^0-70^0-50^0=60^0\)
Xét ΔCAB có \(\dfrac{AC}{sinB}=\dfrac{AB}{sinC}\)
=>\(\dfrac{AB}{sin50}=\dfrac{50}{sin60}\)
=>\(AB=50\cdot\dfrac{sin50}{sin60}\simeq44\left(m\right)\)
Bài 6: Ta có: \(\widehat{BQA}+\widehat{BQP}=180^0\)(hai góc kề bù)
=>\(\widehat{BQP}=180^0-22^0=158^0\)
Xét ΔBQP có \(\widehat{BQP}+\widehat{BPQ}+\widehat{PBQ}=180^0\)
=>\(\widehat{PBQ}=180^0-158^0-15^0=7^0\)
Xét ΔPQB có \(\dfrac{QB}{sinP}=\dfrac{PQ}{sinQBP}\)
=>\(\dfrac{QB}{sin15}=\dfrac{100}{sin7}\)
=>\(QB\simeq212,37\)(m)
Xét ΔQAB vuông tại A có \(sinBQA=\dfrac{BA}{BQ}\)
=>\(BA=BQ\cdot sinBQA\simeq212,37\cdot sin22\simeq79,56\left(m\right)\)
Bài 6: Ta có: ˆBQA+ˆBQP=1800BQA^+BQP^=1800(hai góc kề bù)
=>ˆBQP=1800−220=1580BQP^=1800−220=1580
Xét ΔBQP có ˆBQP+ˆBPQ+ˆPBQ=1800BQP^+BPQ^+PBQ^=1800
=>ˆPBQ=1800−1580−150=70PBQ^=1800−1580−150=70
Xét ΔPQB có QBsin15=100sin7QBsin15=100sin7
=>QB≃212,37QB≃212,37(m)
Xét ΔQAB vuông tại A có