Eamcet Mathematics - Properties of Triangles, Height and Distances Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 20
If in a \(\Delta ABC\) , A=30o , B=45o and a=1, then the values of b and c are respectively
- (a)
\(\sqrt { 2 } ,\frac { \sqrt { 3 } +1 }{ \sqrt { 2 } } \)
- (b)
\(\sqrt { 2 } ,\frac { \sqrt { 3 } -1 }{ \sqrt { 2 } } \)
- (c)
\(\sqrt { 3 } ,\frac { \sqrt { 3 } -1 }{ \sqrt { 2 } } \)
- (d)
\(\sqrt { 2 } ,\frac { \sqrt { 3 } +2 }{ \sqrt { 2 } } \)
Given, A = 30o , B = 45o, a = 1
∴ C = 105o
Using sine rule,
\(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { 1 }{ \left( \frac { 1 }{ 2 } \right) } =\frac { 1 }{ \left( \frac { 1 }{ \sqrt { 2 } } \right) } =\frac { c }{ sin \ 105^{ o } } \Rightarrow b=\sqrt { 2 } \)
and c = 2 sin 105o
\(\therefore \ c=2\left( \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right) =\frac { \sqrt { 3 } +1 }{ \sqrt { 2 } } \)
If A=75o , B=45o , then \(b+c\sqrt { 2 } \) is equal to
- (a)
2a
- (b)
2a+1
- (c)
3a
- (d)
2a-1
Given, A=75o and B=45o
⇒ C = 60o
By sine rule, \(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { a }{ sin{ 75 }^{ 0 } } =\frac { b }{ sin{ 45 }^{ 0 } } =\frac { c }{ sin{ 60 }^{ 0 } } \)
Now, \(\\ b+c\sqrt { 2 } =\frac { sin{ 45 }^{ 0 } }{ sin75^{ 0 } } a+\sqrt { 2 } \frac { sin60^{ 0 } }{ sin75^{ 0 } } a\)
\(=\frac { \frac { 1 }{ \sqrt { 2 } } }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } a+\sqrt { 2 } \frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } a\)
\(=\frac { 2 }{ \sqrt { 3 } +1 } a+\frac { 2\sqrt { 3 } a }{ \sqrt { 3 } +1 } =2a\)
If a2,b2 and c2 are in AP, then cotA,cotB and cotC are in
- (a)
AP
- (b)
GP
- (c)
HP
- (d)
AGP
cot A = \(\frac { cos\ A }{ sin\ A } =\frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 2bcsinA } =\frac { { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } }{ 4\Delta } \)
similarly, cot B = \(\frac { { { a }^{ 2 }+{ c }^{ 2 }-b }^{ 2 } }{ 4\Delta } \)
and cot C = \(\frac { { { a }^{ 2 }+{ { b }^{ 2 }-c }^{ 2 } } }{ 4\Delta } \)
Given that a2, b2 and c2 are in AP.
⇒ -2a2, -2b2, -2c2 are in A p.
⇒ (a2+b2+c2)-2a2, (a2+b2+c2) - 2b2, (a2+b2+c2) - 2c2 are in AP.
⇒ b2+c2-a2, c2+a2-b2, a2+b2-c2 are in AP.
\(\Rightarrow \frac { { b }^{ 2 }{ +{ c }^{ 2 }-a }^{ 2 } }{ 4\Delta } ,\frac { { b }^{ 2 }{ +{ c }^{ 2 }-a }^{ 2 } }{ 4\Delta } ,\frac { { a }^{ 2 }{ +b^{ 2 }-c }^{ 2 } }{ 4\Delta } \) are in AP.
