Solution Of Triangles(11th And 12th > Mathematics ) Questions and Answers
Explanation:-
Answer: Option A. -> 0:
A
Let’s construct the lines and angles which constitute the triangle.
We can see once c and B are fixed the least value it should have in order to form a triangle is b = c sin B. In fact, you can form 2 triangles if b > c sin B and B is acute. If B is obtuse by default b > c ⇒ b > c sin B. Therefore no triangle is possible when b < c sin B irrespective of B being acute or obtuse. Answer is 0 triangles are possible.
Explanation:-
Answer: Option D. -> a=√6, B=105∘, C=15∘:
D
A=60∘⇒B+C=120∘We know, tanB−C2=b−cb+ccotA2⇒tanB−C2=(√3+1)−(√3−1)(√3+1)+(√3−1)cot(602)=22√3(√3)=1∴B−C2=45∘⇒B−C=90∘B+C=120∘∴B=105∘ and C=15∘,From sine rule,a=b sinAsinB=√3+1(sin60sin105)sin105=sin(60∘+45∘)=√32(1√2)+12(1√2)=√3+12√2a=(√3+1)(√3)(2)(√2)(√3+1)2=√6∴a=√6, B=105∘, C=15∘
Option d is correct.
Explanation:-
Answer: Option C. -> 120∘:
C
We know from trigonometric ratios of half angles that tanA2=√(s−b)(s−c)(s−a)(s)
Here we need angle opposite to c i.e., C.
Semiperimeter, S=a+b+c2=3+5+72=152
tanC2=√(s−b)(s−a)(s−c)(s)
=√(s−3)(s−5)(s−7)s=√(15−6)(15−10)(2)4(15−14)(152)
=√9(5)15=√3⇒C2=60∘⇒C=120∘
Explanation:-
Answer: Option C. -> Right angled:
C
Given, a2+b2+c2=ca+ab√3a2+b2+c2−ca−ab√3=0⇒(a√32−b)2+(a2−c)2=0⇒a=2b√3 and a=c2We can see, a2+b2=c2sinB=ba=√32≠1⇒Not isosceles∴Given triangle is a Right angled triangle which is not isosceles.
Option C is correct.
Explanation:-
Answer: Option B. -> √32:
B
In trigonometry one problem can be solved in multiple ways but it is wise to use the best way to get the answer or at least not to use the lengthiest way. Here, we can apply sine rule and get the answer but have a look at the problem again. We are given all the sides. And when sides are given and angles are asked to find ,using cosine rule is a better option.
According to cosine rule -
cosA=b2+c2−a22bc
cosA=(8)2+(7)2−(13)22(8)(7) (On putting values of a,b & c)
cosA=−12
A=2π3
sinA=sin2π3=√32
Explanation:-
Answer: Option B. -> r1=Δs−a,r1=4RsinA2cosB2cosC2:
B
Relation between Exradius and semiperimeter can be given by, r1=Δs−a where r1 is the radius of the circle opposite the angle A. and it's helpful to remember the identity r1=4RsinA2cosB2cosC2