Question

Mọi người giúp tớ 2 bài này với

Answer 1

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 \(sinBQA=\dfrac{BA}{BQ}\)

=>\(BA=BQ\cdot sinBQA\simeq212,37\cdot sin22\simeq79,56\left(m\right)\)

 

Answer 2

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

=>(m)

Xét ΔQAB vuông tại A có