If in \(\Delta ABC,\) \(\Delta ={ a }^{ 2 }-{ \left( b-c \right) }^{ 2 },\) then the value of tanA is
- (a)
\(\frac { 8 }{ 14 } \)
- (b)
\(\frac { 8 }{ 13 } \)
- (c)
\(\frac { 8 }{ 15 } \)
- (d)
\(\frac { 8 }{ 17 } \)
Given, ∆ = a2 - (b-c)2 = (a + b - c)(a - b + c)
⇒ ∆ = [4(s-b)(s-c)]
⇒ s(s-a)(s-b)(s-c) = 16 (s-b)2 (s-c)2
\(\Rightarrow \frac { \left( s-b \right) \left( s-c \right) }{ s\left( s-a \right) } =\frac { 1 }{ 16 } \Rightarrow tan\frac { A }{ 2 } =\frac { 1 }{ 4 } \)
\(\therefore \ tanA=\frac { 2tan\frac { A }{ 2 } }{ 1-tan^{ 2 }\frac { A }{ 2 } } =\frac { 2.\left( \frac { 1 }{ 4 } \right) }{ 1-\frac { 1 }{ 16 } } =\frac { 8 }{ 15 } \)
If in \(\Delta ABC,\) r1=r2+r3+r, then triangle is
- (a)
a right-angled triangle
- (b)
equilateral triangle
- (c)
isoscles triangle
- (d)
None of the above
Given that, r1 = r2 + r3 + r
⇒ r1 - r = r2 + r3
\(\Rightarrow \frac { \Delta }{ s-a } -\frac { \Delta }{ s } =\frac { \Delta }{ s-b } +\frac { \Delta }{ s-c } \)
\(\Rightarrow \frac { s-s+a }{ s\left( s-a \right) } =\frac { s-c+s+b }{ \left( s-b \right) \left( s-c \right) } \)
\(\Rightarrow \frac { a }{ s\left( s-a \right) } =\frac { a }{ \left( s-b \right) \left( s-c \right) } \)
\(\Rightarrow \ s\left( a-b-c \right) +bc=0\)
\(\Rightarrow \ \frac { \left( a+b+c \right) \left( a-b-c \right) }{ 2 } +bc=0\)
\(\Rightarrow \ { a }^{ 2 }-\left( b+c \right) ^{ 2 }+2bc=0\)
\(\Rightarrow { a }^{ 2 }-{ b }^{ 2 }-{ c }^{ 2 }=0\Rightarrow { b }^{ 2 }+{ c }^{ 2 }={ a }^{ 2 }\)
So, ΔABC is right angled triangle.
Three vertical poles of height h1, h2 and h3 at the vertices A,B and C of a \(\Delta ABC\) subtend angles \(\alpha ,\beta \quad and\quad \gamma \) respectively, at the circumcentre of the triangle. If \(cot\alpha ,cot\beta ,cot\gamma \) are in AP, then h1, h2 and h3 are in
- (a)
AP
- (b)
GP
- (c)
AGP
- (d)
HP
We know that, \(\frac { R }{ { h }_{ 1 } } =cot\alpha ,\frac { R }{ { h }_{ 2 } } =cot\beta \)
and R/h3 = cot ⋎ [R is the circumradius]
Since, cot α, cot β, cot ⋎ are in AP.
⇒ \(\frac { 1 }{ { h }_{ 1 } } ,\frac { 1 }{ { h }_{ 2 } } ,\frac { 1 }{ { h }_{ 3 } } \) are in AP.
⇒ h1, h2, h3 are in HP.
Match the Columns
Column I | Column II | ||
---|---|---|---|
A. | \(cot\frac { A }{ 2 } =\frac { b+c }{ a } \) | p. | always right angles |
B. | \(atanA+btanB=(a+b)tan\left( \frac { A+B }{ 2 } \right) \) | q. | always isosceles |
C. | \(acosA=bcosB\) | r. | may be right angled |
D. | \(cosA=\frac { sinB }{ 2sinC } \) | s. | may be right-angled isosceles |
- (a)
A B C D pr, qrs, rs, qrs - (b)
A B C D qp, prs, qs, qsp - (c)
A B C D qr, qps, ps, qpr - (d)
None of the above
A.cot \({{A}\over{2}}={{b+c}\over{2}}\Rightarrow{{cos(A/2)}\over{sin(A/2)}}={{sinB+sinC}\over{sinA}}\)
\(\Rightarrow\) \({{cos(A/2)}\over{sin(A/2)}}={{2sin{{B+C}\over{2}}cos{{B-C}\over{2}}}\over{2sin{{A}\over{2}}cos{{A}\over{2}}}}\)
\(\Rightarrow\) cos \({{A}\over{2}}\)= cos \({{B-C}\over{2}}\)
\(\Rightarrow\) A = B - C
\(\Rightarrow\) A + B + C = 2B = \(\pi\) \(\Rightarrow B={{\pi}\over{2}}\)
If in \(\triangle\)ABC,
a tan A + b tan B = ( a + b) tan \({{1}\over2{}}\) ( A + B )
Rewrite the given relation as
a \(\left[ tan A -tan{{1}\over{2}}(A+B) \right]=b\left[ tan{{1}\over{2}}(A+B)-tanB \right]\)
\(\Rightarrow\) \({{a sin \left( A-{{A+B}\over{2}} \right)}\over{cos A cos{{A+B}\over{2}}}}={{b sin \left({{A+B}\over{2}}-B \right)}\over{cos{{A+B}\over{2}}cosB}}\)
\(\Rightarrow\) \({{2RsinAsin{{1}\over{2}}(A-B)}\over{cos A}}={{2RsinBsin{{1}\over{2}}(A-B)}\over{cos B}}\)
\(\Rightarrow\) sin \({{1}\over{2}}\) ( A - B )( tan A - tan B ) = 0
\(\Rightarrow\) A = B from either factor
a cos A = b cos B
2R sin A cos A = 2R sin B cos B
\(\Rightarrow\) sin 2A = sin 2B
\(\Rightarrow\) 2A = 2B or \(\pi-2B\)
\(\Rightarrow\) A = B or A + B = \(\pi/2\)
\(\Rightarrow\) \(\triangle\)ABC is isosceles or C = \(\pi/2\) . i,e, \(\triangle\)ABC is right angled.
cos A = \({{sin B}\over{2Sin C}}\)
\(\Rightarrow\) \({{{b}^{2}+{c}^{2}-{a}^{2}}\over{2bc}}={{b}\over{2c}}\)
\(\Rightarrow\) b2 + c2 - a2 = b2
\(\Rightarrow\) c2 = a2 \(\Rightarrow\) c = a \(\Rightarrow\) A\(\triangle\)is isosceles
Hence, A \(\rightarrow\) p; r;B \(\rightarrow\)q;r;s;C\(\rightarrow\)r,s;D\(\rightarrow\)q,t,s.
In \(\Delta ABC,\quad if\quad sin\left( \frac { C }{ 2 } \right) =\frac { a-b }{ 2\sqrt { ab } } tan\theta ,\quad then\quad (a-b)sec\theta \) is equal to
- (a)
2a+b
- (b)
2c
- (c)
c
- (d)
c+2
Given that,
sin \(\left( {{C}\over{2}} \right)={{a-b}\over{2\sqrt{ab}}}tan \theta\)
\(\Rightarrow\) tan \(\theta={{2\sqrt{ab}}\over{a-b}}.sin\left({{C}\over{2}} \right)\)
Now, sec2 \(\theta=1\) + tan2 \(\theta=1+{{4ab}\over{(a-b)^{2}}}{sin}^{2}\left( {{C}\over{2}} \right)\)
= \(1+{{2ab}\over{(a-b)^{2}}}(1-cos C)\)
= \({{{(a-b)}^{2}+2ab(1-cos C)}\over{{(a-b)}^{2}}}\)
= \({{{a}^{2}+{b}^{2}-({a}^{2}+{b}^{2}-{c}^{2})}\over{{(a-b)}^{2}}}\) \(\left[ \because cos C={{{a}^{2}+{b}^{2}-{c}^{2}}\over{2ab}} \right]\)
\(\Rightarrow\) sec2 \(\theta={{{c}^{2}}\over{{(a-b)}^{2}}}\)
\(\Rightarrow\) \(sec \theta={{c}\over{(a-b)}}\)
\(\therefore\) \((a-b)sec\theta=c\)
In a \(\Delta ABC,\) a3 cos(B - C)+b3cos(C - A)+c3cos(A - B) is equal to
- (a)
3abc
- (b)
2abc
- (c)
abc
- (d)
5abc
By sine rule, a = 2R sin A, b = 2R sin B, c = 2R sinC
\(\therefore\) a3 cos( B - C ) + b3 cos ( C - A ) + c3 cos ( A - B )
= a3 ( 2R sin A ) cos ( B - C ) + b2 ( 2R sin B ) cos ( C - A ) + c2 ( 2R sin C ) cos ( A - B )
= a2 [ 2R sin ( B + C ) cos ( B - C) ] + b2 [ 2R sin ( A + C ) cos ( C - A )] + c2 [ 2R sin ( A + V ) cos ( A - B )]
= a2 R[ sin2B + sin2C ] + b2 R[ sin2A + sin2C ] + c2R[ sin2A + sin2B ]
= a2R( 2 sinB. cosB + 2 sinC. cosC ) + b2R ( 2 sin A. cos A + 2 sinC. cosC ) + c2R( 2 sin A. cos A + 2 sinB. cos B )
= a2( b cosB + c cosC ) + b2 ( c cosC + a cosA ) + c2 ( a cosA + b cosB)
= ab ( a cosB + b cosA ) + bc ( b cosC + c cosB ) + ac ( c cosA + a cosC )
= abc + abc + abc = 3 abc
The ratio of the areas of two regular octagons, which are respectively inscribed and circumscribed to a circle of radius r is equal to
- (a)
\(sin\left( \frac { \pi }{ 8 } \right) \)
- (b)
\({ sin }^{ 2 }\left( \frac { 3\pi }{ 8 } \right) \)
- (c)
\({ cos }^{ 2 }\left( \frac { \pi }{ 8 } \right) \)
- (d)
\({ tan }^{ 2 }\left( \frac { \pi }{ 8 } \right) \)
Area of inscribed circle of a regular polygon of n sides,
A1 = nr2 tan \({{\pi}\over{n}}\)
Here, n = 8
\(\therefore\) A1 = 8r2 tan \({{\pi}\over{8}}\) ....(i)
Now, circumscribed circle of a regular polygon of n sides,
A2 = \({{n{R}^{2}}\over{2}}.sin{{2\pi}\over{n}}\) [ \(\because\) R = r, given ]
\(\Rightarrow\) A2 = 8r2 sin \({{\pi}\over{8}}.cos{{\pi}\over{8}}\) ....(ii)
From Eqs.(i) and (ii), we get
\({{{A}^{2}}\over{{A}_{1}}}={cos}^{2}{{\pi}\over{8}}\)
If in a \(\Delta ABC,\) the angles are in AP and the lengths of two larger sides are 10 and 9 respectively, then the length of the third side can be
- (a)
\(5+\sqrt { 6 } \)
- (b)
\(2\sqrt { 6 } \)
- (c)
\(3-\sqrt { 6 } \)
- (d)
\(3+\sqrt { 6 } \)
Given that angles are in AP.
A = x - d, B = x, C = x + d
\(\therefore\) x - d + x + x + d = 180°
\(\Rightarrow\) x = 60°
So, two larger angles are B = x = 60° and C.
Now, B = 60°, b = 9 c = 10
Using cosine rule,
cosB = \({{{c}^{2}+{a}^{2}+{b}^{2}}\over{2ac}}\)
\(\Rightarrow\) \({{1}\over{2}}={{100+{a}^{2}-81}\over{2ac}}\)
\(\Rightarrow\) a2 - 10a +19 = 0
\(\Rightarrow\) a = 5 \(\pm\sqrt{6}\)
In \(\Delta ABC,\) if 2a2b2 + 2b2c2=a4 +b4+c4, then \(\angle B\) is equal to
- (a)
45o
- (b)
35o
- (c)
120o
- (d)
30o
Given that,
2a2b2 + 2b2c2 = a4 + b4 + c4
Also, ( a2 - b2 + c2 )2 = a4 + b4 + c4 - 2( a2 b2 + b2c2 - c2a2)
\(\Rightarrow\) ( a2 - b2 + c2)2 = 2c2a2
\(\Rightarrow\) \({{{a}^{2}-{b}^{2}+{c}^{2}}\over{2ca}}=\pm{{1}\over{\sqrt{2}}}\)= cos B
\(\Rightarrow\) cos B = cos 45° \(\Rightarrow\) B = 45°
In \(\Delta ABC,\) A=15o , \(b=10\sqrt { 2 } \) cm, the value of 'a' for which this will be a unique triangle meeting these requirement is
- (a)
\(10\sqrt { 2 } cm\)
- (b)
\(5cm\quad \)
- (c)
\(5\left( \sqrt { 2 } +1 \right) cm\)
- (d)
\(5\left( \sqrt { 2 } -1 \right) cm\)
Given that in \(\triangle\)ABC, A = 15°, b = 10\(\sqrt{2}\)cm and for unique \(\triangle\)ABC, b \(\ge\)a
\(\because\) sinB = \({{b sin A}\over{a}}={{10\sqrt{2}\times{{\sqrt{3}-1}\over{2\sqrt{2}}}}\over{a}}={{5(\sqrt{3}-1)}\over{a}}\)
For \(\triangle\)ABC, sinB \(\le\)1
\(\therefore\) \({{5(\sqrt{3}-1)}\over{a}}\le 1\)
\(\Rightarrow\) \(a\ge5(\sqrt{3}-1)\)
\(\therefore\) a = \(10\sqrt{2}\) cm
In a \(\Delta PQR,\angle R=\frac { \pi }{ 2 } ,if\quad tan\left( \frac { P }{ 2 } \right) \quad and\quad tan\left( \frac { Q }{ 2 } \right) \) are the roots of ax2 + bx + c = 0, \(a\neq 0,\) then
- (a)
b = a + c
- (b)
b = c
- (c)
c = a + b
- (d)
a = b + c
Since, tan \({{P}\over{2}}\) and tan \({{Q}\over{2}}\) are the roots of equation
ax2 + bx + c = 0
\(\therefore\) tan \({{P}\over{2}}+tan{{Q}\over{2}}={{b}\over{a}}\)
and tan \({{P}\over{2}}\)tan \({{Q}\over{2}}={{c}\over{a}}\) } ......(i)
Also, \({{P}\over{2}}+{{Q}\over{2}}+{{R}\over{2}}={{\pi}\over{2}}\) \([\because P+Q+R=\pi ]\)
\(\Rightarrow\) \({{P+Q}\over{2}}={{\pi}\over{2}}={{R}\over{2}}\)
\(\Rightarrow\) \({{P+Q}\over{2}}={{{\pi}\over{4}}}\) \(\left[ \because \angle R = {{\pi}\over{2}},given \right]\)
\(\Rightarrow\) tan \(\left( \frac { P }{ 2 } +\frac { Q }{ 2 } \right) =1\Rightarrow \frac { tan\frac { P }{ 2 } +tan\frac { Q }{ 2 } }{ 1-tan\frac { P }{ 2 } tan\frac { Q }{ 2 } } =1\)
\(\Rightarrow\) \(\frac { -\frac { b }{ a } }{ 1-\frac { c }{ a } } =1\Rightarrow -\frac { b }{ a } =1\frac { c }{ a } \) [ from Eq.(i)]
\(\Rightarrow\) - b = a - c
\(\therefore\) c = a + b
In a \(\Delta ABC,\angle C=\frac { \pi }{ 2, } \) if r is the inradius and R is the circumradius of the \(\Delta ABC,\) then 2(r+R) is equal to
- (a)
c + a
- (b)
a + b + c
- (c)
a + b
- (d)
b + c
We know that, \({{c}\over{sin C}}=2R\)
\(\Rightarrow\) c = 2R [ \(\because\) C = 90° ] ...(i)
and tan \({{C}\over{2}}={{r}\over{s-c}}\Rightarrow tan {{\pi}\over{4}}={{r}\over{s-c}}\)
\(\therefore\) r = s - c
\(\Rightarrow\) r = \({{a+b+c}\over{2}}-c\)
\(\Rightarrow\) a + b - c = 2r ...(ii)
On adding Eqs. (i) and (ii), we get
2 ( r + R ) = a + b
If in a \(\Delta ABC,\) the altitudes from the vertices A,B and C on opposite sides are in HP, then sin A, sin B and sin C are in
- (a)
HP
- (b)
AGP
- (c)
AP
- (d)
GP
In \(\triangle\)BAD,
cos ( 90° - B ) = \({{AD}\over{c}}\Rightarrow\) AD = c sinB
Similarly, BE = a sinC and CF = b sinA
Since, AD, BE, CF are in HP.
So, c sinB, a sinC, sinA are in HP.
\(\Rightarrow{{1}\over{sinC.sinB}},{{1}\over{sinA.sinC}},{{1}\over{sinB.sinA}}\) are in AP.
\(\Rightarrow\) sinA, sinB, sinC are in HP.
The sides of a triangle are \(sin\alpha ,cos\alpha \quad and\quad \sqrt { 1+sin\alpha cos\alpha } \) for some \(0<\alpha <\frac { \pi }{ 2 } \) . Then, the greatest angle of the triangle is
- (a)
60o
- (b)
90o
- (c)
120o
- (d)
150o
Let a = sin \(\alpha,\) b = cos \(\alpha,\) c = \(\sqrt{1+ sin \alpha cos \alpha}\)
Here, we see that the greatest side is C.
\(\therefore\) cosC = \({{{}a^{2}+{b}^{2}-{c}^{2}}\over{2ab}}\)
\(\Rightarrow\) cosC = \({{{sin}^{2}\alpha+{cos}^{2}\alpha-1-sin \alpha.cos \alpha}\over{2 sin \alpha cos \alpha}}\)
\(\Rightarrow\) cosC = \(-{{sin \alpha cos \alpha}\over{2 sin \alpha cos \alpha}}\)
\(\Rightarrow\) cosC = \(-{{1}\over{2}}=cos120°\)
\(\therefore\) \(\angle\)C = 120°
In a \(\Delta ABC,2asin\left( \frac { A-B+C }{ 2 } \right) \) is equal to
- (a)
a2 + b2 - c2
- (b)
c2 + a2 - b2
- (c)
b2 - c2 - a2
- (d)
c2 - a2 - b2
Since,
A + B + C = \(\pi\)
\(\Rightarrow\) A + C = \(\pi\) - B
\(\Rightarrow\) \({{A+B+C}\over{2}}={{\pi}\over{2}}-B\)
2 ca sin \(\left( {{A+B+C}\over{2}} \right)\) 2 ca sin \(\left( {{\pi}\over{2}}-B \right)\) = 2 ca cos B
= 2ac \(\left( {{{a}^{2}+{c}^{2}+{b}^{2}}\over{2ac}} \right)\)
= a2 + c2 - b2
If in a \(\Delta ABC,a=4,b=3,\angle A={ 60 }^{ o },\) then c is the root of the equation
- (a)
c2-3c-7=0
- (b)
c2+3c+7=0
- (c)
c2-3c+7=0
- (d)
c2+3c-7=0
Given that,
a = 4, b = 3 and \(\angle\)A = 60 °
Now, cos 60° = \({{{c}^{2}+9-16}\over{2\times3\times c}}\)
\(\Rightarrow\) \({{1}\over{2}}={{{c}^{2}-7}\over{2\times3c}}\)
C2 - 3C - 7 = 0
In a \(\Delta ABC,tan\frac { A }{ 2 } =\frac { 5 }{ 6 } ,tan\frac { C }{ 2 } =\frac { 2 }{ 5 } ,\) then
- (a)
a,c and b are in AP
- (b)
a,b and c are in AP
- (c)
b,a and c are in AP
- (d)
a,b and c are in GP
Given that,
tan \({{A}\over{2}}={{5}\over{6}}\)and \(tan{{C}\over{2}}={{2}\over{5}}\)
Now, \(tan{{A}\over{2}}.tan{{C}\over{2}}={{5}\over{6}}\times{{2}\over{5}}\)
\(\Rightarrow\) \(\sqrt{{{(s-b)(s-c)}\over{s(s-a)}}}.\sqrt{{{(s-a)(s-b)}\over{s(s-c)}}}={{1}\over{3}}\)
\(\Rightarrow\) \({{s-b}\over{s}}={{1}\over{3}}\)
\(\Rightarrow\) 2s = 3b
\(\Rightarrow\) a + c = 2b [ \(\because\) 2s = a + b + c ]
\(\Rightarrow\) a, b, c are in AP.